Who would’ve thought that an old TV game show could inspire a statistical problem that has tripped up mathematicians and statisticians with Ph.Ds? The Monty Hall problem has confused people for decades. In the game show, Let’s Make a Deal, Monty Hall asks you to guess which closed door a prize is behind. The answer is so puzzling that people often refuse to accept it! The problem occurs because our statistical assumptions are incorrect.
The Monty Hall problem’s baffling solution reminds me of optical illusions where you find it hard to disbelieve your eyes. For the Monty Hall problem, it’s hard to disbelieve your common sense solution even though it is incorrect!
The comparison to optical illusions is apt. Even though I accept that square A and square B are the same color, it just doesn’t seem to be true. Optical illusions remain deceiving even after you understand the truth because your brain’s assessment of the visual data is operating under a false assumption about the image.
I consider the Monty Hall problem to be a statistical illusion. This statistical illusion occurs because your brain’s process for evaluating probabilities in the Monty Hall problem is based on a false assumption. Similar to optical illusions, the illusion can seem more real than the actual answer.
To see through this statistical illusion, we need to carefully break down the Monty Hall problem and identify where we’re making incorrect assumptions. This process emphasizes how crucial it is to check that you’re satisfying the assumptions of a statistical analysis before trusting the results.
What is the Monty Hall Problem?
Monty Hall asks you to choose one of three doors. One of the doors hides a prize and the other two doors have no prize. You state out loud which door you pick, but you don’t open it right away.
At this point, Monty will always opens one of the other two doors that does not have a prize behind it.
At this moment, there are two closed doors, one of which you picked.
The prize is behind one of the closed doors, but you don’t know which one.
Monty asks you, “Do you want to switch doors?”
The majority of people assume that both doors are equally like to have the prize. It appears like the door you chose has a 50/50 chance. Because there is no perceived reason to change, most stick with their initial choice.
Time to shatter this illusion with the truth! If you switch doors, you double your probability of winning!
What!?
How to Solve the Monty Hall problem
When Marilyn vos Savant was asked this question in her Parade magazine column, she gave the correct answer that you should switch doors to have a 66% chance of winning. Her answer was so unbelievable that she received thousands of incredulous letters from readers, many with Ph.D.s! Paul Erdős, a noted mathematician, was swayed only after observing a computer simulation.
It’ll probably be hard for me to illustrate the truth of this solution, right? That turns out to be the easy part. I can show you in the short table below. You just need to be able to count to 6!
It turns out that there are only nine different combinations of choices and outcomes. Therefore, I can just show them all to you and we calculate the percentage for each outcome.
| You Pick | Prize Door | Don’t Switch | Switch |
| 1 | 1 | Win | Lose |
| 1 | 2 | Lose | Win |
| 1 | 3 | Lose | Win |
| 2 | 1 | Lose | Win |
| 2 | 2 | Win | Lose |
| 2 | 3 | Lose | Win |
| 3 | 1 | Lose | Win |
| 3 | 2 | Lose | Win |
| 3 | 3 | Win | Lose |
| 3 Wins (33%) | 6 Wins (66%) |
Here’s how you read the table of outcomes for the Monty Hall problem. Each row shows a different combination of initial door choice, where the prize is located, and the outcomes for when you “Don’t Switch” and “Switch.” Keep in mind that if your initial choice is incorrect, Monty will open the remaining door that does not have the prize.
The first row shows the scenario where you pick door 1 initially and the prize is behind door 1. Because neither closed door has the prize, Monty is free to open either and the result is the same. For this scenario, if you switch you lose; or, if you stick with your original choice, you win.
For the second row, you pick door 1 and the prize is behind door 2. Monty can only open door 3 because otherwise he reveals the prize behind door 2. If you switch from door 1 to door 2, you win. If you stay with door 1, you lose.
The table shows all of the potential situations. We just need to count up the number of wins for each door strategy. The final row shows the total wins and it confirms that you win twice as often when you take up Monty on his offer to switch doors.
Why the Monty Hall Solution Hurts Your Brain
I hope this empirical illustration convinces you that the probability of winning doubles when you switch doors. The tough part is to understand why this happens!
To understand the solution, you first need to understand why your brain is screaming the incorrect solution that it is 50/50. Our brains are using incorrect statistical assumptions for this problem and that’s why we can’t trust our answer.
Typically, we think of probabilities for independent, random events. Flipping a coin is a good example. The probability of a heads is 0.5 and we obtain that simply by dividing the specific outcome by the total number of outcomes. That’s why it feels so right that the final two doors each have a probability of 0.5.
However, for this method to produce the correct answer, the process you are studying must be random and have probabilities that do not change. Unfortunately, the Monty Hall problem does not satisfy either requirement.
Related post: How Probability Theory Can Help You Find More Four-Leaf Clovers
How the Monty Hall Problem Violates the Randomness Assumption
The only random portion of the process is your first choice. When you pick one of the three doors, you truly have a 0.33 probability of picking the correct door. The “Don’t Switch” column in the table verifies this by showing you’ll win 33% of the time if you stick with your initial random choice.
The process stops being random when Monty Hall uses his insider knowledge about the prize’s location. It’s easiest to understand if you think about it from Monty’s point-of-view. When it’s time for him to open a door, there are two doors he can open. If he chose the door using a random process, he’d do something like flip a coin.
However, Monty is constrained because he doesn’t want to reveal the prize. Monty very carefully opens only a door that does not contain the prize. The end result is that the door he doesn’t show you, and lets you switch to, has a higher probability of containing the prize. That’s how the process is neither random nor has constant probabilities.
Here’s how it works.
The probability that your initial door choice is wrong is 0.66. The following sequence is totally deterministic when you choose the wrong door. Therefore, it happens 66% of the time:
- You pick the incorrect door by random chance. The prize is behind one of the other two doors.
- Monty knows the prize location. He opens the only door available to him that does not have the prize.
- By the process of elimination, the prize must be behind the door that he does not open.
Because this process occurs 66% of the time and because it always ends with the prize behind the door that Monty allows you to switch to, the “Switch To” door must have the prize 66% of the time. That matches the table!
Related post: Luck and Statistics: Do You Feel Lucky, Punk?
If Your Assumptions Aren’t Correct, You Can’t Trust the Results
The solution to Monty Hall problem seems weird because our mental assumptions for solving the problem do not match the actual process. Our mental assumptions were based on independent, random events. However, Monty knows the prize location and uses this knowledge to affect the outcomes in a non-random fashion. Once you understand how Monty uses his knowledge to pick a door, the results make sense.
Ensuring that your assumptions are correct is a common task in statistical analyses. If you don’t meet the required assumptions, you can’t trust the results. This includes things like checking the residual plots in regression analysis, assessing the distribution of your data, and even how you collected your data.
For more on this problem, read my follow up post: Revisiting the Monty Hall Problem with Hypothesis Testing.
As for the Monty Hall problem, don’t fret, even expert mathematicians fell victim to this statistical illusion! Learn more about the Fundamentals of Probabilities.
To learn about another probability puzzler, read my post about answering the birthday problem in statistics!
Hello Jim,
The Monty Hall Problem and solution certainly pains the brain.
Here are some interesting ways that I have tried to analyse the Monty Hall Problem, which suggest that there is a 50% chance of winning the car whether a switch is or is not made by the contestant.
At the stage where the contestant has to decide whether to switch or remain with the door first chosen, let’s assume that the contestant had chosen Door 1 and Monty had opened Door 3. At this point, the contestant has to decide whether to switch to Door 2 or remain with the initial choice of Door 1.
Now we stop the game and bring in another contestant that has no knowledge about the previous stage of the game. This new contestant is told that behind Door 1 or Door 2 there is a car. The new contestant is also told that the previous contestant had chosen Door 1 to be where the car is located. The new contestant is then given the choice to remain with the selection of the first contestant or switch to Door 2. As shown below, the new contestant has a 50% chance of winning the car whether a switch is or is not made.
Door Chosen Prize Door Don’t Switch Switch
1 1 Win Lose
1 2 Lose Win
2 1 Lose Win
2 2 Win Lose
According to The Monty Hall Problem : A Statistical Illusion, the first contestant has a 67% chance of winning the car if a switch is made to Door 2.
In the scenario above the second contestant has a 50% chance of winning the car if a switch is made. The question is, why would the second contestant have less chance of winning the car even though the choice that needs to be made at this stage of the game is exactly the same for both the contestants?
Which door the contestant first chooses is irrelevant to the outcome of the game. Regardless which of the three doors the contestant chooses, Monty will eliminate one of the two doors that has no car behind it. So the game of chance does not begin until the contestant has to decide to stay with the door first selected or switch to the other remaining door. The car must be behind one of these two doors and there is an equal probability that the car lies behind one of these doors.
You can look at the game in another way. Let’s assume that the contestant is shown three doors and told that behind one of these doors there is a car. There is a 1 in 3 chance that the car lies behind one of the doors. Monty then says to the contestant, I’m feeling generous, before you make a selection I will increase your chance of winning the car by opening one of the three doors does not have a car behind it. Now there are only two doors remaining and the contestant is asked to select a door. The contestant has a I in 2 chance of winning the car if a switch is or is not made. After the door has been selected, Monty then tells the contestant that a switch can be made. Whether a switch is or is not made, there is a 50% chance of winning the car. Eliminating one door prior to the contestant making the first selection or eliminating a door after the contestant has made the first selection, the contestant has a I in 2 chance of winning the car if a switch is or is not made.
At the stage when the contestant is given the choice to stay or switch after Monty has eliminated one of the three doors, the suggestion is made that the probability of 1/3 for the car being behind the eliminated door must be transferred to the door that the contestant has not chosen. The reason given for this is that there is only a 1/3 probability that the car is behind the door first selected by the contestant and 2/3 probability that the car lies behind the other two doors. While it is true that there is a one in three chance of the car being behind the door first selected by the contestant, that probability changes when one door without the car is eliminated.
With one door eliminated, the odds of the car being behind one of the two remaining doors must be increased. Why should the increased odds be solely attributed to the door that the contestant has not first chosen? When one of the three doors is eliminated by Monty, the 1/3 probability that has been eliminated should be equally distributed between the two remaining doors. With only two doors remaining, the car must be behind one of the two doors.
As the car must be behind one of the two doors remaining to be opened, one half of the 1/3 probability that was attributed to the door eliminated by Monty, i.e. 1/6, should be transferred to the door first selected by the contestant and the other 1/6 should be transferred to the other remaining door. So now the two remaining doors in play at the switch stage each have an equal probability, i.e. 3/6, of hiding the car.
Another way to explain this can be seen in the Table below. The Table shows all of the possible outcomes of the contestants choices if the car is located behind Door 1. The door which Monty opens is shown as Nothing*. The outcome of the contestant’s choice to stay or switch is shown as a win or loss at the right hand half of the table.
Behind Behind Behind Stay Result Switch Result
Door 1 Door 2 Door 3 Door Selected Door Selected
1 2 3 1 2 3
Car Nothing Nothing* W L
Car Nothing* Nothing W L In this case the contestant has selected Door 1. Monty can open door 2 or 3.
Behind Behind Behind Stay Result Switch Result
Door 1 Door 2 Door 3 Door Selected Door Selected
1 2 3 1 2 3
Car Nothing Nothing* L W
In this case the contestant has selected Door 2. Monty must open door 3.
Behind Behind Behind Stay Result Switch Result
Door 1 Door 2 Door 3 Door Selected Door Selected
1 2 3 1 2 3
Car Nothing* Nothing L W
In this case the contestant has selected Door 3. Monty must open door 2.
I believe that the reasoning behind the 67% chance of winning the car when a switch is made does not take into account all of the possible outcomes in the game. The Table above shows all of the eight possible outcomes, four of which provide a win and four of which result in a loss. The number of possible outcomes would be the same if the car were to be located behind door 2 or door 3. Therefore out of 24 possible outcomes, there are 12 outcomes that would result in a win and 12 outcomes that would result in a loss. There is no change in the odds if the contestant decides to switch or stay with the original door selected.
The calculation which demonstrates a 67% chance of winning the car when a switch is made has not taken account all of the possible outcomes in the game. The Table shown in The Monty Hall Problem : A Statistical Illusion, shows 3 wins and 6 losses for a don’t switch and 6 wins and 3 losses for a switch. This Table is based on the premise that there are 9 possible picks in the game with 18 different outcomes. However, this Table does not consider the possible outcomes when the contestant has chosen the door that has the car behind it. in these cases, Monty has two possible doors to open.
In both these cases, if the contestant doesn’t switch that results in a win for a don’t switch and a loss for a switch. The additional win and additional loss would occur in all of the three scenarios where the contestant’s first door selection has the car behind it, i.e. the contestant has chosen the correct door when the car is located behind any of the three doors. So the table shown in the Statistical Illusion should show 6 wins and 6 losses for a switch and 6 wins and 6 losses for a don’t switch. In the game, there are 12 picks(not 9)that could be available to the contestant, with 24(not 18)different outcomes. This is shown in the Table set-out above. If there are 6 wins and 6 losses whether the contestant switches or doesn’t switch, why aren’t the odds of winning or losing the car 50%?
That’s enough, my head hurts!
Hi Ezio,
Trust me, probability calculations and computer simulations all support the fact that switching doubles your chances of winning by increasing it to 67%. What I’d recommend is that, instead trying to devise convoluted scenarios that essentially equate to arguing that 2 + 2 = 5, you should instead focus on trying to understand how and why the correct answers works.
Your scenario involves switching in a new player mid-game. That does not represent the actual game and, therefore, doesn’t address why you should switch in the actual Monty Hall game.
Yes, in both your scenario and the actual Monty Hall scenario, after Monty opens a door, you end up with two doors and the prize is behind one of them. That appears to be 50/50.
However, the big lesson from the Monty Hall problem is that if you understand Monty’s process, you can use that information to determine that the prize is more likely behind one of the doors rather than the other. You’re using that information to improve your odds of winning.
Imagine you have a weighted coin such that heads faces up 2/3 of the time and tails 1/3. Now imagine you have two people. One who knows the coin is weighted and the other who doesn’t. The person who doesn’t have that information will think, there is one head out of two sides of the coin. Hence, it’s 50/50. However, with the knowledge, you know that heads is more likely to occur. If those two people were betting on coin flips, the one with knowledge would win more often! Knowledge is power.
Monty Hall’s process is similar in that it is weighted such that the unopened door Monty offers you is more likely to have the prize.
Moral of the lesson: Just because there is one prize and two outcomes, it doesn’t automatically mean that the chances are 50/50. You must understand the process behind how the prize/heads appears. That’s a good general lesson for life too! 😊
very beautiful solutin. greets <3
I initially felt it had to be 50-50, but changed my mind when I thought about it as follows — involving minimal math.
If I initially were able to pick either one door or a group of two doors, of course I would pick a group of two doors — because it would have twice the chance (i.e., 2/3 rather than 1/3) of containing the winning door. But I’m initially allowed to pick only one door, so I do so.
When Monty reveals what’s behind one of the doors in the group of two that I did not pick, and gives me a chance to switch my choice, he is essentially letting me pick the group of two doors. That group still has a 2/3 chance of containing the winning door, just as it did before he gave me the additional info about one of those doors. So I should definitely switch.
My brain stopped hurting once I thought of it this way!
As I explained in my original post, my issue was trying to rationalize why the answer and explanation you give appears to be counter intuitive. The answer lies in the question itself.
The error in logic is ASSUMING these two questions are the same:
1. What are the odds of the prize being behind Door 1 or Door 2?
2. What are the odds of winning IF I change MY choice?
The answer to #1 is 50/50. This is supported by your table: 9 combinations each having two outcomes (18 in total) but only 9 wins.
The answer to #2, as you explain it, ASSUMES the original choice is incorrect and only holds true IF it is incorrect. IF it is incorrect, then switching to the other door will ALWAYS yield the prize. The odds of winning are totally dependent on the odds of the initial choice being INCORRECT which will happen 2 out of 3 times or 66 PERCENT OF THE TIME!
Hi Kevin,
I think I understand what you’re asking. It wasn’t clear in your first comment.
One thing to realize is that the odds for Monty’s two doors change during the game. Why? Because Monty uses his knowledge to systematically remove a door without a prize.
For your first question, at the beginning of the game, door 1 and door 2 each have a 33% chance of containing the prize. Suppose you pick door 1, that will continue to have a 33% chance of winning.
In this scenario, Monty has two doors (2 & 3) and their combined probability of having the prize is 67%. The door that Monty opens has a 0% chance of containing the prize because he knows where the prize is located, and intentionally won’t reveal it. So, back to your question, Door 2’s ultimate probability depends on what Monty does. If he opens door 2, then it obviously has a 0% chance. If he opens door 3, then door 2 has a 67% chance.
Door 2’s probability changes thanks to Monty and his non-random decisions! Think of it this way. Monty’s two doors have a combined probability of 67%. X is the probability of the unopened door. The opened door has a 0% probability. And the total probability is 67%. So, it becomes an algebra problem. 0 + X = 67%. Hence, X = 67%.
Regarding your 2nd point, yes, for your initial choice, you have a 1/3 chance of picking the correct door and a 2/3 chance of picking the wrong door. Therefore, if you don’t switch, you have a 1/3 chance of winning and a 2/3 chance of losing. Conversely, if you switch, it flips your outcome and, therefore, you have a 2/3 chance of winning and 1/3 chance of losing.
Here is an interesting thing that I read once: the odds depend on whether Monty knows whether he is picking a non-winning door or not. If he does the odds are 66% for the remaining door. If he doesn’t the odds remain the same at 33%?
Hi Jim,
Recently I became aware of the many discussions about the Monty Hall problem. Intuitively I would have said the probability of winning the car was the same if I stayed with Door 1 or switched to Door 2.
I am not a mathematician, and know nothing about probability theory, but I understand the arguments presented about why there is a greater probability of winning by switching from Door 1 to Door 2.
If, on the other hand, I had initially chosen Door 2, a goat was revealed behind Door 3, and all other assumptions remained the same, the same logic would dictate there is a greater probability of winning by switching to Door 1.
In both cases, the probably of winning by switching is the same. Looking at the problem from this perspective would suggest the true probability of winning the car is the same whether or not I switch.
Please help me understand the error in my reasoning…
Hi Kevin,
There’s a variety of reasons why some people have a hard time understanding the correct solution to this problem. I don’t know the exact reason for your confusion.
I’ve recently tried to explain it someone else who was also having a difficult time with it. Perhaps my explanation in this comment will help clarify? If not, reread my article again. You can also look through the comments for various other ways to explain it. Perhaps one of those will work for you.
Kevin Boyle – you’re so close to how it works. 🙂 It doesn’t matter which door you initially pick, or which specific door you switch to. The way the problem works is that you’re essentially choosing between 1 door (the door you initially choose) and 2 doors combined (the 2 doors you didn’t choose). Since Monty *always* opens a door with a goat, you don’t actually gain any useful new information from that red herring act – it’s just a smokescreen and a distraction. The fact is, the door you initially choose has a 1/3 chance of containing the prize, a chance that never changes and can’t change since you chose from among 3 doors, while the other 2 doors *combined* have a 2/3 chance. You already know one of those 2 doors has a goat, so this information being revealed to you isn’t useful. Since that door now has 0 chance of containing the prize, the remaining door – the one you would switch to – must have a 2/3 chance all by itself. Your door: 1/3. The other door: 2/3. Does that make sense?
I used to think like you. But here’s how I have convinced myself. When you get the first choice, select 2 doors in your head (say “A” and “B”). But you say out loud the other door (“C”). There is a 66% chance that either A or B is right. Now the host has to eliminate A or B – but he must eliminate a wrong door – so you are left with only the correct door after he has eliminated the wrong one. So your swapped choice to A (or B if host eliminated A) had a 66% chance of being right in the beginning. My point is to look on the whole game as though you get to choose 2 doors right at the start and the host will kindly eliminate the wrong one of these for you. (but of course you don’t “choose” out loud the pair that you really want at the start).
Hi Stewart, I like that way of thinking about it!
You dont have a 66% chance if you switch. You are simply going from one door of 33% chance yo a different door of 33% chance, initially. You don’t get the statistical advantage of yotalli g the chances of each door. Sine it ends at 2 doors, that is where it starts. Wben you are faced with your pick or a swirch. It is 50/ 50. Assume a different start. Yiu have 3 doors. You pick one. Monte opens no doors and says, do you want to switch? Welk, your first pick only had a 33% chance, so, I am told I should switch. No. 33% for each door. And all Monte does is raise the odds to 50% by eliminating a red herring door. Stay or switch? Same odds. Thanks.
Hi Steve,
Sorry but you’re still missing the point. The process you’re using to calculate the probabilities is based on random events. Monty uses a non-random process that selectively removes a non-prize door from the mix. That intentional selection boosts the odds for the final door.
Look at it this way. After you pick your initial door, there is a 2/3 chance that the prize is behind one of Monty’s two doors. That’s the total probability for those two doors. The door that Monty opens has zero probability of containing the prize. At this point, it’s simple algebra.
X is the probability for that final door for which we’ll solve. 66% is the known total probability for the two doors. 0 is the prize probability for the door that Monty opens. He will never open a prize door and has the knowledge to guarantee that. So:
X + 0 = 66%
Hence, X = 66%
The probability that the prize is behind Monty’s unopened door is 66%.
Yes, the answer doesn’t seem intuitive because your mind leads you astray by using the independent, random events calculations for probabilities, which doesn’t apply. This a proven, solved answer. It’s also backed up by simulation studies. So, instead of doubling down on the wrong answer (which is like insisting 2 + 2 = 5), you should try to understand the correct answer. You are literally arguing that X + 0 = 66% and that, therefore, X must equal 50%! Just mathematically incorrect.
Steve: no, you’re incorrect. Your door has a 33.3% chance, while the other door has a 66.6% chance. Yes, that’s right – the other single door has a 66.6% all by itself. How? Because when Monty opens one of the 3 doors, it *always* contains a goat. This is never random and never changes. When you chose your door, you chose from 3 doors, so that initial choice remains at 33%. Agree? The other 2 doors combined represent a 66% chance. Agree? So when Monty non-randomly opens a losing door, which as you yourself said is just a red herring, the entirety of that 2/3 chance is transferred to the remaining door. Your door: 1/3. The other door: 2/3. See? If you still don’t believe it, you can run this game in computer simulations that prove the results, or even just play the game in real life with a friend and 3 standard playing cards. Record the results accurately, and you’ll find that you only win 33% of the games you stay, but 66% of the games you switch.
u leave out the part where after you pick a door, Monty Hall opens another door that isn’t the prize. Then he gives you a choice to stick with your door or switch. If you switch your odds are 66% in your favor, because you have in essence chosen two doors: the switched door and the one Monty revealed. It may be counterintuitive, but it has been confirmed in computer simulations.
Jim, 1. When you pick a door, you have a 33 % chance. Doesn’t matter which door. By switching, I go to a door that also had a 33% chance. The revealing of a door now takes that up to a 50% chance. Makes no statistical difference whether you stay or switch. The 3rd door in the graph above is the illusion. 2. Assume only 2 doors to start… which is where it ends. 50 % chance. In fact, that is what the reality is. There really are only 2 doors in play. Thanks. Steve
Hi Steve,
There are three doors in play at the start. The point you’re missing is that Monty follows a non-random process that systematically removes a non-prize door from play. That’s why Monty’s final remaining door has a higher probability of having the prize. Computer simulations that play the game thousands of times also bear this out. The correct solution of switching for a 66% chance of winning is confirmed by both probability theory and empirically using computers.
You are thinking about a random process, but that’s not applicable here. That’s the illusion I refer to.
@Jim Frost , This is regarding the 100 door case . In the case where I pick a door and then Monty, knowing fully well where the car is, opens 98 doors, I can see why it makes sense to switch. Probability of the door that I had initially picked having the car is only 1/100 but the probability of me having picked an empty door is 99/100.
But, let’s say that Monty does not know where the car is. He still goes on to open 98 doors and, coincidentally, all these doors are empty. Can you please explain why switching won’t improve the probability in this case ?
Hi,
If you dig back in these comments, you’ll see where I did a similar thing but with three doors. In that case, if you throw out the games where Monty accidentally reveals the prize, the chances do become 50/50 for the final two doors. That’s thanks to redo games. The same would happen with 100 doors.
Here’s a wrinkle to the three door question that has kept me up at night; What if there are two contestants who each pick a door independently, and in this situation one picks door#1 and the other chooses door #3, and it happens that one of them is correct, so Monty opens door #2 to show there is no car there. He then offers each of the two contestants the option to change their pick to either door #1 or door #3. Should they both switch their pick? And if so, which one has really improved their chance of winning?
I have an idea, but would love to hear your explanation and hopefully get a good night’s sleep for once.
OK. I got it now. Once I see it – it becomes obvious! The best convincing arguement I have is that it’s like getting to choose 2 doors instead of one. If you choose (in your head) the “other” two (i.e. not the one you declare out loud) then you have 2 chances of winning. The host is obliged to eliminate one door (of the two you chose in your head) which definitely doesn’t have the car, so your “2 chances” is contained in that one door he didn’t open. Of course, you might have chosen out loud the correct door, and if you swap to your original choice (the one in your head) then you are going to loose – but that is only going to happen 1 out of 3 times.
Excellent explanation, Jim. I used the same analogy to optical illusions just yesterday when explaining this to my 12YO daughter. Along with the missing dollar story, this is one of my favorite math illusions. The setup is easy to explain to a child, yet incredibly difficult to make someone understand the reality. That’s pretty clear from seeing the endless comments explaining why you’re wrong despite being easily proven with a computer simulation or a simple table like you’ve provided in the article.
Honestly, I still sometimes bounce back and forth between understanding and not depending on the day. 😉
I finally wrapped my head around the problem this way: You start with a 1 in 3 chance of being right. Monty could add more wrong doors and you’d still have a 1 in 3 chance of being right. Monty’s removal of a wrong door doesn’t change the likelihood that you’re right, so after he removes a wrong door the remaining door of the original three doors has to have a 2 in 3 chance of being right.
Here’s the problem with the Monty Hall problem as described here: He doesn’t always open another door. If he [i]always[i/] opens a door, then you do double your chances of winning. However, if he opens the door only 2/3 of the time, then you don’t have any improvement in your chances of winning
That’s incorrect. Monty always opens a door. In fact, he systematically opens a door that does not contain the prize, which is why the probability for switching gets inflated. It’s non-random decision that systematically lowers the probability of non-prize doors.
Hi Jim. I’m kinda with Robert, but the error in your table is that there are actually only 6 options. The 2 “goat” doors are in reality only 1 choice. The goat doors are, in probability terms, the same option. So when you work it through, you get 3 wins for each scenario. The only way you will get a deviation from the 50/50 is if you have to preselect “stick” or “swap” BEFORE the host reveals the first goat. Then, if you had chosen “stick”, you become frozen in your first probability and the revelation of a goat door makes not one iota difference – you can’t change your mind. But if you get to chose “stick” or “swap” AFTER the goat door is revealed, then you are in the new probability realm where you have a 50/50 probability – no matter which option you chose.
Hi Stewart,
Thanks for writing!
Unfortunately, what you write is simply incorrect. This is not a matter in which you can have an opinion to side with. It’s a mathematical issue where there is a correct answer versus various incorrect answers. It’s like you’re agreeing with someone who says 2 + 2 = 5! It’s just not correct on a factual level.
As for the table, it having 9 rows is correct. Why?
When you know the contestant’s initial choice and the location of the prize, then the resolution of whether you’ll win or losing by switching and staying is entirely constrained. There are three possibilities that the contestant can choose for their initial door. And there are three possible locations for the prize. Hence, there are 3 X 3 = 9 total scenarios. Those nine scenarios entirely describe the frequencies of winning or losing by switching and stay for all possible games.
Adding more rows as you suggest simply duplicates a subset of the scenarios, which skews the probabilities. What you’re proposing skews the probabilities in fashion that fits with your preconception of the correct answer and, hence, “feels” right. But it’s not!
I think I know a simpler way to explain it. One third of the time, the prize is behind door number one. If you choose door number 1 and switch, you lose, but if you choose either number 2 or 3 and switch, you win, because Monty hall is guaranteed to show you the empty door. There are two doors which make you win if you switch, and only one door which causes you to lose if you switch. That is why you should switch. I didn’t understand you at first either, but I had the common sense to actually draw 3 sets of 3 doors, each with a different prize door, and figure it out myself. Tell anyone who doesn’t believe you to do the same.
Hi Jim. I’m not a mathematician but looking at your table where you list the 9 different scenarios, shouldn’t there be 3 more scenarios? If I choose box 1 and the price is in box 1, then the game show host can either choose 2 or 3, and in each scenario you loose if you switch. Same goes for if you choose box 2 and the price is in box 2 etc…
So basically, I think one needs to count in the possible choices of the game show host which in the end gives a 50/50 ratio in the end and now I can finally sleep :).
Hi Robert,
The table is correct with 9 rows. You can pick three doors and the prize can be behind three doors. Hence, there are 3 X 3 = 9 combinations. Each row equals one of those combinations. If you added more rows you’d be duplicating scenarios and throwing the probabilities off.
In your example, if you pick door 1 and the prize is behind door 1, then you lose if you switch but win if you stay. That’s what the first row shows. Then you simply go through all the rows to figure the outcomes for all combinations of choices and prize locations for both switching and staying. The overall logic in determining the outcomes is that when your choice is also the prize door, you’ll win by switching. Makes sense. Conversely, when your door choice is not the prize door, you win by switching. That also makes sense. Finally, you sum them up and calculate percentages.
The end result is that you win 66% of the time by switching and 33% by staying. Sorry if that costs you some sleep! 🙂
I wonder what are the probabilities when Monty chooses randomly and he chooses an empty door, and of course if Monty chooses a right door, you lose right away, so your chance stays at 33.33%. But what if Monty chooses an empty door and he offers you whether you wanna change, if you stay, does your chance stay at 33.33% or automatically increases to 50% ? and if you change, does the probability change as well ?
comment section is too long, didn’t read them all
the reason why people are arguing is because you didn’t explain well enough about how monty opens the 2nd door. Because he only opens an empty door not random door, that means he’s eliminating 33.33% of chance of getting the wrong door.
let’s think this way, if you choose and stay, your chance keeps at 33.33%, and if you change, your chance would go up to 66.67% comparing to when you first made the choice, because 2/3 of chance you will choose the wrong door, and if you change, you will always get the right door since monty eliminates one for you.
let’s talk about the other scenario, if monty opens a random door after you choose, and if monty chooses a right door and it also eliminates you as well, then your chance stays at 33.33%
Hi Roy, I’m not sure who this is addressed to, but that’s a point that I make very clear throughout my article. Monty’s choice is non-random and affects the probabilities. Unfortunately, some people have a difficult time understanding how one prize behind two doors does NOT equate to 50/50! Thanks for writing!
Michael P you are right. It is always a 50/50 choice when you’re down to 2 doors and one has the prize. It does not matter how you get there. If you solve by guessing the right door it is 1/2. If Monty did not not help then it’s 1/3.
Here the part folks are not getting: if you solve by eliminating goats it is 1/2. If Monty does not help it is 2/3 × 1/2 = 1/3!
Even with 100 doors, taking away 98 with 100% chance of goats leaves 1 goat and 1 car, then …
Hi Robert,
Sorry, but that is incorrect. Just because you have two possibilities and one item does not automatically mean that the probability is 50/50. You need to understand the underlying factors and processes. There are many cases for why it can be something other than 50/50.
Consider, suppose a gambler created a weighted coin to favor heads. In that case, you have one winning outcome (heads) out of two possible outcomes. However, it’s clearly not 50/50 because of the underlying factors.
Or suppose I show you two manufactured parts and say that one has a hidden defect. I don’t tell you which one is good and which is defective. You might think that’s 50/50. Now, imagine you know that part A comes from an assembly with very high quality control and a lower defect rate and part B comes from a line with sloppy quality control and a higher defect rate. In that case, it’s not 50/50 because part B is created by a process that is more likely to create defects.
In the Monty Hall problem, we’re looking at a non-random process that is more likely to leave the prize behind Monty’s door than the door you chose. In previous comments, I’ve explained how that works. If you think about it very carefully, I’m sure you’ll be able to understand it!
You actually need to satisfy very specific conditions for an outcome with two possibilities to be exactly 50/50, and the Monty Hall problem (and many other scenarios) don’t satisfy it. Hint: multiplying 2/3 * 1/2 is not appropriate given what we know about the process so your “solution” entirely misses the mark.
Hey Jim, I think what Richard is saying is that the REAL illusion here is “hanging on” to the 3 door stats. When the “goat” door is opened, it NOW becomes a brand new game so all your statistics should be with just 2 doors as if there was never a 3rd. Same with 100 doors – 98 duds …… NEW game = 50/50 chance. I think Marilyn vos Savant knows this 😉
Hi Michael,
Uh oh, you’re falling for the illusion that two doors automatically mean it’s 50/50!
You can’t just restart a new game with two doors being 50/50 because Monty has interacted with the doors in a non-random manner. You need to incorporate that when you’re working with the final two doors.
Imagine that there are 100 doors. You choose 1 and Monty has the remaining 99. You know that his 99 doors had a 99% probability of collectively containing the prize at the outset, while your lone door only had a 1/100 chance of having the prize.
Of those 99, he systematically opens 98 of them and intentionally does not reveal the prize. He opened his doors but carefully avoided revealing the prize. Surely you can see that his final, unopened door almost certainly contains the prize. If you can’t picture that, put yourself in Monty’s shoes. Imagine you have 99 doors. You know where the prize is located. Your job is to open 98 of the doors but not reveal the prize. You’ll be very careful to avoid that one door with the prize, which means it will be the last remaining one that you don’t open.
Conversely, your door choice won’t magically increase from 1% to 50%. Its probability does not change from its initial 1%.
Monty’s process was a non-random interaction that affects the outcome. You need to incorporate that knowledge to correctly understand the probabilities at play. Simply resetting to two doors each with a 50/50 chance throws out all that process knowledge and leads you astray!
Ignoring psychology, I wonder if the “solution “ is ignoring the new information. Whichever door the host reveals, you can now rule out one of the three initial possibilities, I.e. that you picked that particular one with the goat. Therefore you would need to delete 3 of the rows in the table showing loosing staying with the original choice and winning if you switch, leaving 3 and 3.
The wrong solution looks so right, but upon further considerations we can arrive at the established solution. The wrong solution using accounting of all scenarios is as follows (copying to Excel should make this again more readable):
True Door Contestant Choice Round 1 Host Eliminates Contestant Choice Round 2 Strategy Outcome
1 1 2 1 Stay Win
1 1 3 1 Stay Win
1 2 3 2 Stay Lose
1 3 2 3 Stay Lose
2 1 3 1 Stay Lose
2 2 1 2 Stay Win
2 2 3 2 Stay Win
2 3 1 3 Stay Lose
3 1 2 1 Stay Lose
3 2 1 2 Stay Lose
3 3 1 3 Stay Win
3 3 2 3 Stay Win
1 1 2 3 Switch Lose
1 1 3 2 Switch Lose
1 2 3 1 Switch Win
1 3 2 1 Switch Win
2 1 3 2 Switch Win
2 2 1 3 Switch Lose
2 2 3 1 Switch Lose
2 3 1 2 Switch Win
3 1 2 3 Switch Win
3 2 1 3 Switch Win
3 3 1 2 Switch Lose
3 3 2 1 Switch Lose
Under each strategy, we would count an equal number of success. Thus it appears to be 50/50 chance of regardless of choice…
…but we also know that the probability of the scenario where the contestant picks the correct door in Round 1 is 1/3, thus the two options for the Host must total in probability to that equal to each of the scenarios where the constest picks the wrong door (i.e. 1/3). Upon accounting for the probability of each scenario:
True Door Contestant Choice Round 1 Host Eliminates Probability of this row Contestant Choice Round 2 Strategy Outcome Outcome N Weighted_value
1 1 2 0.166666667 1 Stay Win 1 0.166666667
1 1 3 0.166666667 1 Stay Win 1 0.166666667
1 2 3 0.333333333 2 Stay Lose 0 0
1 3 2 0.333333333 3 Stay Lose 0 0
2 1 3 0.333333333 1 Stay Lose 0 0
2 2 1 0.166666667 2 Stay Win 1 0.166666667
2 2 3 0.166666667 2 Stay Win 1 0.166666667
2 3 1 0.333333333 3 Stay Lose 0 0
3 1 2 0.333333333 1 Stay Lose 0 0
3 2 1 0.333333333 2 Stay Lose 0 0
3 3 1 0.166666667 3 Stay Win 1 0.166666667
3 3 2 0.166666667 3 Stay Win 1 0.166666667
1 1 2 0.166666667 3 Switch Lose 0 0
1 1 3 0.166666667 2 Switch Lose 0 0
1 2 3 0.333333333 1 Switch Win 1 0.333333333
1 3 2 0.333333333 1 Switch Win 1 0.333333333
2 1 3 0.333333333 2 Switch Win 1 0.333333333
2 2 1 0.166666667 3 Switch Lose 0 0
2 2 3 0.166666667 1 Switch Lose 0 0
2 3 1 0.333333333 2 Switch Win 1 0.333333333
3 1 2 0.333333333 3 Switch Win 1 0.333333333
3 2 1 0.333333333 3 Switch Win 1 0.333333333
3 3 1 0.166666667 2 Switch Lose 0 0
3 3 2 0.166666667 1 Switch Lose 0 0
We arrive at staying strategy having 1/3 success rate and switching strategy having 2/3 success rate. I think it is helpful to consider the probability for each scenario with the “two door” choice for the Host to help convince the “yet to be” convinced.
I was reading through this whole thread and from the very beginning, I was waiting for someone to point out what I was thinking.
There are only in fact 2 choices all along and not 3 because Monty will ALWAYS open one of the two doors which you haven’t chosen, which is always one of the two booby-prize door. Therefore, you only had the choice of two doors from the start, as the same one is ALWAYS discarded.
After this, it is a matter of guessing the psychology of the host, Monty.
It simply turns into a mind game, which is the point of the problem.
It has nothing to do with statistics.
The question Monty forces you to ask is: Do you feel that he is on your side or not?
Is he routing for you to win or does he want to make a fool of you and trick you into losing?
To make it clearer. You have a Nazi who gives you the choice of three doors. Two leads to your death, (which he wants), the other leads to freedom, which is is obliged out of a warped sense of honour to, honour, even though he actually wants to kill you.
After you have chosen and when he opens one of the doors which leads to your death, you must assume that his intention is to get you to switch. (Unless it’s a double bluff?).
Hi Reb, it actually is about statistics and probability, but psychology also plays a role. For starters, it’s a fair game. Monty will open a door that doesn’t contain the prize. And he won’t cheat about the prize’s real location. So, no need to invoke the Nazis!
By understanding the process, you can figure out you double your chances of winning by switching. Psychology comes into play in a different manner than you suggest. Many people incorrectly think the chances are 50/50. They are a bit invested in their original choice and they see no reason to switch, so they stay and actually reduce their chances!
Hi Jim, you have amazing patience do deal with all of these comments! I’m hung up on the optical illusion of square a and b. I just can’t see them being the same.
Hi Michael, they look different to me too! To see the truth, you can create a paper cut out to reveal only those two squares. Or use an image/photo editing application to sample the colors from those squares. Once an illusion takes hold, it’s hard to see anything else!
Jim, I think I solved our dilemma to “What is the probability of winning if the coin says guess red?”
You roll the die and cover it so I don’t see it. I toss the coin and it says guess red. One could think that the probability guessing correctly is 1/3 because that was the probability of the die being red. But that would contradict the fact that the die result is fixed and does not change. Also, it’s incorrect since there is only one correct probability of both matching. If I had tossed the coin behind my back then I don’t know that it says guess red. Then I uncover the die to see it is red. Now the same logic would say the probability is 1/2 of guessing correctly. It can’t be both. Solution: once the die is rolled and the coin tossed, the results are determined, giving us the fourth “fact” I described previously. So, the probability is either 1 when the die is red or 0 when it is green. This is the result for a single game.
What about the long-run? Remember I asked you what was in the numerator and denominator to get 1/3. I dug deeper into the data and found the answer. The question means that the denominator must be “coin red” instances and the numerator is “die and coin red” instances. Using my posted simulation data, I got 0.348, or approximately 1/3. I had calculated a fraction with “die red” instances as the denominator but the same numerator, yielding 0.507, approximately 1/2.
Each answers correctly a different question and that is why you didn’t see anything wrong with my equations and data but correctly said it didn’t answer your question. Your question is about the probability after the coin is tossed and mine was before the coin is tossed. Whew!
Anyway, thank you again. Your switching to an analogy with the die was a great idea to address my original question:
“If I don’t know that the earth is the third planet from the sun, then what is the probability that earth is the third planet from the sun?
With your analogy, we looked at two situations: a) I knew the distribution of the colors on the die and b) I did not know the distribution (hence, the coin tossing).
I learned so much from you and the discussion including the following.
L1. When guessing whether an event occurred, the probability of guessing correctly depends on the probability the guesser uses.
L2. The probability the guesser uses depends on the information the guesser has.
L3. When guessing, if no information is available to prefer one option, then each is treated equally likely.
L4. Once an event occurs, it’s probability is 1 (when it has the feature of interest) or 0 (when it doesn’t).
L5. The probability of a fact is 1, while the probability of guessing whether a fact is true depends on the probability distribution used to guess.
L6. Probability of fact being true is not the same as probability of guessing a fact.
L7. Since you advocate using simulations, I also learned how to do probability simulations in Excel, thanks to you.
L8. The long-run probability is the same as the single event probability before the event occurs, but not after it occurs.
L9. The probability of sequential events changes as events occur and as the order changes.
Applying my learnings to my original question, if the guesser only knows there are eight planets (sorry, Pluto!), then we assume each planet is equally likely to be the third planet and guess with 1/8 probability for each. Earth is already the third planet, just as the die was already rolled, when the guessing is to be done.
Thus,1/8 = P(earth is third planet)*P(guessing earth is third planet) = P(earth is third planet)*1/8. So P(earth is third planet) = 1 even when not known.
L10. Of course, I learned the long-run probability for the Monty Hall Problem when Monty always opens a door without the car behind it.
On to other puzzles!
Hi Georg,
I think most of the lessons learned are right on, but with a few clarifications below.
So, the joint probability of the coin showing red and the die showing read is unambiguously 1/6. It’s the joint probability of two random, independent events just happening to occur by chance. It’s simply 1/2 * 1/3 = 1/6. There’s no dilemma or mystery there. There is no cause-effect relationship. It’s just two events occurring (as defined by us) entirely by chance.
But the larger issue is that you shouldn’t be using a coin to decide what to do! Understand the system and then make the choice that gives you the best chances of winning. Do not introduce unnecessary random elements!
As for L8, you’re still misunderstanding long-run vs. individual probabilities. They’re the same in this case regardless of the order. You’re too focused on the order of events. With the die example, you can guess the outcome (green or red) before you roll the die, while you’re rolling the die, or after the die was rolled by someone else and hidden from you. In all three cases, the probabilities are the same both for an individual die roll or a series of die rolls. Regardless of the order, when you guess green, you have a probability of 2/3 and when you guess red it’s 1/3.
At any rate, I’m glad it was helpful! Yes, definitely time for other puzzles!
Jim, I hope this reaches you in time so you can replace the previous long response with this succinct one.
I see four cases which I think are facts—you can tell me which ones are not. In each case the question is “What is the probability of winning with coin saying red?”
1. Both die and coin distributions are relevant. The die roll and coin toss will be done simultaneously. The probability of winning with red coin is P(die red)*P(coin red) = 1/3*1/2 = 1/6.
2. Only coin distribution relevant. This your example: the die is rolled, the result fixed and unchanging. Now try to match it by tossing the coin and getting red. Probability of winning is P(die red given die)*P(coin red) = 1*1/2 = 1/2.
3. Only die distribution relevant. Toss the coin first, the result is red and unchanging. Now try to match it with the roll of the die. Probability of winning is P(coin red given coin red)*P(die red) = 1*1/3 = 1/3.
4. Neither die nor coin distributions are relevant. Roll the die, the result is fixed and unchanging; toss the coin, the result is fixed and unchanging. The probability of winning if the coin says red is P(die red given die red)*P(coin red given coin red) = 1*1 or P(die red given die green)*P(coin red given coin red) = 0*1 = 0.
Cases 1 and 2 are confirmed by simulations for both the long run and single games.
Case 3 I didn’t mentioned previously because it was not the example. This is what I understand when you ask what is the probability of winning when the coin says red. Which means the coin must be red (already tossed). And for the probability to be 1/3, then die must not have been rolled. But simulation confirms it for both a single game and multiple games.
Case 4 applies only to a single game and each single combination with coin red in a simulation.
What do you think? Does this clarify our disagreement?
Hi Georg,
I’m again getting to the point where I’m realizing it’s not worth my time trying to sort this out. I (and others) have worked through multiple points a number of times with you, yet you’re stuck on the same issues. It’s not a productive conversation. I respect your willingness to dive into this. You said in an earlier comment that you were used to pushing on things. That can be good. However, sometimes you need to be receptive to things as well. I feel like you’re not taking what I’m saying onboard but instead just rephrasing your previous arguments.
I’m not going to go through all your questions. (See above.) And you still haven’t answered my question about the mechanism about how a coin flip improves your odds of winning! That’s because they’re completely unrelated.
I did look at your point 1. Yes, the probability of the coin telling you to pick red and getting red on the die is 1/6. That’s the probability of those two specific outcomes occurring in the same game purely by chance because they are independent events. There is no relationship between the coin and the die but they’ll occasionally agree by chance. By chance, you’ll get the red coin side and the red die to occur for the same game 1/6. But that’s not improving your odds of winning. It’s not illuminating the process. It’s not improving your chances of winning. It’s just the probability of two random events occurring in the same game. That’s not helpful.
Think of it this way. The coin will tell you to pick Red 50% of the time. However, based on the properties of the die, you should pick Red 100% of the time. Ditch that die!
At any rate, it’s not our disagreement. It’s you disagreeing with probability theory! Or misinterpreting what your results are telling, as I illustrate above.
For the Monty Hall problem, the key is to understand the process and know that it produces a situation where switching causes you to win 2/3 of the time. Knowing that you, the player needs to switch. The contestant should NOT flip a coin to help them decide. As I’ve indicated multiple times, that just introduces an irrelevant random factor that diminishes their chances of winning.
So, we’ll have to leave it there.
Georg – let’s say I decide to flip a coin to determine whether or not I jump off a cliff. Heads, I don’t jump; tails, I do jump. Would you say the odds are 50/50 that I’d survive the jump if the coin comes up tails? Do you think the coin flip has anything to do at all with my odds of survival if I end up jumping?
Obviously not, right? So why is this situation any different? If your coin flip tells you to choose red, then you’ve chosen the color that gives you 1/3 chance of winning. Simple as that.
You have been overcomplicating this problem from the beginning, and I can’t begin to understand why. It’s a very simple problem with a non-intuitive answer, which is what makes it interesting. Just stick with the parameters and quit introducing unneeded complexity that serves no purpose other than to change the problem. All you’re accomplishing by doing that is sabotaging your own chance to understand the problem AS WRITTEN.
Jim, thank you very much for these last two responses—I was not expecting them and greatly appreciate them, your time and patience.
I’ve struggled to understand how you get 1/3 probability when the coin says guess red.
I am hoping there is one more lesson you can teach us all about the 1/3 probability when the coin says red:
You explain that the probability distribution of the die is the only physical reality that you use when the coin says red, despite that distribution in a single game no longer applying and only the coin’s probability distribution applies. Why does this calculation completely exclude the coin’s probability distribution? And how is the calculation done to get 1/3 or what simulation can I do?
(Notes. You implied the 1/3 probability applied to the outcomes split and not combined as I did to get 1/2. Unfortunately, splitting the outcomes does not result in 1/3 probability of winning when the coin says red. In my simulations, 1/6 of the red coin results are wins as 1/6 = die red 1/3 of the time x 1/2 the coin will say red.
It can’t be after the coin is flipped because that probability is 0 if the die is green and 1 if it is red. Reversing the order, coin flip first then roll the die, does get 1/3 probability of wining when the coin says red, but that is a different game.)
Thanks.
Hi Georg,
I stand by my answer. I guarantee you it’s correct. But I don’t have anything to add about it. I’ve explained it fully. I don’t know where your calculations and simulations have gone wrong, but there’s something not right in them.
Instead, I’ll ask you a question.
How do you imagine that a coin toss changes the probability for the die roll? What’s the mechanism that you’re envisioning? We know that a die with four green and two red sides will come up red 1/3 of the time. That’s based on the physical properties of the die. How does the coin toss change that? You’re saying that if the coin toss tells you to pick red, you’ll then have a 1/2 chance of winning rather than 1/3.
Of course, the answer is that the coin toss doesn’t affect the die roll. They’re completely independent processes that don’t influence each other. But I’m curious what you think the connection is?
Jim, that’s very kind of you to not only respond but provide a detail explanation. I appreciate that.
When you said my 1/2 answer was incorrect and 1/3 is correct, I didn’t know whether I had lost my mind. I think only half because I’m still missing something.
Initially I said “And if I don’t know how many sides are green and how many are red, I might choose to flip a coin to guess. Then, my chance of guessing correctly is 1/2.” (Side note: I used this because in Monty Hall most contestants don’t know the long-run probability so they would guess between two doors.)
I did a simulation 1000 times of trying to match the die result (1 to 4=green, 5,6=red) with the coin result (head=green, tail=red) which is a win and got 519 wins (coin matches the die), or 0.519 which is close to 0.5. We agree on the long-run chances when combining cases.
But my simulation didn’t agree with the 1/3 probability for either red or green die color. Here’s what I got:
662 die green cases with 345 coin green matches for 0.521 wins
338 die red cases with 174 coin red matches for 0.515 wins.
So I must be misunderstanding what you mean by “When the coin tells you to switch, you have a 0.67 probability of winning. When it tells you stay, you’ll have a 0.33 chance.” Staying or switching was not part of my simulation because to stay or switch I must have selected something first.
What do I select first so that using the coin to stay or switch results in a 1/3 probability of winning?
More to learn–thanks, Jim.
Hi Georg,
I think we’ve bridged some of the misunderstanding from before. However, there’s still clearly some left. It’s hard to have this type of conversation in a comments section. So, I think we’ll have to wrap this up given the difficulties and time involved. But I did want to answer your question before we set this aside.
First, I don’t understand why you’re trying to link a coin flip to the die or Monty Hall outcome. I don’t recommend it. As I mentioned, assuming you want to optimize your chances of winning, you’ll want to understand the underlying probabilities of the system (die or Monty Hall) and use that to make your decision. You maximize your chance of winning at 0.67 by always choosing green with our die or switching in the Monty Hall problem. By introducing a random factor into your decision, you’re only reducing your chances of winning. That’s true for these examples but also a generally good rule of thumb for life! Understand the underlying system and then make decisions based on that knowledge without introducing irrelevant, random factors (error)!
As for the 0.33 probability of winning via the coin toss, here’s how it works. Imagine you go against my advice and decide to use a coin toss to decide to choose green/red (die) or switch/stay (Monty Hall). I’ll stick with the die for simplicity. Imagine you decide that when you get a heads, you’ll pick red. Tails and you pick green. The die still has 4 green sides and two red sides.
When you flip the coin and you get heads, you’ll pick red. Regardless of what led you to pick red, anytime you pick red you’ll have 1/3 chance of winning. That’s because there are 2 red sides out of 6 total sides (2/6 = 1/3). The reason this works is because it doesn’t matter why you’re picking red. It just matters that you are picking red. You could pick red because of a coin toss, throwing darts at board, by playing pin the tail on the donkey, letting someone else tell you what to do, drawing a random card from a deck, using a Ouija board, because it’s your favorite color, etc. None of that matters.
The physical properties of the die dictate that red only occurs 1/3 of the time. Therefore, anytime you pick red for whatever reason, you have 1/3 chance of being correct. It all goes back to the properties of the die, which don’t change based on how you make your choice. You pick red and you only have a probability of 0.33 period. 2 red sides out of 6 total sides. That’s unchanging. The coin toss doesn’t change the physical reality of the die.
That’s why I say, understand the system (die, Monty Hall, life), and use that knowledge without introducing irrelevant factors. There’s just no reason at all to pick red in this scenario. And in the Monty Hall problem, there’s no reason to stay with your original choice.
Jim, I am so sorry. Please accept my apologies to you and your readers.
I’m retired and got interest in puzzles during this pandemic and I came across Monty Hall and your blog. I guess my college habits returned of pressing and pressing to learn more and in depth.
And I have learned a lot from you and you have been so patient. With your interactive approach I thought it would be great to learn quickly and more. But I’ve been hogging your time and failing to see some of what you’ve been teaching. Sorry abut that.
I know I don’t have the right and I completely understand if you refuse, but would you please tell me how you got the 1/3 probability?
Don’t worry I will leave you and your readers alone and move on to other sources of information.
Thanks again and sorry again
Hi Georg,
Thanks, I appreciate that. And I think I did misunderstand what you were saying.
I think I now understand what you’re say. Maybe. Are you figuring the total win probability overall when you combine both the heads and tails outcomes? I was splitting the outcomes down to each separate case of switching or staying. If that’s the case, we’re both correct.
When the coin tells you to switch, you have a 0.67 probability of winning. When it tells you stay, you’ll have a 0.33 chance. That’s what I was saying.
However, combining both cases and looking at the overall wins, yes, you’d expect to win 50% of the time. That’s because half the time you switch and half the time you’ll stay. That’s what you were saying (I think).
Basically, by introducing randomness into your decision making, you’re reducing your probability of winning from 0.67 to 0.50. That reduction occurs because the randomness eliminates the benefits you gain by understanding the die (or the Monty Hall game). The coin tells you to do the right thing half the time and the wrong thing have the time. In this case, that balances out to 50/50. For other pairs of probabilities, it wouldn’t necessarily work out to 50/50, but it does with 0.67 and 0.33.
Interestingly, in the real world, most people choose to stay with their original door choice. It’s a psychological thing where they see no reason to switch (mistakenly thinking it’s 50/50), so they stay, reducing their chances of winning to 0.33! So, using a coin toss like that actually breaks up that consistently wrong choice to stay and improves their chances.
But, again, the best case is to NOT use the coin toss. Use your knowledge about the process to decide to switch (or choose green on our die) and maximize your win probability at 0.67! There’s no need to introduce a random factor into your decision making.
I was surprised when you said that my calculation of winning when tossing a coin was wrong, that it is 1/3 and not 1/2. So, I checked again the formula P(die red)*P(coin red) for after the die has been tossed, I did several examples myself using prime numbers (1, 2, 3, and 5) as green and nonprime as red on a die, and then did a simulation of 1000 times. They all confirmed the 1/2 probability.
The good thing is I now understand that for the Monty Hall Problem you have consistently been answering with the long-run probability. I should have seen that since you tell people to play the game multiple times and to do simulations which clearly shows the 2/3 probability of winning by always switching in the long-run.
Somehow we have to be clearer as to which question we are answering, so we don’t confuse each other or think we are contradicting each other.
Georg,
This will be my last reply to you. I’ve trimmed your comment down immensely. You’re going down a long, dark rabbit hole leading to the wrong conclusions–and I won’t subject the other readers to it.
No, flipping a coin to decide which door to choose does NOT make the probability 50/50. Your calculations are incorrect. And, as I’ve explained to you MANY times, the probability of 2/3 switching & 1/3 staying are for BOTH individual games and the long run. You’re trying to make a distinction that doesn’t exist.
You are the only one confusing yourself by throwing in irrelevant things like coin tosses and long run vs. an individual game. I’ve taken plenty of my personal time to explain how it works, but you are insistently ignoring it. Consequently, I will not waste any more of my time discussing this with you when you’re refusing to learn.
Georg, you aren’t the only one to do this, but I just don’t understand why some people insist on changing the parameters of the problem, and then stating, “See? It doesn’t really work.” Yeah, if you flip a coin, etc – that’s not the problem that was presented to you! Jim said you know how many green and red sides there are, and you’re not flipping a coin! Irrelevant! Stick to the problem, and you’ll come to understand it. Keep changing things, and you’ll never understand it. You’re just confusing yourself.
Indeed! He’s chasing his own tail in circles with irrelevant distractions like coin flips and trying find a distinction between the probabilities for long run vs. individual games.
Jim, thanks. Both the example and key phrases clarified the issue. But let me confirm.
The difference is between 1) What is the probability of a result after it occurs? and 2) What is the probability of guessing what a result is after it occurs?
And if I don’t know how many sides are green and how many are red, I might choose to flip a coin to guess. Then, my chance of guessing correctly is 1/2.
That explains why I was confused. I was asking #1 and you were responding to #2.
Hi George,
You’re over thinking this a bit.
You don’t know the result, but you do know the probabilities associated with the die. Consequently, as long as you use that die, you can use the same probabilities. That’s true whether you guess before or after the die has been rolled. Either way, the probability of green is still 2/3 and the probability of red is still 1/3. Don’t overcomplicate it. The key point is that you don’t know the outcome in either scenario, so you need to rely on probabilities to make your guess.
You’re confusing yourself because you’re adding unnecessary complications!
No, your coin toss scenario is incorrect. The applicable probabilities are based on the properties of the die. Using a 50/50 coin doesn’t change how the die works. Answer this question. Given the same die characteristics as before (4 green/2 red), suppose you flip a coin, and it says to guess red. Would you do that even knowing the properties of the die? Of course not.
Now, if you didn’t know the properties of the die, and you used a coin toss because you didn’t have anything better, it would lead you astray. If the coin tells you to choose red, you might think you have a 50% chance of winning, but you actually only have a 33%. That error occurs because the coin toss doesn’t reflect the true probabilities in play. The coin toss can’t magically change the properties of the die!
Again, don’t over complicate things.
Hi Jim, it’s using the contestant’s lack of knowledge that confuses me. Why aren’t just the facts used to calculate the probability rather than someone’s ignorance of the facts?
Georg,
Let’s walk through an analogy to help you understand.
Imagine I have a six-sided fair die. Four sides are colored green, and two are colored red. I roll the die and keep it out of your sight. I don’t change the results after rolling it. There is a definite correct answer for which color is showing, but you don’t know what it is. However, you do know the characteristics of the die (four green sides vs. two red sides).
Now, you need to guess which color is showing. Answer these questions for me.
What color will you guess and why? What are your chances of guessing correctly if you say “green?” What’s your chance if you say “red?”
To Georg:
You are missing a point. When we play the Monty Hall game in the long run, everytime those three steps that you mention are fulfilled, not sometimes yes and sometimes no. In each individual game the car was put behind a door, the contestant selected a door and then Monty opened a door. What you are saying is like if in some of the trials any of the steps would not occur.
Note that in your example if the statistics show that the initial probability to have a fatal accident is 10% but after passing the first intersection it is only 8%, that can only be true if 2% of the travelers didn’t manage to get past it.
The Monty Hall game is like if every traveler that once went through the road always passed the three intersections, but still 10% of them end having an accident. So, that can only mean that the 10% risk to have the accident exists after the intersections. In the same way, the 2/3 probability of winning by switching exists after Monty reveals a goat.
If you don’t understand why, note that since the host knows the positions and must always reveal a goat from the two doors that the player did not choose, then he is forced to reveal one specific door when the player’s has a goat, that is which has the other goat. Instead, when the player’s door has the car, the host is free to reveal any of the other two, we don’t which in advance, they are equally likely for us, because both would have goats in that case.
For example, if you start picking door 1 and the host opens door 2, it is easier that the reason why he did that is because the car is in door 3 rather than in door 1, because if it was in door 3 it is sure that he would have opened door 2, as he wouldn’t have had another option, while if it was in door 1 maybe he wouldn’t have preferred to reveal door 2 but rather door 3 in that case. Assuming it is 1/2 likely that he reveals 2 or 3 in that case, then it is only half as probable that the car will be in door 1 this time.
This is better seen going to the extremes. Let’s keep the example that you pick door 1 and the host reveals door 2, so only doors #1 and #3 remain closed, but consider these two scenarios:
1) If it was a rule that when you pick the correct option the host must reveal from the others which has the highest numeration, then this time it would be a certainty that the car is in #3, because if it was in #1 he would have revealed #3 for sure, not #2. So at this point you would have 100% probability to win by switching and 0% by staying.
2) On the other hand, if it was a rule that when you pick the correct option the host must reveal from the others which has the lowest numeration, then this time the chances would be even for both door 1 and door 3. That’s because if the car was in #1, the host would have been forced to reveal #2, and also if it was behind #3 he would have been forced to reveal #2. Therefore, both staying and switching would be equally likely -> 50% chance for each.
As you see, the switching strategy can only vary from 50% probability to 100% probability depending on the extreme, and the staying strategy can only vary from 0% to 50%.
The Monty Hall game is a midpoint of those two extremes, because we are not sure if he will prefer the lowest or the highest numbered door, so the probabilities of the strategies must be between those extremes. The probability of winning by switching must be more than 50% but lower than 100%, and the probability of winning by staying must be more than 0% but lower than 50%.
Wow! That is awesome and well explained. I was going to type a long rant about how it had to be 50/50 because one of the doors is eliminated by the first choice, but once it ‘clicked’ that THAT choice is only going to be right 33% of the time, it made sense! Thank you
Jim, you’re the expert but I hope you can explain your answer a little more. This is what I understand you are saying.
Before any game starts there are nine possible combinations. After Monty opens a door in a single game, there is only one combination possible. For three of those, switching wins 0% of the time and for six of them switching wins 100% of the time. The proportion 6/9 = 2/3 is the long-run probability of winning by switching. For a single game after Monty opens a door, the 0% or 100% or long-run 2/3 probability are NOT used to determine the probability of winning by switching.
Instead, the 9 combinations are reduced to only 3 that have the door the contestant selected. In two of those three combinations the contestant would win by switching, getting the 2/3 chance you mentioned of winning by switching.
I understand why six combinations are eliminated. I thought two more would also be eliminated because of where the car is. But is your reason for not eliminating two more combinations simply because the contestant doesn’t which to eliminate? If that’s the reason, can you explain more why the contestant not knowing is used to calculate the probability?
Hi Georg,
It’s really very straightforward. The contestant does not know where the car is located so they can’t narrow their situation down to a single row! The contestant would need to know the car’s location to narrow it down to one row. Because the contestant doesn’t know where the car is located, they’re forced to make a random choice.
When they’ve picked a door, they can only narrow it down to the 3 rows in the table that correspond to their initial door choice. Of those three rows, two win by switching and one wins by staying. Hence, they have a 2/3 chance of winning by switching.
And, again, that 2/3 chance of winning applies to an individual game or a series of games. Probability 101.
Jim, I don’t think that’s true according to your table of the nine possibilities. Let’s say you are playing one game, and the car is placed behind door 1 and you select door 1, and Monty opens door 2. Then you are in the possibility shown in row 1 of the table. What is the probability if winning by switching?
Georg, in that very specific scenario, you lose by switching, as the table indicates. Of course, you won’t know that row 1 applies to you specifically because you won’t know that the car is behind door 1.
Instead, all you know is that you’ve picked door 1 and Monty has opened door 2. You don’t know the prize door. Consequently, you need to consider rows 1 -3 because they apply to your choice of door 1. (After choosing door 1, the other six rows don’t apply to you.) Two of those three rows have you winning by switching while only 1 has you winning by staying. Hence, you have a 2/3 chance of winning by switching.
Alfonso, I think I see why you don’t know why the distinction: your example of the roads only has the long-run probability and doesn’t include what happens as you take one of the roads.
Let me complete your example. Suppose road B is 5 miles long with 3 intersections in the first 3 miles. Assume there is a 10% probability of being in an accident with 2% at the first intersection, 3% at the second and 5% at the third. We both agree that in the long-run and before you start down road B, there is a 10% chance of an accident. But after you pass the first intersection safely, the probability is now 8%, and after passing safely the second intersection it is 5% and after passing safely the third intersection it is 0%.
In Monty Hall, the first intersection is putting the car behind a door, the second intersection is the contestant selecting a door, and the third intersection is Monty opening a door. The probabilities change as events occur.
That is why I was asking if the Monty Hall problem is about the probability in a single game after passing the intersections or about the probability for multiple games or in the long-run?
Hi Georg,
I’m going to skip the accident scenario because that wasn’t mine.
Instead, I want to address your question about whether it’s a probability for a single game or the long run proportion of wins of a series of games. As I mentioned earlier, it’s both.
The probability of winning by switching applies to single games. If you switch and you play a single game, your probability of winning that single game is 0.67. Or you can look at it as an odds ratio where your odds of winning are double (0.66/0.33 = 2) when you switch versus staying.
Additionally, it applies to a series of games. If you played 100 games, you’d expect to win 67 by switching and 33 by staying.
I hope that clears that up!
To Georg:
I don’t know why that kind of doubt comes in the first place.
When we talk about playing multiple times, we are implicitly assuming that each game is independent of the previous ones, as if it were the first you were doing. The previous results are not going to influence the newers. So, if switching tends to win more times, it is because in each of the individual games there is something that makes that strategy more plausible to be correct than the other. The bigger results in the long run are just a consequence of that.
It is the same that you would intuitively apply in other aspects of life. Imagine that you are traveling and there is a point in which there are only two possible roads to take, let’s call them A and B. You are informed that on road A there have been multiple accidents in recent years, while on road B there has only been one accident. Those statistics required multiple trials to be registered, but if you only need to travel through one of them once, you would intuitively assume that B is safer and take it.
Jim, I agree with what you say: Switching will win 2/3 of the time while staying wins 1/3 of the time when the contestant plays multiple games.
This is your description of the game when the contestant must decide:
At this moment, there are two closed doors, one of which you picked.
The prize is behind one of the closed doors, but you don’t know which one.
Monty asks you, “Do you want to switch doors?
So the contestant isn’t asking what he should do if playing the game mulitple times. The contestant is being asked to stay or switch in a single game AFTER the prize is placed behind a door and the contestant selects a door and Monty opens a door.
That is why I asked about a single game, because I know the long-run probability is 2/3 that the prize will be behind the other door and therefore always switching over many games wins 2/3 of the time.
Is the Monty Hall problem about the probability in a single game or about the probability for multiple games or in the long-run?
Hi Georg,
I hope you realize that probabilities apply whether you’re playing one game or multiple games. Switching gives you twice the chances for winning even when you play just one game.
Hi Georg,
Say A is the winner and B and C are the losers
If you choose A and switch, you lose.
If you choose B and switch, you win.
If you choose C and switch, you win.
If you choose A and don’t switch, you win.
If you choose B and don’t switch, you lose.
If you choose C and don’t switch, you lose.
Thus, if you switch, you win 2/3 of the time. If you stay, you win 1/3 of the time. If you choose to switch or stay at random or with a coin toss, you win 1/2 the time.
Thanks. Although, as I point out to Georg, if you base your decision on a coin toss for whether to switch or stay, the probabilities remain 0.67 and 0.33, respectively. How you make your decision doesn’t affect the probability, just the decision itself.
Thanks! Just thought of two more, thought I’d add them here aswell:
Say instead you were asked to choose to keep the chosen door or open both the remaining doors. You would of course choose the other two doors, since the chance of one having the prize is 2/3. Now, Monty showing which of the doors doesn’t have the prize doesn’t change that. It’s like by showing which of the two doors is redundant, he is combining the two remaining doors into one. Figuratively speaking, he has taken the eggs from both baskets and put them all into one basket. The 2/3 chance that they shared between them has all been placed in the single remaining door. It is as though he has simply swapped the two doors out for one big door.
or
Probability is all about how much you know (like if I flip a coin and see it is heads, then for me the chance of it being tails is 0, while for you who hasn’t seen the coin, it is still 50:50).
You know that if you picked the right door, Monty is pointing at a wrong door, so switching gives a wrong door.
You know that if you picked the wrong door, Monty is pointing at the right door, so switching gives you the right door.
So for you, the chance of having picked the right door is 1/3 and the chance of having picked the wrong door is 2/3.
Thus from your perspective, the chance of switching and losing is 1/3 and the chance of switching and winning is 2/3.
Thanks again!
Jim, you must admit that if the contestant flips a coin when there are two doors left to stay or switch, the probability is 50% of winning by switching. One reason the contestant might flip a coin is that he most likely doesn’t know the long-run probability of 2/3 of the time the prize is in the remaining unopened door.
I aslo noted that at that point in the game, the problem does not state how the contestant decides whether to switch or stay. So, without any assumptions about how that decision is made, there are lots of different answers.
My question is: When there are two doors left in a single game, what decision process does result in 2/3 probability of winning by switching?
Hi Georg,
Keep in mind that you should not confuse the probability of getting heads/tails in a coin toss with the outcome of the game. Yes, the coin toss is 50/50. But the outcome of the game is not.
It does not matter how contestants decide to stay or switch. They can flip a coin or use any other method. Switching will win 2/3 of the time while staying wins 1/3 of the time. Because this probability is unchanging in this game, contestants should always switch.
However, suppose a contestant plays multiple games and uses a coin toss to decide whether to stay or switch. If they get a “heads,” they switch. Tails and they stay.
For half the coin tosses they’ll get a heads and switch. In that subset of games, they’ll win 2/3 of the time.
For the other half of the coin tosses where they get tails, they’ll stay with the original door. In that subset of games, they’ll win 1/3 of the time.
Hi Jim,
I was just talking about this problem and came across this excellent explanation, but I’m shocked that some are still arguing against this bit of easy maths. I had some fun trying to come up with some explanations anyone can understand, might help to clear out any trolls still trying to argue otherwise. I’ll post my write up here, hope it helps:
For those who still don’t understand the Monty Hall problem:
Switching gives a 2/3 chance of winning. This is absolutely correct.
I’ll word the explanation in three different ways, all pretty much say the same thing. But if you aren’t interested in how it could be worded and just want the best explanation, you can just skip to the last two paragraphs.
Firstly:
Say you choose a door.
A) There is a 1/3 chance of the winner being the chosen door.
B) There is a 2/3 chance of it being one of other two remaining doors.
However, once one of the remaining doors is exposed, B now becomes ‘There is a 2/3 of the winner being the remaining door’.
Monty is telling you which of the other doors has the 2/3 chance.
He doesn’t tell you anything about the door you’ve picked, but he does tell you in which of the other two doors the 2/3 chance lies.
There is a 1/3 of the winner being chosen door, or a 2/3 of being one of other two. Once one is exposed, other door becomes 2/3. (He’s telling you which of the other doors has the 2/3 chance)
or:
If you picked right, the chance of the other door being right is now 0.
If you picked wrong, the others doors chance is now 1.
So if you pick the wrong door (2/3 chance) and switch, you win. If you pick the right door (1/3 chance) and switch, you lose.
So if you pick the wrong door, he’s telling you the right one. If you picked the right door, he’s telling you nothing.
Now if we say the chance of switching is 50%, we get this:
A: Choose win door + Don’t switch = Win
P = 1/3*0.5 = 1/6
B: Choose lose door + Don’t Switch = Lose
P = 2/3*0.5 = 2/6
C: Choose win door + Switch = Lose
P = 1/3*0.5 = 1/6
D: Choose lose door + Switch = Win
P = 2/3*0.5 = 2/6
Thus, if you choose randomly whether to switch or not, the chance of winning is 1/6+2/6=1/2 and the chance of losing is similarly 1/2, as it was meant to be.
However, if you only insist on switching, we now only have C and D.
P(C) (ie. Losing) = 1/3
P(D) (ie. Winning) = 2/3
Thus, if you switch, you will win more often than losing.
or:
Instead of Monty opening door A, think of it as him pointing to door B. Now when you pick the right door, he points towards either of the other two doors, either way you lose. But when you pick the wrong door, which has a much higher 2/3 chance, he will then point towards the winning door, thus giving you the win if you switch.
or in other words:
Let A be the right door and B and C be the wrong doors.
If you point to A, he will point to B or C, both wrong.
If you point to B, he’ll point to A, the winner.
If you point to C, he’ll point to A, the winner.
So 2/3 of the time, he’ll point to the winner.
—
But whichever way you look at it, it all boils down to the same thing:
A) If you choose the right door (1/3 chance) then switch, you’ll lose.
B) If you choose the wrong door (2/3 chance) and switch, you’ll win.
Choosing the right door first has a 1/3 chance while choosing the wrong door first has a 2/3 chance. Thus the chance of A is 1/3 and the chance of B is 2/3. Thus the chance of losing is 1/3 and the chance of winning is 2/3.
What Monty does is, he messes with B. Without him, B should have said ‘you’ll have a 50% chance to win and 50% chance to lose’, rather than ‘you’ll win’.
Thus, if you choose the wrong door, Monty will give you the right one, and make switching give you a 100% chance of winning. He cuts out the ‘choose wrong door > switch > lose’ possibility, increasing the chance of winning when you switch.
(And for anyone who asks, I’m no specialist, just a normal Maths graduate)
Hi Yusuf,
Thanks for posting this! I saw your 2nd comment with the correct wording and tried to incorporate it into the first comment so it’s all in one place rather than split between two. Please let me know if there’s any changes to make.
Hi Jim,
I agree with Steve in this one. I get that Monty has a privileged and use it to decide whether he opens the door or not. When the person in charge of choosing one door picks the wrong one, this is the one without the prize, it’s easy to see that Monty will open all the other doors except for the one with the prize behind it.
But let’s imagine a situation where the person picks the right door since the beginning (without having this knowledge, of course) and that’s door number 3. Now, Monty can either open door 1 or door 2 because none of them has the prize: in that case, he has two possibilities and both of them lead to “lose” if the person switch doors.
So, in conclusion, the table you made isn’t wrong, but it hides the fact that when the door picked is also the one with the prizes, we have another ramification of the problem, but both with the same result – the person either loses if changes door or wins if keep the initial decision.
Here’s a little graphic demonstration of my argumentation, I will represent the doors by ( ), the prize with $, the choice with * and the door that was opened by Monty is marked with an X.
3 DOORS SCENARIO
(*) ( ) ($) → (*) (X) ($)
( ) (*) ($) → (X) (*) ($)
( ) ( ) (*$) → (X) ( ) (*$) OR ( ) (X) (*$)
And the same applies for when we have the prize behind door number 1 and door number 2.
This all leads to a 50% probability of winning when switching and also an equal 50% probability of losing with the same action.
Hi Luiza,
Keep in mind that Monty has two choices in the situation that you describe. However, he can only choose one of them. He can’t choose both. So, ultimately, it’s just one. You can say each one has half a chance and together it totals one.
And, no, what you conclude with is incorrect. It’s not 50/50. You have 2/3 chances of winning if you switch. 1/3 if you don’t switch. That’s been proven with both probability theory as well as computer simulations that play the game thousands of times.
I read your comments in that thread and you are completely wrong. Play the game in real life, “as Monty plays it”, using cards. Have a friend play the role of Monty. No computers, no “rigging”. Do 100 samples of switching, and 100 examples of not switching. Be honest, since this isn’t a blind test. What are the results? Something like 33/66, right? So what’s more likely, that the explanation here is wrong or that the game is “rigged”, or that you’re wrong or not understanding how the problem works?
You wouldn’t be the first Ph.D. to get this problem incorrect. The OP, Marilyn vos Savant, was riddled with letters from teachers and professors from around the nation telling her how wrong she was, only to end up with egg on their collective faces. This problem has now been popularly around for decades and I promise you haven’t discovered anything new about it. It works exactly as described in Jim’s article. Just think it through carefully, and you should be able to arrive at why it’s true.
Hi Maggie and Jim,
I believe Maggie is correct here in that “The act of opening the door is a non-event – it does not add any knowledge that wasn’t already known to the player at the start of the game.”
Think of this as a dataset problem with 2 queries instead of a statistics problem solved by summing up the Wins and Losses of the 9 different outcomes in the table. If you unpivot the 9 rows, you actually end up with 18 different possible outcomes, 6 for each Door. (When you unpivot the table, the Switch and Unswitch columns are transformed to one column with Switch status and one column with Win status, and the row count is doubled) After you pick the first door, you now have 6 potential outcomes for that door; therefore 6 rows. After Monte selects a door with the goat, you are left with 4 potential outcomes, not the original 6. The two outcomes that were possible, are not longer included in the dataset and must be removed as possible outcomes (this is key). Note that at the start you always knew that two possible outcomes would be removed and you knew that those rows would not represent a prize door.
Now you are left with 4 possible outcomes for your second query: Door X can win or lose and door Y can win or lose. Those are the four possible outcomes that occur when you make your second choice and these are the same four possibilities you always will have no matter what door you choose at the start of the sequence.
The sequence has three steps: (1) You choose a door, (2) Monte chooses a door with a goat that is not your door, and then (3) you choose one of 2 doors remaining as possible choices. The first two steps do not ever affect the outcome and in this case with 3 doors, they do not affect probability of a win.
Hi Steve,
Opening the door provides valuable information. Specifically, you know for certain that the car is not behind that door! In terms of probabilities, we know that there is a 1/3 chance the car is behind the door the contestant picks at the beginning. And there’s a 2/3 chance that it is behind Monty’s set of two doors. When Monty opens one of his doors, he does so with full knowledge that it won’t reveal the car. So, the probability that it contains the car is zero, hence the full 2/3 probability applies to Monty’s unopened door.
The reason why the probability transfers like that is becomes Monty’s choice is not random. Hence, our natural inclination for using the rules that apply to independent, random events do not work here!
This benefit of switching has been shown using probability theory as well as computer simulations. Those simulations simply play out thousands of games rather than using complex mathematics. You can even try it empirically yourself. Read my other post about the Monty Hall problem for more details.
Please visit https://www.access-programmers.co.uk/forums/threads/estimating-the-age-of-the-universe-and-getting-it-all-wrong.321070/ and look at posts in that thread starting from 17. Don’t let the title of that thread fool you. My posts are made in that forum under the name “The_Doc_Man” – and that is not totally pretentious. I have a Ph.D. in Analytical Chemistry and studied probability intensely.
I also choose to dispute your analysis because by omitting the host’s actions, you change the nature of that game. It is not the same game at all. I have a spreadsheet (in PDF format) showing that if you play the game as Monty Hall plays it, there is no advantage to switching.
I apologize that I didn’t send my first comment to this thread, but to your general business contact point. That was my error.
Hi Richard,
I’ve taken a quick look at your table. While I haven’t studied it in depth, there’s clearly an error in it. You show the courses of action and outcomes for the prize being behind door 1. Disregarding the ineligible outcomes (host revealing the prize or the contestant’s door), your table should have 6 outcomes/rows.
The reason why it should be six is because we’re looking at the prize being behind one door (1), contestant has a choice of three doors (3), and the contestant can switch or don’t switch (2). That works out to 1 X 3 X 2 = 6. That’s 6 possible outcomes per door. In my table, I have three rows per door and include both the switch and don’t switch options in a row. So, my table has 3 rows per door, but you can split them out (as you do in your table).
Instead, you have 8 outcomes/rows for one door. Consequently, your table is incorrect, but I don’t have time to work out the nature of the error.
I did read the discussion at the link. I see that the other site owner also disputes your claims and suggests that you try it empirically. I’ll point out that not only does probability theory (when applied correctly) support switching but so do computer simulations that simply play the game thousands of times. I’ll add my voice to the other site owner who suggests that you try it empirically. I write about doing this in my other post about the Monty Hall problem. As he says, you’ll quickly see the benefits of switching!
Happy New Year!
Right, for n doors it is a probability of (n-1)/n . A larger number of doors than three does seem to be more illuminating
Jim, the table does not show all possible outcomes. Note that there should be 12 rows in the table and 1 more column (representing which door the host opens). The Monty Hall problem is not a problem at all. When the player chooses a door at the start, he/she already knows with 100% certainty that the other two doors contain at least 1 door with no prize behind it. The act of opening the door is a non-event – it does not add any knowledge that wasn’t already known to the player at the start of the game.
Hi Maggie,
The table is correct with nine rows. You can pick three doors and the prize can be behind three doors. 3 X 3 = 9. Hence, there are nine possibilities. And. there’s no need for a host column. The way it works is very simple. Switching doors will flip your outcome. If you win by staying (your door = prize door), then you lose by switching. And the other way around. If you lose by staying (your door ≠ prize), then you win by switching.
So, all the information you need is the door you choose and the door that has the prize.
I hope that’s clear now!
Although I figure you would probably still have gotten a majority of the disagreements you got, you
should state the _rules_ of the game _in the problem_, rather than having the problem just describe
what is happening for the current contestant, and burying the rules in your reasoning for the solution.
(For example, is it the case that, if the contestant picks a non-prize door then Monty has a
1/2 chance of opening that door? If so, that simply didn’t happen to the current contestant.)
The relevant rules are
Monty will ALWAYS open exactly 1 door.
That door will NEVER be the door the contestant chose.
Monty _knows where the prize is_, and will NEVER open that door.
If that door is the one the contestant chose, then Monty will choose 50-50 which other door to open.
After Monty opens a door, Monty will ALWAYS offer a switch.
.
The main ones are, Monty won’t open the contestant’s door and won’t open the door with the prize.
If instead Monty
opens a random door other than the contestant’s door, without regard to where the prize is
or
opens a random non-prize door, without regard to whether-or-not it’s the contestant’s door
, then switch-or-not won’t matter.
You are clearly well aware of the “random door other than the contestant’s door” part of that,
but I’m not sure you’re aware of the “random non-prize door” part, and in any case,
those should be excluded _in the problem statement_, rather than just in your solution.
(Incidentally, the “random door other than the contestant’s door” corresponds to the
Monty Fall problem, as Ronald mentioned roughly 2 years ago: The idea there is,
Monty slipped and accidentally opened a door, which happened to have not hide the prize.)
When one clearly states the
Monty will open a door chosen uniformly-at-random from the
1 or 2 that are neither the contestant’s door nor the prize-door
rules, I have have 2 other things you might start with for people who still disagree.
(If those were _not_ both clearly stated, then I would begin by giving those rules, acknowledging that
going against either or both can easily make the 1/2 answer correct, and checking whether-or-not their
disagreement and/or confusion and/or lack of understanding, as the case may be
still applies when those are the rules for the game.)
frequentist approach:
Consider 600 shows in which the contestant choose Door 1.
Q(1): Among those 600 shows, in roughly how many of them is the prize behind Door 1?
Q(2): Among those 600 shows, in roughly how many of them is the prize behind Door 2?
Q(3): Among those 600 shows, in roughly how many of them is the prize behind Door 3?
Q(1,2): Among the roughly [insert your answer to Q(1) here] shows in which
the prize is behind Door 1, in roughly how many does Monty open Door 2?
Q(1,3): Among the roughly [insert your answer to Q(1) here] shows in which
the prize is behind Door 1, in roughly how many does Monty open Door 3?
Q(2,2): Among the roughly [insert your answer to Q(2) here] shows in which
the prize is behind Door 2, in roughly how many does Monty open Door 2?
Q(2,3): Among the roughly [insert your answer to Q(2) here] shows in which
the prize is behind Door 2, in roughly how many does Monty open Door 3?
Q(3,2): Among the roughly [insert your answer to Q(3) here] shows in which
the prize is behind Door 3, in roughly how many does Monty open Door 2?
Q(3,3): Among the roughly [insert your answer to Q(3) here] shows in which
the prize is behind Door 3, in roughly how many does Monty open Door 3?
Q(_,3): Among the 600 shows (no restriction on where the prize is),
in roughly how many of them does Monty open door 3?
Q(stay): Among the roughly [insert your answer to Q(_,3) here] shows in
which Monty opens Door 3, in roughly how many is the prize behind Door 1?
Q(switch): Among the roughly [insert your answer to Q(_,3) here] shows in
which Monty opens Door 3, in roughly how many is the prize behind Door 2?
betting approach:
Jim and Mike are watching the show on TV.
The contestant chooses a door and, before Monty chooses a door, the show goes to commercial break.
During the break, Jim offers Mike the following bet.
If the prize is behind the door the contestant chose, then Mike pays Jim $7.
If the prize is behind either of the other 2 doors, then Jim pays Mike $4.
Should Mike accept this bet?
The commercial break ends, and Monty chooses a door, revealing that it does not hide the prize.
Before either of the remaining 2 doors are opened, the show goes to commercial break again.
During the break, Jim offers Mike the following bet.
If the prize is behind the door the contestant chose, then Jim pays Mike $6.
If the prize is behind the door the contestant can switch to, then Mike pays Jim $5.
Should Mike accept this bet?
Jim, my friend and I tried a simulation. We grabbed several coins from our bags. Then we took one and flipped it behind our backs in one room of our apartment so we didn’t see the result. Then we went to the kitchen and we each flipped a coin 20 times and wrote down each of our results. We repeated that five times flipping the coins we don’t know the result of in different parts of our apartment. We had a total of 200 results for two coin flips. When we checked the results of the coins we didn’t know the result of, 102 times they were tails and 98 times they were heads. But when we calculated the probabilities, all 102 times of course we did not get two heads so the probability was 0. And we got exactly 50% of the time two heads when the first coin was heads. What did we do wrong—how do get 1/4 when we don’t know the result of the first flip?
Hi again Kamala,
Please read my last reply to you. You have a fundamental misunderstanding about joint probabilities and conditional probabilities.
Yes, there is always a 50/50 chance of heads vs tails for an individual coin flip. No surprise there. But, you’re misunderstanding the 0.25 joint probability for two heads.
Thanks for explaining the difference, Jim. I’m looking forward to your post on these topics.
So how do I prove that for the conditional probability, it’s still 1/4 for the person who doesn’t know what the first coin toss was but 1/2 for the person who knows it’s heads and 0 for the person who knows it’s tails? What formula can I use or can I do a simulation? Or how can I show it with coins? This would make a great puzzle to use on my friends! Thanks.
Hi Kamala,
I’m afraid you’re misunderstanding what joint probability and conditional probability each mean.
The joint probability that two coin tosses will produce two heads is 0.25. That probability applies to the outcome for both tosses, not one.
The conditional probability P (H2 | H1), is the probability that the 2nd coin toss will be a heads given that the first coin toss was a heads. That probability is 0.5 because coin tosses are independent events. One coin toss doesn’t affect the next. For independent events P (H2 | H1) = P (H2) = 0.5. This probability applies to one coin toss.
I have posts about all these concepts that you should read up on. Or get a book or take a class. I mean this in a friendly way, but you have some fundamental misconceptions about these concepts that I cannot fully address in the comments section. Unfortunately, I don’t have a book myself on this topic, but I plan to someday!
Jim, thanks for the explanation and the links (they helped). I understood everything up to “With conditional probabilities, you need to know what the condition is to be able to calculate the probability.” I went back to find the conditional probability formula: P(A|B) = (PA and B)/P(B). So, how do I get what a person knows or doesn’t know into the formula to get the answer of 1/4? Or, do I have to use a different formula depending on what a person knows?
Hi Kamala,
Well, the probability of getting two heads is a joint probability rather than a conditional probability. Joint probabilities assess the probability of events occurring together. That’s different than a conditional probability where you’re assuming one event has occurred and determining the probability of the other event.
For conditional probabilities, you’re assuming an event has occurred. Basically that’s where you’re starting at.
Jim,
Yet, my point was that you do not need the subjective component and I showed why and how to eliminate it using objective probabilities to get the exact same probabilities you did. But you are choosing not to read that suggestion.
Why not? If you find flaws with it, just tell me what they are. I’m happy to learn more.
I addressed it when I said:
By the way, when you say, “we need to calculate the probabilities given the information that the guesser has, not the person with complete knowledge,” that is exactly what is meant by subjectivity. It’s from each contestant’s point of view.
Your own comments incorporate subjectivity.
We’re done.
Jim, we used a coin tossing example to learn about conditional probabilities. You won if you got two heads in a row. Since the chances of getting heads once is 50-50, then getting two heads in a row is 1/2×1/2=1/4. For conditional probabilities we saw that the chances changed after the first toss. If it was heads then two heads in a row had a 50% chance but if it was tails, then there is 0% chance. I thought that this was true even if you didn’t know what the first toss was. You still knew there were only two choices, and so you knew that it was either 50% or 0%.
But you are saying that if I don’t know if the first toss is heads or tails, then the chance it is heads is 50%. Since the chance of heads on the second toss is also 50%, the chances of two heads in a row is still 1/4, even after the first toss as long as I don’t know what it is. Is this what you are saying
Hi Kamala,
Sometime in the near future, I’ll be writing a post specifically about conditional probabilities. For now, I have a post where I discuss conditional and other types of probabilities in contingency tables.
As for your question, what you write first is correct.
For starters, if you ask about the probability of getting two heads before flipping any coins, that’s a joint probability. And, you’re correct that it’s 25%.
Conditional probabilities assume that an event has occurred and given that occurrence, what is the probability of a different event occurring. As you’ve found, it’s certainly possible for the probability to change when the initial event changes. If you get that first head, then the conditional probability of getting two heads is 50%. However, if that first toss is a tails, then it’s impossible to get two heads.
With conditional probabilities, you need to know what the condition is to be able to calculate the probability.
There is some subjectivity to this too. If you don’t know what the first coin toss is, then you can only go by the probability that it’s 50% heads. So, yes, the best that you can calculate it is without the knowledge is 1/4 for two heads. Of course, if you gain information about that first toss, you can improve your probability calculations. What’s changing is the information you have, which affects the quality of your calculations.
Put another way, if you don’t know the outcome of the first coin toss, you don’t have the information necessary to calculate the conditional probability. However, you do have the information to calculate the joint probability.
Keith Hawley, you can also get that when the host randomly reveals a door and just by chance it results to have a goat the probabilies cannot be 1/3 vs 2/3 realizing that assuming that you end with absurd conclusions.
First, note that since the host is doing it randomly, it would be the same if you (the contestant) were who made the revelation instead of the host. By the end both are doing it without knowledge so the results shouldn’t tend to be different. For example, you could pick door 1 and then decide to reveal door 2. But in this way what you are doing is basically selecting which two doors will remain closed for the second part. I mean, selecting door 1 and then revealing door 2 is the same as picking both door 1 and door 3 at once, and then discarding the other. So, if door 2 results to have a goat, which one do you think is which should have 2/3 probabilities of having the car, door 1 or door 3? Note that the doors don’t “know”, if you picked them at once or one after the other. The result would have been the same if you had first declared door 1 as your staying option and number 3 as your switching one, or viceversa. So, neither of them can be more likely than the other.
Another issue is that each of the three original objects is a different content by its own, that has 1/3 chance of being selected at the beginning and 2/3 chance of not being selected. For example, the three could be “white goat”, “gray goat” and “car”. (In this case we are differentiating the goats by their colors, but even if they have the same color they are each a different entity).
So, let’s take for example that you pick a door and then Monty randomly reveals one of the others, which results to have the gray goat. Now, if you think that you have 2/3 chance to win the car if you switch just because you had 2/3 chance to pick a content that was not the car, then you should also agree that you have 2/3 chance to get the white goat if you switch, because you also had 2/3 chance to start picking a content that was not the white goat. But of course, both things cannot be true at the same time.
You may wonder if this same reasoning can be extrapolated to the original game. The answer is no. Since in that game the host is forced to not reveal the car, then it is sure that when your door has the gray goat the host will reveal the white goat, but when yours has the car it is not sure because he could reveal the gray goat in that case. So, once the white goat is revealed it is easier that it occurred because your door has the gray goat rather than the car.
Keith Hawley.
Related to your last post (we are talking about the random revelation): since each individual game you play is independent of the others, then if in the long run switching is not advantageous respect of staying, it is because in each of the single trials it is not advantageous either. It wouldn’t make sense that in the first attempt it helps, and then suddenly loses its relevance when the same game with the same conditions is repeated more than once.
To play a single time can be seen as randomly selecting one game from the list of multiple attempts you could make. As only 1/2 of them are ones in which you win by switching, then you are only 1/2 likely to be playing one of those in which switching wins, not 2/3.
Now, you are probably wondering why the chances are 1/2 for both strategies and don’t remain 1/3 vs 2/3. It is because the revelation of the goat is, in some way, “difficult to occur” if the car is not in your door, and therefore if it occurred it is difficult that the reason was because your door has not the car, so their original 2/3 chances decreased. With the word “difficult” I am meaning that it is not something that necesarily occurs in that case.
Note that you could make it difficult enough to turn it impossible. For example, suppose this variation of the game in which Monty knows the locations but only reveals a goat and offers the switch if the player started picking the car door, because his intention is to make him switch so he loses. If the player started picking a goat, the host deliberately ends the game. In this way, it is impossible to win by switching; despite the player still has 2/3 chance to pick a wrong door at first, if a goat is revealed it is a certain indication that he is under the 1/3 cases when he has picked the car door (a subset), so the chances of winning by staying are now 100%. And obviously this is true regardless of if he is repeating the game one time or more.
Hi, Chris.
Sorry, but your reasoning is incorrect. It is true that you always end with two doors, but the first selection determines which will be the staying one and which will be the switching one, and the most important thing: it makes the switching one to be correct more frequently than the other.
To see it better, imagine the same game starting with 3 doors but instead of only one contestant, there are two contestants A and B that want to win the car. One chooses first and then the other must choose a different door. But the trick is that contestant B will have the advantage of already knowing where the car is, while A won’t. There are two cases depending on who chooses first:
1) If person B is who chooses first, it is obvious that he/she will win with 100% probability. Knowing the locations prevent that person from failing. Moreover, since the second person cannot repeat the first’s selection, the poor A will always select a wrong door for sure. If we reveal the third door (which neither of them selected) it is sure that we will find a goat in it, and we could do it everytime they play. So, at that point two doors will remain closed: A’s door and B’s door, but that does not mean that each will have 50% chance to have the car. We know that it is 100% sure that B’s door is which has it.
2) Now, if A is who chooses first (randomly), then that person can get the car with 1/3 probability (1 out of 3 attempts on average). But when A starts failing (2 out of 3 times), then B will automatically win, because knowing the car’s location prevent that person from failing if it is in any of the two that he/she can choose, so B still has advantage. Same as before, if we reveal the third door we will find a goat in it for sure, and so two doors remain closed: A’s and B’s, but if you had to bet who won, would you say it is indifferent? We know that A’s tends to be correct 1 out of 3 times, and B’s 2 out of 3.
If you note, the Monty Hall problem is equivalent to case 2). The contestant’s selection is equivalent to A’s selection, because he/she chose randomly. The other door the host avoided to reveal (the switching one) is equivalent to B’s, first because the host knows the positions, and second because the host cannot reveal the car and neither the contestant’s selection, so everytime the contestant has failed (which is more likely), the other door the host avoids to reveal is which has the car, in the same way that B chooses the correct everytime that A fails. The revealed door with the goat is equivalent to the one that neither A nor B chose.
Tina, since the car is 100% sure to be behind one of the three doors, then if it has 1/3 chance to be in yours, it must have 2/3 chance to be in one of the other two ones. Where else would it be in the 2/3 cases in which it is not in your door?
There is one minor point and I promise to move on!
I had asked before if these probabilities were subjective and you said no. So, I am not completely clear on how there is a subjective aspect to nonsubjective probabilities.
Let me explain my view and you can tell me where I am wrong or missing something. “The die shows 4. Now, what’s the chance that contestant A will guess correctly (1/6) and the chance for contestant B (1/3).” As I see it, the only way for the contestants’ guesses to be assigned probabilities is if they use a random process where the sample space and probability distribution are given/assumed. For example, contestant B can make his guess by rolling a different fair die and guess the result. Similarly, contestant B can also roll a fair die with the faces numbered {2,2,4,4,6,6}. Each random process yields exactly the probabilities you state. Furthermore, if the contestants accidently use the other’s die, then the probabilities are reversed—even though the information they have are different.
Thus, the probability of guessing correctly or incorrectly is not particular to what the person knows but is the critical aspect of the random process used. Granted, contestant A would choose a different random process than contestant B given the information they have. But anyone, regardless of what they know, has the same probability of guessing correctly as contestant A if they use the same or equivalent random process as contestant A.
Rather than viewing this as a subjective aspect, view it as all nonsubjective probability calculations: they are conditional on the probability distribution used. Perhaps a minor change to your statement “we need to calculate the probabilities given the information that the guesser has, not the person with complete knowledge.” Consider stating it this way: “we need to calculate the probabilities given the random process the guesser uses which may depend on what information the guesser has.” And perhaps add parenthetically, the probabilities are the same for anyone using that random process even if they have different information.
Kicab,
I’ve always said that there’s a degree of subjectivity to probability. Your ability to predict the outcome depends on the amount of information you have.
I’m not going to slog through your long text, but the answer is simple.
For the first contestant with no information, their set of possibilities is {1, 2, 3, 4, 5, 6}. We know the die shows a 4, but they don’t know it. So, they’ll pick one of those 6 numbers. Hence, they have 1/6 chance of being correct.
The second contestant is told that it’s an even number. Their set is {2, 4, 6}. Again, we know it’s a 4, but they don’t. So, they’ll pick one of three numbers randomly, producing a 1/3 chance of being correct.
It’s really that straightforward.
By the way, when you say, “we need to calculate the probabilities given the information that the guesser has, not the person with complete knowledge,” that is exactly what is meant by subjectivity. It’s from each contestant’s point of view.
Because you’re going around in circles, I will not approve any more comments about this aspect.
Jim, when you went back to the die example, I realized that we have been talking about different things. No wonder we were getting frustrated! Probably my not being clearer was my lifetime education from 1st grade to graduate degrees in US public schools (😉 LOL).
Your discussion is about what you stated in the die example. “Imagine that a person is running a game rolls a die and it shows a 4. There are two people guessing who can’t see the result. One contestant has no information. Their guess has a 1/6 chance of being correct. However, the other person is told that it’s an even number. Their guess has a 1/3 chance of being correct.” I completely agree (of course, assuming that the contestants randomly guess with each of their choices equally likely) and agree that that is “how statisticians have studied probabilities.” In addition, even the person who rolled the die and knows the result, would agree that contestant 1’s random choice has a 1/6 probability of being correct and 1/3 probability for contestant 2.
My questions and comments have been about the probabilities when the game continues—and this is consistent with your comment about information. In this example, imagine the contestants have guessed but each still does not know that the actual result was a 4:
1. Contestant 1’s guess is 5. The probability that he guessed correctly is 0.
2. Contestant 2’s guess is 4. The probability that she guessed correctly is 1.
Are we in agreement now?
Hi Kicab,
That is correct from one very specific view point, the person who rolled the die and knows the answers. They’ll know whether the contestant’s guess is correct.
But there’s also a subjective aspect to probabilities that you’re overlooking. The contestants don’t have that knowledge and can only make guesses based on the information they have. And, again, we’re talking about probabilities that their guesses are correct. The die shows 4. That’s a set thing. Now, what’s the chance that contestant A will guess correctly (1/6) and the chance for contestant B (1/3). The reason for the different probabilities is that they have differing amounts of information. Certainly you can see that the person with less information will have a lower chance of guessing correctly than the person with more information? And we’re talking about from their perspective. Just because the die roller (or Monty in the MH problem) knows the answer doesn’t mean that the contestant knows the answer. To solve these types of problems, we need to calculate the probabilities given the information that the guesser has, not the person with complete knowledge.
You can also look at this probability as a long run relative frequency. So, while an individual’s guess is right or wrong, if you run the game multiple times, what’s the percentage of times that the contestants will get it right? That percentage varies based on the knowledge they have. In the die rolling example, contestants with no information will be correct 16.67% (1/6) of the time. While contestants knowing whether it is an even or odd number will be correct 33% (1.3) of the time.
Now, we’ve thoroughly beat this dead horse into the ground. No more on this, OK?
From my understanding of statistics, the probability should be the same for contestant and (either) Monty.
Your understanding is incorrect. As I explained in my previous reply, the probability that each person’s guess is correct depends on the amount of knowledge they have. As you gain knowledge, you can have better guesses. Better guesses have higher probabilities of being correct. With perfect knowledge like Monty’s, it’s not even a guess.
Imagine that a person is running a game rolls a die and it shows a 4. There are two people guessing who can’t see the result. One contestant has no information. Their guess has a 1/6 chance of being correct. However, the other person is told that it’s an even number. Their guess has a 1/3 chance of being correct. The person who rolled the die knows the number with 100% certainty.
Three different people. Three different amounts of information, And three different probabilities even though they’re all guessing about one set die roll. The point is that even though the die roll result is fixed at a particular value, it’s the probability of guessing correctly that changes.
Now I’ve explained this multiple times. Please move on.
Hi Jim
To be fair, I have already said I agree with exactly what you are saying here!
In my 27th September post I explained how I used exactly that logic on my 120 sets of doors proof for Knowledge Monty to understand Random Monty and said
‘To summarise, if the game is played multiple times and Monty has no knowledge and opens doors at random, then some games will have to be ignored if a car is revealed, and overall there is statistically no benefit in the contestant switching doors.’
My confusion now relates solely to the odds in a single game where a goat is revealed, but we’ve come to the end of the road on that and I’m clearly starting to annoy others too now.
Bye!
Keith, I just explained that. And the probability for a single game is the same as a group of games. It was just easier to illustrate with 999 games. But the probability for a single game is still 50/50!
That’s a superb and nicely condensed explanation of the difference between the 2 types of games! I hope Keith understands that one, lol. 😉
Thanks! And I sure hope so too!
Hi Jim
You really must have infinite patience. Just as well, as I must query a couple of points you make before I quit.
Firstly though, just to clarify, yes I totally accept that there is no double-counting in your simulations and that they are correct.
In my original suggestion for proof of the MH problem nearly two years ago (good grief, how did that happen!) I reasoned as follows:
Once the contestant has picked their door there are just three possibilities left for the remaining two doors:
Goat/goat
Car/goat
Goat/car
In the first case Monty obviously reveals a goat and the remaining door hides a goat so the contestant will loose if they switch.
In the second case we know he reveals a goat and so the remaining door hides a car and the contestant will win if they switch
The third case gives the same result – switching wins a car
So swapping doubles the chances.
The assumption is that he knows where the prizes are, otherwise he could reveal a car and spoil the game, and I accept that is how it would normally be played.
My conjecture however is that we are specifically told he does not reveal a car, he reveals a goat. Once we know that, then the outcomes for each of the three options is exactly the same even though the goat was just revealed by chance. How could they be different? No outcomes are discarded to alter the odds, it is just the same three options with exactly the same results.
Swapping still doubles the chances.
You say ‘The solution for Random Monty is really simple as there is one prize and two unopened doors = 50/50’ I really don’t understand why you say this.
Surely the basic premise of the MH problem is that the chance of the contestant’s door hiding a car is always 33%? That is locked in right at the beginning as soon as they have picked a door.
Nothing Monty subsequently does can alter it, and this fact is often the basis for dispelling the doubters who think it does mysteriously change to 50%.
The other 67% must therefore be the with remaining door as it has nowhere else to go.
Finally, you are saying that if Monty picks a goat at random then the odds are 50/50 and there is no benefit in switching but if he picks a goat intentionally then switching doubles the chance.
If that is true then the only way the problem can work is if the contestant knows WHY the door has been picked. Your final paragraph seems to confirm you agree with that.
But if you read the MH problem all it says is that Monty opens a door and reveals a goat. No information is given about why he chooses this door. Even if he did know where the prizes were, there is absolutely nothing to suggest the contestant knows why he has opened it. How could they know? All they see is an open door, and a goat and they are asked whether they want to stick or switch. You say knowledge is power, but they just don’t have that knowledge.
This, to me, is another way of showing that the odds must be the same whether Monty reveals a goat by chance or knowledge, otherwise it means the contestant requires information they do not have and the whole thing just won’t work.
I know we have gone round in endless circles but I hope one of the three arguments I have made might be convincing. If not, then I guess we’ll sadly have to draw a line there and thanks again for the entertainment your site provides. Sorry to have been an irritating distraction!
Hi Keith,
This really is my last attempt! I’ve enjoyed our back and forth overall. But, we are going around in circles. Here’s one more time! And, it’s instructive to highlight why Random Monty and Knowledge Monty has different solutions. Spoilers. You seem to have a blind spot for the effect that the 1/3 of cases where Random Monty reveals the prize has on the outcome!
For both of these, I’m going to work through an imaginary set of 999 games.
Random Monty
After playing the imaginary set of 999 games, the contestant’s initial choice is correct 1/3 or 333 times. Monty accidentally reveals the prize 1/3 (333) times. And, 1/3 of the time (333 games), Monty reveals a goat. Now, we exclude the times that Monty reveals the prize. So, we’re down to 666 games. Of those 333/666 (50%) are cases where staying wins because the contestant’s original choice was correct. And the other 333/666 (50%) is where switching wins. Hence, it’s 50/50.
Knowledge Monty
Here’s why it doesn’t work out that way for the original Monty Hall problem. In this set of 999 games, the contestant chooses the correct door 333 times. For the other 666 games, the car is behind one of Monty’s two doors. Because Monty knows which one the car is behind, he always chooses the door with the goat. Consequently, for this set of 666 games, the car is always behind the unopened door. Consequently, the contestant wins 333/999 (1/3) by staying and 666/999 (2/3) for switching.
I can’t make it any simpler than that to understand! I hope it’s clear now, but if it’s not, I’m going ask you to think about it more on your own. I’m out of different ways to explain it!
Jim, the confusion I have is when you mix verb tenses. For example you equate “probability of where the car is located” with “probability of a getting a number on a fair die.” The former is past tense (located) while the latter is future tense (getting). My question was about the past events, as we agree about future events’ probabilities.
Let’s use your example: “For example, someone rolls a die. It shows a 4 but that is hidden from the contestant…the probability of whether it was a 4 (which is always 1/6).” Suppose the person who rolls the die can see the result as 4. For that person the probability is 1 that the result is a 4. Correct?
If correct then, my point #2 correctly describes how you see the probabilities: #2 After the die is rolled and results in a 4, we have two probabilities. The probability is 1 for everyone who knows the result and the probability 1/6 for everyone who does not know the result.
Is this what you are saying—or am I still missing something?
Hi Kicab,
I’m not sure if English is your native language but both are present tense. Both of those are cases where there is an outcome but the contestant doesn’t know what it is. However, by gaining knowledge, they can increase their probability of making the correct guess.
No, the problem is not the tenses that are used. I can see that you have basic misunderstanding of how probability works. I don’t mean that as an insult, but I highly recommend that you take a course in probability or read a book about it. I am not able to teach you about probabilities from the ground up.
And please understand that it is not “how I see probabilities.” It is how statisticians have studied probabilities. Yes, there can be a subjective aspect to probability. Keep in mind that the uncertainty is in the people’s assessment or guess of the outcome. The outcome itself can be a specific, set value (e.g., the die has a 4). For a person who observes the outcome, their assessment of the outcome has a probability of 100%. However, for someone who hasn’t observed it and needs to guess the outcome, the probability associated with their guess can certainly be less than 100%. As that 2nd person gains more knowledge, the probability associated with their guess can increase. There is no inconsistency there.
Hi Jim
Many thanks for giving up even more of your time to try and convince me, I really appreciate it.
Since my last post I’ve been trying to understand why we disagree and I think we are actually both right in our own way, it’s just that we are looking at slightly different slants on the problem.
Firstly, and importantly, I now agree totally with what you have been saying!
I know you feel I have not been listening to what your reasoning was, but that is really not the case. I pointed out what I thought was a discrepancy with your 12 option table and that it was effectively double-counting the results where the contestant had picked a car.
The computer programs were more difficult to decipher without seeing the raw data, but I felt the same was happening there too.
What finally made me understand, just before I read your latest comment strangely, was thinking about the solution to the basic MH problem which I posted 10th Sept.
Here if you remember I suggested using 120 sets of doors. After the contestant had selected one from each set they were removed from the room temporarily. This was to emphasise the 40 cars and 80 goats behind them could not possibly be affected by anything Monty did.
My lightbulb moment was when I realised that if Monty chooses randomly he would pick around 40 cars from the 120 doors he opened randomly. These would have to be eliminated from the results and so the corresponding 40 goats from both the contestant’s and the remaining unopened doors would have to be removed too. This leaves 40 cars and 40 goats behind both sets of the remaining doors and so clearly no benefit to switching. Apologies for taking so long to get it!
To summarise, if the game is played multiple times and Monty has no knowledge and opens doors at random, then some games will have to be ignored if a car is revealed, and overall there is statistically no benefit in the contestant switching doors.
I was looking at the question in a slightly different way, which to my mind was more the problem as it appears when first presented on your site:
A SINGLE game, two goats and one car, three doors, contestant chooses a door, Monty opens door to reveal a goat, should you stick or switch? That was all the information you were given.
There was no indication at that stage as to whether Monty knew where the prizes were, so no clue as to whether it was by chance or knowledge that he revealed a goat, but still you were asked to come up with your answer as to whether the contestant should stay or switch.
My argument, in these circumstances, is still that all that matters is that a goat has been revealed.
Monty has not revealed a car and so no outcomes need to be ignored, so consequently there is nothing to skew the odds.
Surely the chances for the doors can only be 33% – 0% – 67% so the contestant should switch.
Or another way of looking at it, if Monty’s knowledge was relevant to the outcome of this single game then how can people possibly make a considered judgement on whether to stick or switch without knowing whether he knows!
I really hope that you will be able to agree with me now I have shown more precisely the question I was actually trying to address.
If not, just ignore me and I’ll go away!
Hi Keith,
First, I’m so glad you’ve seen the light! About the simulations, it was just playing the game as you defined it. You can see the code and follow the steps. So, no double counting. That’s something you really just need to accept.
The solution for Random Monty is really as simple as there is one prize and two unopened doors = 50/50. That’s not the case for the real Monty Hall problem because those cases where he would’ve revealed the car become cases where the car is behind the unopened door, and thus stacking the odds in favor of the unopened door relative to the random case.
I can’t agree that we were both right though, unless you mean that you were just resolving the original MH problem. The probability for Random Monty is 50/50 for a single game or a series of game.
And, the probability of winning by switching for the original MH problem is 0.67 for a single game or a series of games. So, I’m not sure what distinction you’re trying to draw there.
In the original MH problem, you’re told that MH will open a door that has a goat. It’s not outright stated but to ensure that happens, he must have knowledge of where the car is and will avoid it. That’s part of the trick to the solution. So, what you’re describing there is just the original MH problem. And, yes, 2/3 for switching is the right answer for that. But, I don’t think we disagreed about that?
And, I’m reading your 4th from last paragraph and it looks like you’re still getting it wrong. “All that matters is that a goat has been revealed . . .” No, it’s the process that counts. If Monty chooses a door randomly and it is a goat, then the probability of winning by staying vs. switching is 50/50. Period. The reason why is in my 2nd paragraph of this comment.
Finally, if the contestants don’t know how Monty chooses a door, then you’re correct. They don’t have the information necessary to help them improve their odds. Knowledge is power. And, as I mentioned early in our conversation, that’s one thing I love about this problem. We’ve come full circle!
Jim, I may not have been clear. I completely agree that “So, when you guess one of the numbers, you have a 1/6 chance of guessing correctly.” But the issue is not “when you guess” but AFTER you guess, what is the probability that you guessed correctly.
You explain “It’s like you roll a die but you don’t see the results. Sure, the die has a specific result but you don’t know what it is.”
Are you saying this is how it works?
1. Before the fair die is rolled, the probability of getting a 4 is 1/6. This probability is the same for everyone regardless of what anyone knows or does not know.
2. After the die is rolled and results in a 4, we have two probabilities. The probability is 1 for everyone who knows the result and the probability 1/6 for everyone who does not know the result.
3. In general, if p is the probability of any event before it occurs, then after the event occurs the probability is 1 for those who know the result and p for those who do not know the result.
Hi Kicab,
I think you’re mixing up two probabilities. One is the probability of where the car is located. Or, the probability of a getting a number on a fair die.
The other probability is of guessing which door has the prize or what number the die will show. They’re related but different probabilities.
The probability of an event occurring (e.g., car behind door 1, die showing a 4) is fixed in these examples. However, the probability of guessing the correct answer can change based on information.
For example, someone rolls a die. It shows a 4 but that is hidden from the contestant.
Now we’re interested in the probability of the contestant guessing correctly, not the probability of whether it was a 4 (which is always 1/6).
Let’s say the contestant makes a random guess. That too will have a 1/6 probability of being correct. Imagine now that the person rolling the die says that the number is an even number. That narrows it down to 2, 4, 6. The contestant’s chance of guessing correctly has changed from 1/6 to 1/3 even though the number 4 remains constant. It’s the information available that changed. You can make a more informed guess, which improves your chances.
Hi Jim
I don’t expect a reply to this, but having enjoyed your website so much over the years I didn’t want to leave on a discordant note.
I totally accept you know far more about statistics than I do. I’m just a curious engineer who likes to try and deal with problems logically.
Like all the Monty sceptics, of course I think I’m right with my related query, but I am actually quite open to being convinced I’m wrong. I just wanted to know where the fault in my thinking was.
I wanted to try and understand why, after years of you patiently explaining to the Monty Hall disbelievers that the chances of the contestant’s initial choice were 33% and nothing Monty did could possibly change that to 50%, you seem to be saying to me that is just what happens if Monty reveals a goat by chance.
Anyway, best wishes for your website. There is still clearly a steady stream of doubters out there who have yet to see the light and need convincing.
Hi Keith,
Alright, let me try one more time. But, please, take the simulation seriously. It is literally playing the game 100,000 times using the rules and scenarios that you described yourself. Monty chooses a door randomly after the player. He has no knowledge about the prize location so he can accidentally reveal the prize door. Then we look at only cases where he randomly reveals a goat. In those scenarios, what are the chances for the player to win by staying versus switching?
I really want you to take those results at face value. It is telling you that the answer of what really happens is 50/50. Do not reject it just because it doesn’t agree with your preconceived notion.
Now, I’ll try explaining the why of it, whereas the simulation just reports what happened.
I’ve tried two different ways already. I’ve explained how in the original Monty Hall problem, you have 2/3 chance of winning by switching because Monty has a zero probability of opening the prize door, which transfers the full 2/3 chance to the unopened door. However, when he chooses randomly there is a chance that he’ll open the prize door, hence the final door doesn’t get the 2/3 probability. It only gets a 1/3 probability. Yes, I know we’re considering only cases where he reveals a goat, but you still need to model the probability for the entire problem before getting to the more specific cases.
So, the outcomes for all cases are 1/3 chance the prize is behind the door Monty opens, 1/3 behind the door he doesn’t open, and 1/3 behind the contestant’s choice. Because we’re excluding cases where Monty reveals the prize, we’re only considering 2/3 of the total outcomes and 1/3 favor staying and 1/3 favor switching. And, from the full set of outcomes, we know that there’s an equal probability that the prize is behind the contestant’s choice vs. Monty’s unopened door. Because they’re equal, it must be a 50/50 outcome in the reduced set of outcomes.
I also showed this to you in the table format, which shows how it’s 1/3, 1/,3 and 1/3 for the full set of outcomes and 50/50 in the reduced set.
ALL of that is stuff I’ve explained to you earlier and really why I didn’t want to comment more. I’m just spending time retyping what I’ve already typed and you’ve rejected out of hand. And please DO NOT say that I haven’t tried showing you the solution. I have! So, that’s source of my frustration.
I do have a little bit of a different angle to add in addition to the above. But, it’s really just an elaboration of how the above works
In what we’re calling the random Monty scenario, when he reveals a goat, there are two doors where the prize could be–the contestant’s choice or Monty’s unopened door. Because both Monty and the contestant are choosing randomly, the probability is equal for those two doors. Two doors and one prize. Hence, 50/50. It’s really that simple.
The more interesting question is why that’s NOT the case for the original Monty Hall problem where he has perfect knowledge and never reveals the prize. In that scenario, he’s not making random choices. He intentionally opens a door with no car, which ends up stacking the odds in favor of his unopened door having the car. We don’t have that 1/3 of cases where he accidentally reveals the prize. Instead, those cases don’t happen because he’ll intentionally open the door without the prize, meaning that there are more outcomes where Monty’s unopened door has the prize than if he were acting randomly. This results in the prize being behind his unopened door 2/3 of the time. But that only happens when he knows where the car is and can avoid opening it by accident.
So, even though you don’t want to consider the cases where he accidentally reveals the prize, the fact that they happen in the random Monty problem impacts the outcomes!
I sincerely hope that it is clear now.
I was basically where you are Keith. I can only say what helped me understand.
If you forget everything your thought of previously – I think some explanations are actually a distraction – and just consider the same game with 100 doors.
In a 100 door version the contestant has a 1 in 100 chance of choosing the right door.
Then – whether random or not – does anything Monty do actually change the fact that the contestant had a 1 in 100 chance? It doesn’t.
So then what if Monty closes 98 doors at random? There is a 1 in 100 chance the car is in the remaining door. And still a 1 in 100 chance it’s in the contestant’s door.
But if not random then the 99 out of 100 chance it was in a door other than the door the contestant has chosen becomes a 99 out of 100 chance it’s in the remaining door. All because the chance the contestant picked the right door at the beginning is still 1 in 100.
Hi Jim
I know I had said I would not be posting again, and I realise you must be totally fed up of me by now, but after brooding over your last comment for days I really have to respond as I do feel your points were rather unfair.
I have presented my proof in various ways and have repeatedly asked you to show any flaws in my logic, but you have repeatedly refused to do so. Not once have you shown where I am saying 2+2=5. If you are right, then I must be wrong, but you have never shown what flawed assumptions I am making.
I have pointed out where I think your 12 option table is wrong (My comment of 13th September and again in more detail 16th September) but you have not responded to that point. I believe your computer runs are also wrong for the same reason.
But let’s keep it simple.
The contestant’s door has a 33% chance of a car
The Monty Hall problem simply states Monty opens a door and reveals a goat. Whether by chance, or because he knew where the prizes were, cannot alter the fact that once a goat is revealed there is, obviously, zero% chance it is a car.
The remaining door must therefore have a 67% chance of a car
My challenge to you Jim, is to show me what is wrong with that logic.
No more computer simulations, just good old rationale.
Unless you can show it is wrong, then it must be correct.
Hi Keith,
I almost didn’t reply to you, but then I realized that you should at least know why I’m not discussing this with you further.
I see no reason to continue because you simply reject my explanations out of hand.
I created and ran a simulation of the exact scenario you were interested in. That simulation shows it is 50/50 when Monty chooses a random door and you only consider cases where he reveals a goat. You rejected that and are claiming it’s wrong when it is really just because you don’t like the answer. You say “no more computer simulations,” but the simulation is absolute proof of the answer because it actually played the game 100,000 times following the specific rules you described! You should be thinking, OK, now I know it’s 50/50, but why is that the case? Instead, the simulation is wrong.
And, I have explained to you using two different methods for why it is 50/50. Here, you’re not just saying I’m wrong, but you’re actually saying that I’ve refused to explain it at all. Now that’s taking rejecting evidence to a whole new level.
So, I really have no incentive to continue. It’s not an intellectually honest conversation. Why would I waste more of my time? I won’t be goaded into that with a “challenge.”
I have nearly 30 years of professional experience as a statistician. That doesn’t make me infallible by any means. But, you might consider that I actually do know what I’m talking about rather than just rejecting it out of hand!
We don’t always have enough information to ascertain whether something is true, or know what the outcome of an experiment will be, so we might say, loosely, that an event is probable or improbable. Why? If it’s probable that it’s going to rain tomorrow we know it’s wise to take an umbrella. Probability as a branch of mathematics attempts to make this more precise by assigning a numerical value (between 0 and 1, or 0% and 100%) to the likelihood that something is true or will be true, when in fact we don’t know if it’s true or will be true. If we say there’s a 50% chance of rain tomorrow, I may or may not have to have an umbrella; if there’s a 99% chance I almost certainly will. The whole reason for probability’s development as a branch of mathematics is that we can’t know everything, but it’s good to be able to predict what’s likely to happen, or likely to have happened, so we know what else we may need to do. The Monty Hall problem is a good example: if we want to win the car, it’s good to be able to calculate what choice of door makes that most probable. Mathematical probability, if you ever decide to take a course in it, is really interesting, in part, because people smarter than most of us have been able to come up with results that are precise in ways most of us wouldn’t have been able to think of on our own, or think of in a non-analytic, intuitive manner.
One of the points I was trying to make is that different people may assign completely different probabilities to the same thing depending on what they know. One person may have more accurate knowledge of something than another person, and thus be able to make a more accurate assessment of whether something is probable.
But perhaps you’re looking for something more, such as exactly how we assign a number between 0 and 1 to something. I suspect that’s what may be bothering you. As an example, why exactly do we say the contestant in the Monty Hall problem has a 2/3 chance of finding the car behind door #2 after Monty reveals the goat behind door #3? A computer simulation, or an actual run of trials, may convince you that you’re better off switching to door #2, but the question is: why is this the case? That’s what mathematical probability can answer. As I said, the first time I saw the problem I incorrectly thought switching doors wouldn’t make any difference; only when I thought of it more carefully did I see what my mistake was. I would have liked to have been correct from the first, but there’s tremendous satisfaction in the moment you can say, “Aha! Now I see it! I see what I didn’t get initially!” So, let me ask you: do you accept that switching to door #2 in the Monty Hall problem is correct only because computer simulations of a large number of trials shows it to be the case, or have you had that “Aha!” moment when the math behind it all makes sense? Because that “Aha!” moment is really what you want.
Or did I confuse you with something else I said?
Tell me what definition of probability you are using, so I can understand your explanations. Thanks.
You did not antagonize me — if you had I wouldn’t have bothered with a reply.
I have two comments to make in response to what you’ve written. When I first saw the Monty Hall problem I came up with the wrong answer. I think the reason for that is that I analyzed it only by considering that we knew a goat was behind door #3. What I didn’t consider was how we obtained that knowledge (that Monty always opens a door with a goat). That makes a difference. We always know that whatever door we pick at least one of the remaining doors has a goat behind it, so Monty’s shtick of opening a remaining door that has a goat behind it is a red herring distracting us from the fact that the car has a 1/3 chance of being behind door #1, and 2/3 chance of being behind either doors #2 or #3. On the other hand, if we were to accidentally hear a goat behind door #3, we’d simply conclude that the car is either behind door #1 or door #2, and that the probability must be 50/50 since whether we initially picked door #1 or door #2 has no bearing on where the car actually is.
Second comment: I disagree with your statement that, “Once the coin is tossed and ‘it winds up heads,’ the probability is 1 that it is heads even if you don’t know the result.” I think the proper way to think of this is to say the probability is 1 only to someone who knows the result and 50/50 to someone who doesn’t. Our calculation of probabilities depends on what we know and what we don’t; probabilities are there as a predictive tool and our predictions change according to what we know. If I toss a coin in the air and conceal the result once it lands, you don’t know the result so all you can say is that there is an equal chance of it being heads or tails. Only when you look can you say the probability is 1 for heads and 0 for tails, or vice versa. So knowing the result makes all the difference.
Joe, I am so sorry I antagonized you. I had no intention of doing that and in fact I thought I was agreeing with you on a couple of issues.
So let me start over. There are two cases.
Case 1 Single Game: In this case you are correct that the probability of winning by switching is 1/2 but only if the contestant decides to switch by randomly choosing where switching and staying are equally likely. There is nothing in the problem statement that says the contestant cannot do this, so it is one solution for the single-game case. This 50-50 odds is better than what you get in Las Vegas. These odds are guaranteed for every single game. Keith was also correct in saying it doesn’t matter how the opened door revealing a goat was chosen in this case. If another probability distribution is chosen for switching or staying, then the probability of winning by switching will not be 50% nor guaranteed for each game.
Case 2 Multiple Games (Or long-run): As Jim has alluded to by his table of 9 possibilities, encouraging readers to do simulations, and using the phrase “in N games,” the long-run odds can be improved. If you are playing the game many times, then when Monty knowingly always opens a door with a goat behind it, the best strategy is to always (not randomly) switch. Now you should win 2/3 of the time. This is not the probability in a single game; only the long-run probability.
As an aside, you state “If you toss a coin you’d say it’s 50/50 whether it comes up heads or tails. If it winds up heads, the probability was still only 50/50 before the coin was tossed.” That’s true. But not true after the coin was tossed. Once the coin is tossed and “it winds up heads,” the probability is 1 that it is heads even if you don’t know the result.
I never stated that “the selection of the door for the car is independent of the contestant’s selection of a door.” Of course the selection is dependent upon the selection — it IS the selection! What I said was that the location of the car is independent of what the contestant’s first guess is, so the probability of it being behind a certain door is independent of that guess.
I also never said the “initial choice of door #2 makes THE CAR MORE LIKELY to be behind door #2.” Rather, I pointed to the absurdity of thinking the initial choice would make the car more or less likely to be behind either door, and hence the necessity of a 50/50 probability of it being behind either door #1 or door #2.
I can’t understand why you would say, in speaking of what the probability is about, that “It is not where the car is but which door the contestant chooses.” It is, rather, precisely the probability of where the car is that concerns us; the contestant will guess according to how he calculates that probability.
You also state, “Since the probability is 0 of the car being behind that door after it was opened it was also 0 before the door was opened.” That’s not correct. The whole point of using probabilities here is to give us the best possible guess as to what the real state of affairs is when we don’t know what that state is. When all three doors are closed, all we can say is that each door as a 1/3 chance of having the car behind it. If we open all three doors, we will see that two of them do not reveal a car. At that point we know the car has 0% probability of being behind one of those two doors since we can see it’s not there. But that doesn’t mean we were wrong to assign a 1/3 probability to the car being behind either one in the beginning when all the doors were closed and before we had knowledge of which door the car was actually behind. If you toss a coin you’d say it’s 50/50 whether it comes up heads or tails. If it winds up heads, the probability was still only 50/50 before the coin was tossed.
Chris, you are right but only (as I previously commented) if you randomly decide to stay or switch with each option equally likely. If it is not a random decision or you use some other probability distribution than half each, the probability is not 1/2.
The question I posed to all is: What probability distribution for staying and switching will yield a 2/3 probability of winning by switching?
Hi Kicab,
You’d use a binomial distribution with a 0.66 probability for the event (winning the car) to occur by switching. That’ll describe probability of X wins in N games. Conversely, you’d use the binomial distribution with a probability of 0.33 for winning by staying.
Joe, you are correct that the selection of the door for the car is independent of the contestant’s selection of a door. That is why Jim’s table describes nine possibilities (3 car doors x 3 contestant doors), each equally likely.
But the issue is that AFTER the car is placed behind a door, it never moves; the same with the goats. When you say “initial choice of door #2 makes THE CAR MORE LIKELY to be behind door #2,” you are misstating what the probability is about. It is not where the car is but which door the contestant chooses.
Consider the door that is opened to reveal a goat. The goat never moved, so it was behind that door before the door was opened. Since the probability is 0 of the car being behind that door after it was opened it was also 0 before the door was opened.
Question: Given that analysis, what are the probabilities for the other two doors?
Joe’s description sounds correct to me.
Kicab, some times you have to update probabilities as you gain information. Imagine you have a deck of cards and you pull out one card at random but you can’t see its face. The dealer asks you to state the probability that it has heart on it. Of course you’d say 25%.
You keep that card set aside but don’t look at it. Now imagine that the dealer reveals a number of cards but not the entire deck. Amongst those cards, there is a greater proportion of hearts than the other suits. At this point, if you were asked to state the probability of your card being a heart, it would be lower than 25% because you observed relatively more hearts in the revealed cards. Your card hasn’t changed but your probability of hearts has changed given the additional information you acquired.
The same thing happens in the Monty Hall problem. By understanding that Monty has information and uses it to act non-randomly, you can better estimate the probability for the door. So, think about it as being able to come up with a better estimate of the probability given additional information. The car doesn’t need to move or the card doesn’t need to change suits for you to need to change the probability. Sometimes the available information changes, which helps you better evaluate the situation.
Hey to everyone
I just saw this problem a couple hours ago watching 21 and I think I can explain it simply.
The answer is actually 50/50
Basically your first choice is irrelevant.
Host will always open a goat.
Your actual choice is just the final choice between the 2 doors of which 1 of the 2 doors will always be the car.
The chance of getting correct the first time is in fact 1/3 but that does not matter.
Your actual choice is between 2 doors every time.
If you want to do it in the way where by the host does not know which door has the car then you would have to allow for having a 0% chance of selecting the car when the host opens the door with the car and you are forced to still pick between your original or remaining door.
But like I said you don’t need to worry about that as the host changes the odds to 1/2 and makes you choose again which is where the odds stay.
Think of it this way. Suppose you initially guess door #1. You then accidentally hear a goat bleat behind door #3. You would conclude the car is either behind door #1 or #2. And had you initially guessed door #2 and heard the goat bleat behind door #3 you would reach the same conclusion.
But the probability of the car being behind either door #1 or door #2 is independent of whatever your initial guess was — wherever it is has nothing to do with whether you initially guessed door #1 or door #2. Which means you can’t say an initial choice of door #1 makes the car more likely to be behind door #2 any more than an initial choice of door #2 makes the car more likely to be behind door #1. That being the case, you’d have to say the odds are 50/50. Make sense?
Keith,
Jim’s simulations show exactly what happens in the long-run (playing the game many, many times). In other words, what are the probabilities BEFORE a game is played. If Monty always opens a door with a goat behind it, then 2/3 of these numerous games result in the car behind the remaining unopened door. If Monty randomly selects a door to open, then 1/2 the time the car will be behind that door.
You ask does it matter whether Monty’s choice to open a door is random or not. It does matter in the long run. It does not matter in each single game.
For a probabilistic answer, the problem requires assumptions about random events and their probability distributions. This classical solution makes two assumptions: the door for the car and the contestant’s door are selected randomly with each door equally likely. But the question by the contestant in the puzzle is about the single game he is playing. It is asked AFTER the doors for the car and by the contestant have been selected and a door is opened showing a goat.
At this point in this (and every) single game there are two cases.
Case 1: As Jim has stated in his description, there are no more random events. The question now is “What is the probability of a specific outcome for a nonrandom event?”
Case 2: Assume that the contestant randomly chooses to switch or stay with each choice equally likely. (We can make this assumption just as we made the others since none are included in the problem statement). Now what is the probability of winning?
As a bonus question, what probability distribution can the contestant use to randomly decide to stay or switch that will result in a 2/3 chance of winning by switching in this single game?
Hi Jim
You say I’m incorrect, but do not explain why. What is the fault in my logic?
This will sadly be my last post but I will give the proof in one final way:
The contestant’s initial door will ALWAYS be 33% chance that it’s a car
The opened door with a goat will obviously ALWAYS be 0% chance
How can the final door therefore EVER be anything but a 67% chance?
What part of that do you think is wrong?
Hi Keith, I’ve actually explained three different ways! In some cases, it’s not possible to figure out why someone thinks 2 + 2 = 5. However, I can show you why 2 + 2 = 4.
For this problem, I’ve explained why using probability, tables, and several simulations.
You ask, why is it wrong, and I’ve showed you multiple times. I’m not going to try explaining further . . . horse and water and all. BTW, having a contestant choose a door randomly doesn’t prove anything at all about what happens when Monty chooses a door randomly. I honestly don’t understand what you’re aiming at with that.
But, if I were you, I’d start by focusing on my 2nd simulation – the one that considers only the cases where Monty (or the contestant) reveals a goat. The simulation models the exact process and doesn’t make any assumptions. It just runs it 100,000 times and reports what happened. It shows that for a random Monty without knowledge, who can pick the door with the car, but we only consider cases where he reveals a goat, the answer for switching or staying is definitively 50/50.
Now you know what the answer really is. The next step is to figure out why. At that point, consider the tables I’ve created as well as the probability explanations I’ve provided.
Hi Jim
Sorry my comments have been so convoluted.
Ironically, since posting my last comment I realised there is a proof so short and simple I can’t believe I never thought of it before:
The contestant choses a door.
Before Monty has chance to open a door, a lady in the audience leaps onto the stage.
She does not know where the prizes are, but opens one of the other doors at random to reveal a goat.
Should the contestant stick or switch?
Obviously they should still switch (33% chance with original door, 0% with open door, 67% chance with remaining door)
If instead, Monty had NOT known where the prizes were, but had revealed a goat by chance?
Exactly the same odds.
If Monty HAD known where the prizes were and revealed a goat?
Exactly the same odds.
ALL that matters is that a goat was revealed.
Hi Keith,
Sorry, but that’s just not correct. I’ve created another simulation. The process of choosing doors randomly is the same. Prize door is chosen randomly. Contestant picks one door randomly. Monty opens one of the two remaining doors randomly–it can be the prize.
However, in this simulation, the program counts the times that Monty opens a door with a goat and discards all cases where he reveals the car. Among those case where he reveals a goat, the program counts the number of times the contestant will win by staying vs. switching.
As you can see, the probability is essentially 50/50 in the output below where I run it 100,000 times. So, now I’ve shown you this in table form and simulation form and I hope you can accept it. As I’ve described several times, this works because by picking randomly, the full 2/3 probability doesn’t transfer to the unopened door. That affects the probability. If that’s not clear, there’s really nothing more for me to stay I’m afraid! I’ve illustrated the solution two different ways and explained why it works that way! Output and simulation program below.
Hi Mike
I agree totally with everything you say in your second paragraph about the chances of winning – that is exactly what I have been trying to say all along!
I originally posted an additional way of proving the MH problem and then, just as an aside, questioned people’s assumptions about Monty’s knowledge.
I thought Jim would agree to be honest but he didn’t, and we have so far failed to reach a resolution.
Sorry if you find it an unnecessary distraction. Can’t say I blame you.
Keith, respectfully, in my opinion you are hopelessly overthinking this problem, and strangely introducing an element that isn’t even part of the problem – Monty DOES know where the prizes are; he always opens a door to reveal a goat. ALWAYS. That IS the problem. Why are you trying to add this additional element that frankly ruins the fun of the problem as originally stated?
Even if he didn’t know where the prize was, and opened his door randomly (sometimes revealing the prize), during those times when he reveals a goat, the remaining door still has a 2/3rds chance of containing the prize. The math at this point doesn’t change just because Monty wasn’t guaranteed to reveal a goat from the beginning. Those times he opened the door to reveal the prize, obviously, renders the whole problem moot, and the contestant simply had a 1/3 chance to win with his or her initial choice. I don’t see what’s remotely interesting about exploring this avenue, I’m sorry.
Hi Mike and Keith,
I’ll answer both here. Mike, I do think it’s an interesting question about the effects of Monty’s knowledge because show how the problem works out when Monty doesn’t have knowledge highlights why the original Monty Hall problem works out as it does.
Keith, if you could please make your comments shorter, they’d be easier to read! I believe I have shown how the two versions are different. For one, I disagree that the lose scenarios are irrelevant. Those definite lose scenarios just don’t exist in the original game. Monty’s lack of knowledge will force some contestants to lose. When he has knowledge, contestants always have a chance of winning. But, if you look at ONLY the cases where Monty does not reveal the prize, I’ve answered that as well. Both my table and simulation agree that you’ll have a 50/50 chance of winning/losing by either switching or staying. There is no advantage to switching. That’s another difference from the original game.
So, I believe I’ve answered your question about that by demonstrating how it works using two different methods!
Hi Jim
Sorry for the rather tardy response but I’ve been away for a couple of days.
It’s great that others have been drawn into the debate. I hope they don’t mind if l just respond to you at the moment. Frankly, walking whilst breathing is about the limit of my multitasking abilities these days, so it could get rather confusing.
I should also like to emphasise to any bewildered onlookers that I agree totally with the basic Monty Hall problem: that switching doors doubles your chances. I am just trying to get to the bottom of a very specific question re the relevance of Monty’s actions because I’m a pedantic and persistent old twit.
Thanks for doing the computer program – sorry I’m turning out to be rather high maintenance!
It’s a fair while since I did any programming, and even then it was just in Basic, but I don’t doubt the accuracy of what it is showing for what has been programmed.
My query, as I’ve said, is solely related to those instances where Monty has revealed a door, and whether the chances of winning are doubled whether he knew where the prizes were or not.
The ‘Lose’ outcomes are therefore not relevant, but if we ignore those it is still showing equal outcomes for stay or switch. Can I come back to that later please?
The only way I’ll be prepared to believe I’m wrong re my specific query is if you can point out to me anything I’m missing when I repeat what I have already asked, but not yet had answered.
Is there any fault in the following logic?
Let’s assume Monty does not know where the prizes are.
The contestant chooses a door (1/3 chance of car)
Monty opens a door at random, and we know he reveals a goat (zero chance of car) *
Then the final door must have a 2/3 chance of a car.
Obviously the game would be over if he had revealed a car by chance but the problem says specifically ‘Monty opens one of the other two doors, and there is no prize (or car) behind it’ so we know for certain that is not what has happened
Another way of putting it would be to use the reasoning I used in my November 2019 comments:
The contestant chooses a door. There are then only 3 possibilities for the other 2 doors
1 Goat/goat
2 Goat/car
3 Car/goat
Let’s first assume Monty knows where the prizes are.
In case 1 Monty would reveal a goat whichever door he opened and the contestant would lose if they switched
In case 2 he reveals the goat and the contestant would win if they switched
In case 3 he again reveals the goat and the contestant again wins if they switch.
Now assume he does not know where the prizes are
In case 1 as before Monty would reveal a goat whichever door he opened and the contestant would lose if they switched
In case 2 he reveals the goat by random (remember any game where he reveals a car is ignored because, to repeat again, the problem says specifically ‘Monty opens one of the other two doors, and there is no prize (or car) behind it’ so that is not what has happened, and contestant wins by switching
In case 3 he again reveals a goat by random and again the contestant wins by switching.
Surely this shows exactly the same odds as when he knew where the prizes were, unless again you can show a fault with my logic?
What the 12 option table from your earlier reply is saying, I believe, is that there are 4 possible outcomes from these 3 options:
It is giving case 1 two different outcomes
Firstly he opens the door to reveal one goat
Or, secondly, he could open the other door to reveal the other goat.
These are statistically two different possible outcomes, but they could not happen in the same game!
So either line 1 or line 2 of your table should be removed.
Similarly, because you show the options for the contestant initially picking all three doors, lines 6 and 7 are duplicates as are lines 11 and 12
If you remove those three duplicated lines the odds came back to 3/6 in favour of switching; the same as given in your original proof with 9 possible outcomes when presenting the Monty Hall problem.
I’m pretty sure that the computer simulation you showed must be doing the same thing, but I’m afraid I’m not conversant enough with the computer language used to be able to check that.
“So, when you guess one of the numbers, you have a 1/6 chance of guessing correctly.” That’s correct. But the probability that the die has a specific result is 0 or 1. Isn’t there a difference between what the specific die result is and what one guesses it to be?
Keith, what I like about Jim’s tables is that they implicitly state the two requirements for calculating the probability of an event’s outcome: 1) the event must be random and 2) there must be a probability distribution for the possible outcomes. Jim’s initial table with nine rows, shows that the events of selecting the door for the car and the contestant selecting a door are random events with each door equally likely in each event. The other tables show the same thing including what happens if the opened door is also a random event with each of the two doors equally likely to be opened.
These tables show that under the two assumptions of randomness and probability distribution and the assumption that Monty knowingly opens a door with a goat behind it, the probability of winning using the switching strategy is 2/3 in the long-run or before the game starts.
You are asking is the probability of winning by switching affected by whether the door was opened deliberately to reveal a goat or by chance. You can get the right answer by simply a) determining after the door is opened which following event is assumed to be random and b) what is the probability distribution of that event’s outcomes. Jim’s initial description provides a clue.
There is only a 1/3 chance the car is NOT behind those two, since it must then be behind the remaining door, and that probability, as we know, is 1/3.
The probability that the car is not behind two particular doors has to be the same as the probability that it IS behind the remaining door since those are just two descriptions of exactly the same event.
I believe Tina is correct, any two doors have a 66percent of NOT having the car
Hi Trudy,
Keep in mind that the probabilities for the event occurring and not occurring must sum to 1. Consequently, because the probability for the prize being behind either of Monty’s two doors is 2/3, then the probability of the prize being behind neither of his doors must be 1 – 2/3 = 1/3.
Before opening any door, there is a 66 percent chance the car is behind any two doors and 66 percent chance the car is NOT behind those two.
Hi Jim
I appreciate the time and trouble you took to produce your tables and send your reply, especially on a Sunday!
I’m afraid though that I partly disagree with your conclusions, which means I’m probably about to appear an idiot, but it wouldn’t be the first time, so here goes….
Just to be clear first, I totally agree that, if playing the game for real, Monty needs to know where the prizes are to avoid revealing the car and spoiling the game.
The contentious point still I think is, if we know he has revealed a goat, as the original problem states, does swapping doors double the chances of winning a car even if he didn’t know before opening the door where the car was.
Your second table seemed at first glance to make prefect sense but the 50/50% conclusion did not.
It took me a long time to work out why I think it’s wrong.
If you look at the table you give in your original explanation of the problem and its solution, above the comments section, you list 9 possible scenarios. You total up the results to show you win a car for 3 of these if you stay with the original door and 6 if you swap.
In the second table you did for me you give 12 scenarios, with both switching and staying both giving 6 wins each, which is obviously different from the table just mentioned, and that doesn’t seem to make sense.
If the contestant picks the correct door initially you have counted Monty picking either of the two remaining doors as two different options in the table with 12 options, whereas in the table with 9 scenarios they are treated as one. As the outcome of picking either door is the same (swapping means losing the car) then surely that is the correct interpretation. Removing these duplicated outcomes brings the split back to 3 wins if you stick and 6 if you swap.
Another way of looking at it would be, if the 12 options table was correct, would it not give exactly the same list of 12 scenarios if Monty had known where the prizes were, and we know that answer is certainly not 6 wins for both staying or switching.
Or, one final way of looking at it:
I originally gave the logic as follows:
Let’s assume Monty does not know where the prizes are.
The contestant chooses a door (1/3 chance of car)
Monty opens a door at random, and we know he reveals a goat.
Then the final door must have a 2/3 chance of a car.
Again excuse the blunt way of putting it, but if that is not true, then what part of my reasoning is incorrect?
Hi Keith,
Hey, what else are you going to do on a Sunday afternoon other than working through Monty Hall scenarios!
Monty’s knowledge is the fundamental factor behind how the original Monty Hall problem works. The reason that the door you can switch to has 2/3 chance of winning is because of Monty’s two doors, there’s 0 probability that he’ll open the door with the car. Hence, the full 2/3 chance transfers to his unopened door. The only reason there’s a zero probability of opening the door with the car is because he knows where the car is located and will intentionally avoid it. So, Monty’s knowledge is baked right into the solution. It drives the process. Another way to think about it is that we’re comparing scenarios where he has perfect knowledge and it drives his choices versus another scenario where he chooses randomly. Surely there must be differences in outcomes between intentional actions using knowledge vs. random choices? And the table shows that’s the case.
I might need to revisit my tables to be sure I’m getting them correct. I’ll need to ponder that down the road. However, I do know that my tables for the random Monty scenario reflect the correct probabilities. I’ve used a simulation program to simulate the game as follows.
I ran it 100,000 times and here are the results.
The output shows that each potential outcome has 1/3 chance of occurring.
I used Statistics101 to create this simulation. It’s giftware, so you can try it yourself. Here’s the script I wrote that you can run with Statistics101: RandomMontyHallProblem.
Below is a screenshot of the script so you can peruse it!
Thanks for your reply Jim. Whilst I follow and understand everything you say I’m afraid I still don’t accept Monty needs to know where the prizes are as the problem is worded.
When communicating online like this, without the subtle benefits of body language or nuances in your voice, it is all too easy to come across as confrontational or even aggressive, so apologies if I come across as such now. That is not my intention – one of the things I like about your website is that you do not tolerate rudeness or unpleasantness.
Let’s assume Monty does not know where the prizes are.
The contestant chooses a door (1/3 chance of car)
Monty opens a door at random, and we know he reveals a goat.
Then the final door must have a 2/3 chance of a car.
Surely that is all there is to it?
Now I obviously accept that in practice, if he does not know where the prizes are, then he would reveal a car roughly a third of the time. But in the problem we are considering we specifically know that this possibility has not arisen.
I realise we are perhaps more in the realms of pedantry and semantics, rather than statistics, but I still maintain the fact Monty reveals a goat is all that matters, not whether he knew or not.
Hi Keith,
I’m always open to politic discussion and your comments definitely come across as that! Funnily enough, this Monty Hall post is the one that generates the most rude comments of any post that I have.
So, to explore your question, I’ve simulated that Monty Hall problem but assuming that Monty acts randomly. He has no special knowledge. This is an interesting question because we can observe how his knowledge changes the outcomes. I’ve written about the importance of him using his knowledge to affect the outcomes, and now we’ll see what happens when he doesn’t have that knowledge.
What I did was I came up with two tables to capture all possible outcomes of the game. For any given game, there are 6 possible combinations of door choices (3 * 2 * 1 = 6). And the car has an equal chance of being behind any of the doors.
For the first table, I don’t remove the occasions where Monty reveals the car by chance. That will happen 1/3 of the time and forces the contestant to lose. So, I don’t think you can just discard those. Imagine it from the contestants point of view. Monty tells him one of the door contains the prize. The contestant picks one and then Monty opens the door with the car. Monty tells the contestant the s/he can either keep his door (without the prize) or switch to the other door, which also doesn’t have the prize. That contestant won’t be happy! That’s a Hobson’s choice.
In this table, green represents the contestant’s choice while yellow represents Monty’s choice. Red is the door that Monty doesn’t open. The status column indicates how the contestant wins (Stay or Switch) and the cases where the contestant is forced to Lose. Then by counting up the instances of each outcome and dividing by all outcomes, we get the probability. Notice in the table the rows where the car is behind Monty’s choice (yellow), winning is impossible for the contestant, which corresponds to Lose. The winning status of the other rows depends on which door the contestant picked and where the car is located–either the contestants door (green, stay) or the unopened door (red, switch).
In this scenario, contestants win by staying 1/3 of the time, win by switching 1/3 of the time, and lose 1/3 of the time.
Now, let’s look at your specific question. Only those scenarios where Monty doesn’t choose the door with the prize. Again, I don’t think that’s a good approach because it’s a scenario where the constant cannot win. But, here you go! I literally went through the first table and removed the rows where Monty picks the winning door.
In this scenario, the contestants win 50% of the time by staying and 50% of the time by switching. Of course, that’s excluding the 1/3 of the time they have no chance of winning!
Based on all that, I think it’s clear that Monty’s knowledge matters. Obviously, there’s no part of the probability formula where you input whether Monty has perfect knowledge or not. Instead, it shows up in the effect of his knowledge. Namely, that the probability of him opening the door with car is 0, which means that his unopened door has the original 2/3 probability for his group of two doors.
I love this problem because it’s a great analogy for life. Sometimes we don’t know the correct answer to a real-life scenario. However, by understanding how a process works and using that information to inform our choices, we can improve our chances of coming out ahead! That’s particularly true when the process is not random and we can use the patterns to our advantage.
Since discovering this website nearly two years ago I have regularly come back to peruse the latest comments. The most satisfying perhaps are those from people who are initially sceptical, but are prepared to keep an open mind, eventually understand, and are gracious enough to admit they were wrong.
Many are clearly willing to be convinced but still can’t quite get their heads round it. The way the answer seems initially unbelievable is of course part of its appeal, but does make it all the more satisfying when you do finally get it.
Amongst the confused there is often the belief that the problem is somehow split into two different parts, and that once Monty has opened his door then the odds of the initial door hiding a car have changed.
Various people have offered different explanations for why this cannot be the case, but I would like to add the following explanation as it tries to move away from the use of ‘probability’ and ‘chance’ and instead concentrates on the numbers of physical objects, which some might find easier to relate to.
Imagine that rather than one set of three doors there are 120 sets. Happily we can ignore the logistical, financial, health & safety and animal welfare issues that doing this for real would involve.
Going for 100 sets of doors might have seemed more sensible, but I want a number that will divide by 3 and still give a round number.
So we have a total of 120 hidden cars and 240 hidden goats.
The contestant now picks one door from each of the 120 sets. Or 120 contestants can each pick a door from a different set, it doesn’t matter and would speed things up a bit.
Since there is a probability of 1in 3 of picking a car (yes, I know I said I’d try and move away from probabilities, but frankly if you don’t understand with this bit then perhaps this website isn’t for you) then you would expect the 120 chosen doors to be hiding around 40 cars. For simplicity let’s assume it is exactly 40 to make things easier to follow. The other 80 doors obviously hide goats.
OK, let’s assume those 120 doors and the objects behind them are now temporarily removed from the room and lined up outside, out of the way.
Monty now needs to pick one door from each of the remaining 120 pairs of doors. This will reveal 120 goats.
We no longer need these goats and so they can be released back into the wild, returned to their owners, or put up for adoption, hopefully by people who will not further traumatise them by making them hide behind random doors all day.
Their corresponding doors can also be taken away as they are now just cluttering the place up.
We can now bring back into the room the originally picked 120 doors and their hidden contents.
Right, now this is where it starts getting exciting. Well, perhaps not as exciting as white-water rafting, or bungee jumping , but everything is relative and this is a statistics website after all.
So, unless some of the goats were pregnant and have messed everything up by giving birth in the meantime, this first set of doors will still be hiding 40 cars and 80 goats. The point being, nothing Monty has been doing in the meantime could possibly have had any effect on these numbers.
We can now do the maths (sorry, I’m British, can’t bring myself to say math)
We started off with 120 cars and 240 goats.
Monty revealed 120 goats by opening his 120 doors which leaves 120 cars and 120 goats still hidden.
The doors initially chosen hid 40 cars and 80 goats
The final set of doors must therefore hide 80 cars and 40 goats.
If the contestant(s) stick with their original door, they will win a car 40 times out of 120
If they decide to swap doors they will win a car 80 times out of 120.
Swapping doubles the chance of winning.
Hopefully that makes it easier for at least some people to understand.
A question for you Jim please:
It is generally assumed in the comments that Monty knows where the car is and it is this fact that alters the odds. In my original posting 12th November 2019, where I suggest the game only has three possible outcomes, I also assumed he knew (although it ultimately makes no difference to the logic I used)
Thinking further about it however, the only relevant fact is that you are told he opens one of the two remaining doors and reveals a goat. Surely whether by chance or deliberately does not matter? The odds will now be the same regardless.
If he did not know where the prizes are, and revealed a car, then obviously the game is over (you win a goat whether you stick or change) so those outcomes are taken off the board by default. It is this that alters the odds.
Conversely, if he chose a door, but did not reveal what was behind it, it would mean swapping doors had no benefit, but this is not what happens.
Once he has revealed a goat however (whether by chance or knowledge) you know that whilst the odds of the contestant’s original door hiding a car still remain at 1in 3, then the odds of it being behind the final door must be 2 in 3, as it has nowhere else to go.
Would you agree?
Hi Keith,
Thanks so much for your insights and your new way of looking at this problem! I, too, have found the thread of comments to be interesting. The different points of view have given insight into different stumbling blocks and different ways to explain the solution.
There are several reasons why I love the Monty Hall Problem so much. For one thing, it really highlights the fact that people in general are not good with probabilities. There’s that sense of “what the heck?!” (to be polite) upon learning the correct answer. It just feels so wrong. Indeed, many people refuse to accept it. I’ve had college professors write and say that many of their statistics students refuse to believe it. Another point, which I don’t write about in the post, is the tendency to stick with the door you chose. When people think it’s a 50/50 choice, they really have no reason not to switch. But, according to other research I’ve read, most people will stick with their original door. They want to go with their original choice. So, that’s another way in which their intuition leads them astray.
Yet another reason I love this problem relates to your question–Monty’s knowledge. By understanding how Monty operates, that he has knowledge of the prize’s location and will avoid it, you can use this information to improve your odds of winning. And, yes, it is Monty’s knowledge that underlies why it’s better to switch. Remember, that Monty’s two doors together have a 2/3 chance of hiding the prize. The reason why the final door that he offers to you retains that full 2/3 probability is because there is ZERO chance that he’ll open a door with the prize. So, the full 2/3 probability goes to the unopened door. Why is there a zero probability of Monty opening the door? Because of his knowledge! That full probability wouldn’t transfer if he didn’t know where the prize was and could, thereby, always avoid it. So, that’s the underlying factor for why switching doubles your chances of winning!
You mention the possibility of Monty opening the door with the car if he didn’t know where it was. That leaves the contestant losing whether they stay or switch. You mention that those outcomes are taken off the board, but they really aren’t. In that scenario, Monty’s lack of knowledge locks-in the outcome where the contestant loses. There’s no way to win. However, if Monty operates using knowledge to avoid the car, the contestant retains the chance of winning. So, all in all, Monty’s knowledge and his informed actions increase the chances of winning by not locking contestants into a losing scenario.
You could say, there’s a 50/50 chance that the prize is behind the remaining door after Monty opens his… IF Monty hadn’t know where the prize was. BUT, this all means that about 50% of the time, Monty must deliberately avoid one of his two doors of choice. THAT is what enhances the probability that the prize is behind the remaining door. It’s enhanced by the average number of times Monty in ignorance would have gotten it wrong.
The word “probability” implies an objective property of the doors, and that’s what fools the mind. If probability is objective reality here, then how is there a 100% probability of Monty picking the right door if asked, while only a 33% probability of you doing so? Because it’s not a property of the doors or the prize.
So replace the word “probability”, or the word “chance”, with the word “certainty”. Now it’s a lot more intuitive, right? Certainty is based on how much you know, and what you know changes throughout the game. You know with 33.3% certainty where the prize is, Monty knows with 100% certainly.
Monty tips you off. Monty increases your certainty of where the prize is from 33.3% to 66.7%, because he affords you two doors for your one door.
At the start, you’re not actually picking a door. You’re randomly throwing out one door — in the guise of picking it — with hopes that it’s not the one with the prize. Doesn’t matter which door, just throw one out, sorry. You’re actually randomly picking the other two doors for the play, because you’re vying for the remaining 66.7% certainty that the prize is behind one of two remaining doors. Then Monty is going to throw one of those two remaining doors out out — no hoping there this time, because you know for certain he picked one without the prize. Good intel!
So you’re randomly throwing out one door and then waiting on Monty. He will tell you which of the two doors you can rest your remaining two-thirds certainty upon. (You already threw out one third of your certainty, while crossing your fingers, in order to place your bet on the other two as one.) Monty made one door worth two in terms of your level of certainly about it.
(Restated from my reply elsewhere.)
“The probability of selecting the door with the prize is:
–1/3 for the contestant
–1/3 for Monty”
No, it’s 0 percent for Monty, because he knows with 100 percent certainty where the prize is. Thus with his insider knowledge, he has the ability to tip you off using some of what he knows.
The word “probability” implies an objective property of the doors, and that’s what fools the mind. If probability is objective reality here, then how is there a 100% probability of Monty picking the right door if asked, while only a 33% probability of you doing so? Because it’s not a property of the doors you pick from.
So replace the word “probability”, or the word “chance”, with the word “certainty”. Now it’s a lot more intuitive, right? Certainty is based on how much you know, and what you know changes throughout the game. You know with 33.3% certainty where the prize is, Monty knows with 100% certainly.
Monty tips you off. Monty increases your certainty of where the prize is from 33.3% to 66.7%.
At the start, you’re not actually picking a door. You’re randomly throwing out one door — in the guise of picking it — with hopes that it’s not the one with the prize. Doesn’t matter which door, just throw one out, sorry. You’re actually randomly picking the other two doors for the play. Then Monty is going to throw one of those out — no hoping there this time, because you know for certain he picked one without the prize. Good intel!
So you’re randomly throwing out one door and then waiting on Monty. He tells you which of the remaining two doors you can have your remaining two-thirds certainty about. (You already threw out one third of your certainty.) He made one door worth two in terms of your changing level of certainly about it.
Think of it like this: Monty does NOT have a 33.3% chance of *knowingly* opening a door to nothing. He has a 100% chance. You now know that Monty’s door has a 0% chance of having a prize. By “spoiling” the unknown status of his door, he’s tipped you off to a NEW statistical reality. Remember, he can’t have picked your door to open. And he can’t pick a door with the prize.
Let this sink in: Thanks to Monty, opening the remaining door at the end of the game is effectively opening BOTH remaining doors at the same time. Before Monty acted, there was 66.7% chance of finding the prize when you open both remaining doors, right? Monty with his insider knowledge has not changed that reality for you. There is, and there will always be a 66.7% chance to you that your door has no prize.
That. Will. Not. Change.
Monty has revealed his door to have a zero percent chance. The other door is now worth two doors, informationally, thanks to Monty. You don’t have to choose from two remaining doors. Both are now the same door. You’re doubling your chances by going to the door Monty knowingly did not pick. Monty has rigged the scene by bestowing his door’s probability to the remaining door. Remember that unlike you, Monty is not guessing.
The illusion is the idea that, at the end of the game, before opening your own door, you will open one remaining door to see if the prize is there. What has really happened is that, at the end of the game, to reveal what’s behind them, you with Monty reveal what’s behind TWO remaining doors at the same time — behind two doors with one look. 1/3rd chance plus 1/3rd chance the prize will be there when you open TWO remaining doors, right? That’s what you and Monty are doing together. (And he’s not guessing, he’s giving you information, moving statistical information from one door to the other door, by zeroing out his door, as it were.)
Kicab – you are overthinking this and adding parameters that aren’t part of the problem. There is no “before” or “after” the prize is placed behind a door – the prize is always behind one door. Also, Monty has a probability of 1 – he knows where the prize is at all times.
When the contestant selects 1 door from a choice of 3 doors, his chances are 1 in 3, or 33.3%, correct? Therefore, the other 2 doors combined have a chance of 66.6%. Nothing that happens afterwards changes any of those odds – how could it? Your eventual choice will be to stick with the door you chose – 33.3% – or switch doors – 66.6%. One of the other doors has been opened to reveal a goat, but this changes nothing, all it does is transfer that door’s 33.3% chance to the remaining door. Think it through and see if that makes sense to you.
If you had 100 doors, and you chose 1 for a 1% chance, then Monty opened 98 other doors, leaving one door remaining, wouldn’t you switch?
Jim, to confirm.
The probability of selecting the door with the prize is:
–1/3 for the contestant
–1/3 for Monty
After selecting a door, the probability that the door selected has the car behind it is:
–1/3 for the contestant
–0 or 1 for Monty.
Is this what you are saying?
You are absolutely correct that when the contestant is selecting/guessing which door, he has a one-third chance of guessing correctly. After he has guessed, it is no longer one-third chance that he guessed correctly. Not the difference between before guessing and after guessing, i.e., guessing vs. guessed.
Hi Kicab, sorry but what you write is incorrect. The probability that the contestant’s initial choice is correct remains at 1/3 until the very end when Monty reveals the location of the prize.
Mike–the 33.3% chance is BEFORE the prize is placed behind the door. AFTER it is placed behind the door, the chance is 0 or 1. The contestant has a 33.3% chance of selecting the door containing the prize if the door is selected randomly with each door having an equal chance. But AFTER the contestant has selected a door, what is the chance contains the prize?
Hi Kicab, no, that’s not how it works. The contestant doesn’t know where the prize is located. It’s like you roll a die but you don’t see the results. Sure, the die has a specific result but you don’t know what it is. So, when you guess one of the numbers, you have a 1/6 chance of guessing correctly. Similarly, when you guess one of three doors, you have a 1/3 chance of guessing correctly and a 2/3 chance of guessing incorrectly.
However, Monty has seen the result, and that’s why there is a zero probability of him opening the door with the prize.
Tina – if there are 3 doors, each door has a 33.3% chance of containing the prize. So it follows that 2 doors will have a combined chance of 66.6% that one of them contains the prize. How is it possible for 2 of the doors to have a 66.6% chance of NOT containing the prize? That math just doesn’t compute.
Why do we automatically assume the chance is with the contestant? Or to phrase it better, why do we automatically assume there is a 66% chance that the car is behind one of the 2 doors and not that there is a 66% chance that the car ISN’T behind one of the 2 doors? I understand the probability aspect of it but never understood why it’s automatically assumed the probability is in the contestant’s favor. The same way there is a 33% chance the car is not behind the door chosen, there is a 33% chance the car IS behind it, and a 66% chance the car is not behind the other 2.
Let’s say you pick door # 1 and Monty gives you 2 choices:
(a) You win if the prize is behind door # 1
(b) You win if the prize is NOT behind door # 1
That’s what this is and put this way it’s obvious that (b) is twice as good as (a).
There would technically be 12 games if you split it down that far, but all 12 games by that count are not equally likely to take place. Looking at your 12 games spreadsheet, the simulated player chooses the correct door as their first pick 6/12 times, when in reality they would only have a 1/3 chance of choosing the correct door as their first pick. The right door games should be combined.
1/3 chance player chooses wrong door 1 (and therefore wrong door 2 is then opened)
1/3 chance player chooses wrong door 2 (and therefore wrong door 1 is then opened)
1/3 chance player chooses right door (1/6 chance wrong door 1 is then opened, 1/6 chance wrong door 2 is then opened)
Ah, thanks Jim! I think it was a combination of trying to debug the code for #3 above and #2 above that finally got me there.
You bet, Matthew!!
My comment refers to Saeed’s earlier comment regarding the 66percent chance being the chance that the initial guess is wrong.
That’s exactly my thinking. I posted a comment similar to yours but can’t find it at the moment. I think the “correct” explanation disregards a number of possible games because they don’t impact the outcome, but still keeps the initial starting point at a 1/3 chance. I’m not sure you can do both of these things. I’m with you – you know the game so you know at some point where you will end up. Also, Monty makes a non-random choice as to which door he reveals – except in the case where you already have the right door, where his choice really is random. I think these scenarios should be counted (making for 12 possible “games”, with the additional games being those of the type where switching makes you lose) or the starting probability needs to change to 1/2 to reflect what you know about how the overall game actually works.
I also explained what I was thinking in a seperate post (https://www.matthewdegeorge.com/2021/07/08/monty-hall-problem-another-guy-who-thinks-hes-solved-it/). There is a comment on my post correct me – but I’m still thinking about if I agree with the comment.
Hi Matthew,
There are multiple ways to explain this problem. I and others have gone through them in this post and throughout the many comments. So, I’ll briefly discuss several ways of looking at this problem. For more details on them, look through the comments.
1) You pick one door randomly which has 1/3 chance. Monty’s two doors logically has 2/3 chance. Monty opens one of his two doors. However, he won’t open a door with the prize. So, the probability that the door Monty opens will have the prize is zero. Because the probability that the opened door will have the prize is zero, Monty’s unopened door retains the full 2/3 probability for those two doors (2/3 – 0 = 2/3).
2) You can also think about it as a non-random process, which becomes more obvious when you increase the number of doors. Imagine that we have 100 doors instead of 3. There is still one prize. You pick one door and Monty gets the other 99 doors. Now, Monty will open 98 of his doors except one. He won’t open the door with the prize. After he opens 98 doors without the prize, he offers you the chance to switch to his one unopened door. Would you take it then? The chance the prize is behind the door that he offers you is 99%! The same principle is true with three doors. However, the nature of how it affects the probabilities is less obvious with fewer doors.
3) Finally, try it yourself! Read my other post about an experiment my daughter and I did by playing the Monty Hall problem 100 times and performing a hypothesis test. When you see the process from Monty’s perspective, it becomes more obvious how and why it works out as it does. It’s also an easy problem to simulate with computers. These simulations run the game thousands of times and they always indicate that your chance of winning by switching is 2/3.
Matt – no doors are removed; 1 door is revealed. You’re still choosing from 3 doors.
Another way to look at it is that yes, you have two choices: stay or switch. That may seem like a 50/50 choice. But it isn’t, because one choice gives you 1 door while the other choice essentially gives you 2 doors. Wouldn’t you rather take your chances with 2 doors instead of just 1? The fact that 1 of the 2 doors was opened to reveal a goat doesn’t mean anything – you’ve always known that 1 of the 2 doors is 100% guaranteed to have a goat. In fact knowing which door has the goat also tells you which door has a 2/3 chance of winning, while your chosen door only has a 1/3 chance.
Opening a door doesn’t change the fact that you chose your door from 3 choices, it only indicates that since it has a 0% chance of winning, the remaining door must have a 2/3 chance.
Your explanation makes a lot of sense.
What I have trouble with is saying it’s 1/3 chance to begin with. You know the game, you know regardless of what you pick, one door is being removed so it was always a game of 2 doors. I accept that smarter people have weighed in and have a consensus but no one has ever been able to explain that. Before it starts, Monty asks 1 out of 3 doors but he’s immediately going to pick a door strategically to reduce it to 1in2. Isn’t it always 1in2?
When you make your initial choice you have a 33% chance of it being the car. So on average it will be one of the other doors 66 % off the time. Once one door is removed there’s still a 66% chance your pick was wrong. Makes sense I think.
I think there is an illusion in 33-66 calculation. The 33% chance of winning if you stay with your choice is correct, but chance of 66% winning if you switch is not correct. It is, in fact, chance of your pick not winning which by illusion considered as switching your pick.
This problem has been modeled many times by computer, and the percentage always works out to be 66%. It’s a settled question.
Yes – the penny finally dropped when I overcame my assumption that a new game/rules/statistics had been created by his door opening.
It’s so interesting – we run 2 parallel concepts in our mind simultaneously. First (and the strongest and most compelling and hard to dismiss) is the FACT that ‘the car is now behind 1 of 2 doors’. But the second concept is the STATISTIC that ‘the car is 2/3 more likely to be behind the door we should switch to’. I’ve reflected on this, and decided that when FACTS conflict with STATISTICS, facts generally win our cognitive beliefs hands down.
That’s exactly why Monty opens a door, to introduce a red herring to make you think everything has changed and been reset, when in fact revealing a goat behind one door (a fact that you already knew) changes absolutely nothing, except that the remaining unopened and unchosen door has now adopted the 1/3 chance of the door that was opened, and now has a 2/3 chance of containing the prize. Your brain will do everything it can to reject this because “it doesn’t make common sense”, but as I saw elsewhere, you confirmed this for yourself by actually running the experiment. 🙂 The truth is, it’s one game from start to finish, and your initially chosen door was chosen from 3 doors, so nothing that happens afterwards can alter the odds from 1/3 for that door.
This explanation makes the most sense. And YET my brain has STILL taken so long to accept it!! I had to do 20 game runs to really accept it, and the result was so clear. Staying = I won 7/20 times. Switching = I won 16/20 times. Using a random generator, exactly replicating the scenario.
Goddamn brain.
Yes Frederick! That is my view too. By the host effectively removing one option, it has created a new ‘game’. A new game (1 of 2 doors has the car) with new statistical assumptions (the car has 50% chance of being behind either remaining door). The statistics of the first 3-door scenario CANNOT be applied to the new statistic of the new 2-door scenario. New game = new rules = new statistics. I would love to be convinced to switch, truly, but I just can’t see it!!
Okay; thanks for taking the time to respond.
Having rethought this, I reimagined it in a way that made more sense to me.
Specifically, if we look at this from Hall’s perspective, should I pick the correct door (1/3 probability), Hall has two choices of door to open; if I pick an incorrect one (2/3 probability), Hall only has one choice.
Ergo, there is about a 67% chance Hall only has one choice, and it makes more sense to select the other door.
Or I have managed to come to the right conclusion for flawed reasons.
Hi, Frederick.
This is easy to understand making an analogy with the lottery. I don’t think you believe you have 50% chance to guess the number that will win the jackpot of the lottery. To make the experiment, you don’t actually have to play the lottery; all you have to do is to guess the number. You tell me the number you thought and then the day when they give the results you don’t see them but you let me do it for you. If I see that your number was not the winner, I will say the winning number out loud to you, but if luckily yours was the winner, then I will say any other incorrect that I can think of.
In this way, the winning number will always be one of the two that one of us said: the one that you thought of first (at random), or the one that I said later (already knowing the results). It is like in Monty Hall when there are only two doors remaining. But do you think that each of them has 1/2 chance of being the winner? Or do you think that it was easier for me to be the one who said the right one?
This is basically what happens at Monty Hall. Since the host knows the positions and never discards the door with the prize, then other door he leaves closed is like the number I’d say, and the revealed doors are like the ones neither of us said.
The difference is that in Monty Hall you have the opportunity to switch your guess to the other option that I would say.
Because if you pick one its 1/3. If you get to pick a second one, that’s 2/3.
Okay; I have little doubt that I have managed to totally screw up the issues around this, but one begin with a 1/3 chance and, as presented, one selects Door 1. Hall obviously can only open either Door 2 or 3; selecting 1 would remove the whole point of the show. The absolute only thing to be sure of then is that the prize is not behind the door he opened and must be behind the two untouched doors.
At that point, the contestant effectively begins with a completely different and new scenario and faces just two choices, at least as I interpret it. Again, this definitely could be wrong.
Please explain to me why it is wrong to see this as an entirely new choice, or perhaps, in a slightly different situation, how it differs from the scenario where the contestant approaches the three doors at the beginning, but, before having to chose, Hall reveals that Door 3 is definitely not the door. Thus the contestant faces just a 50/50 choice.
You are a great teacher Jim! Thank you
Does this line of thought make sense –
Anytime you made your choice, there’s a 66% chance that you picked the goat.
Monty comes along and tells you that you can count out at-least this one door.
Now you are still left with a 66% chance that the door you had picked has a goat behind it. Why? Because that chance was locked in when you made the choice.
Monty pointing out a door with a goat behind it can’t affect that %
So basically, now you have a selection with a 66% chance of it being goat with just one other choice. Which means this other choice has (1-0.66) = 33% chance of it not being goat (as a choice can only be a goat or a car, this means 33% chance of not being a goat is the same as 66% chance of being a car)
Thus, if you switch, you go from
**66%(33%) chance of selecting a goat(car) to 33%(66%) chance of not selecting a goat(car)**
Did this make sense?
Thanks!
Before the game starts there are nine combinations. After the car is placed behind a door, there are three combinations. After the contestant selects a door, there is one combination.
Hi Nqobzin! It’s because the choice is really between 3 doors, not 2, even after Monty opens a door to reveal a goat. Like I said elsewhere, the opening of a door by Monty is a “red herring” – it’s meant to make you think that things have changed, but in reality it doesn’t change a single thing. Think of it this way: you choose your door from 3 doors = 1/3 chance, right? That means the other two doors combined have a 2/3 chance. When Monty opens a door to reveal a goat, he hasn’t given you any new information whatsoever – you always knew there was at least one goat behind at least one of the 2 doors you didn’t choose. By showing you a goat, Monty effectively transfers that door’s 1/3 chance to the remaining door that you didn’t choose, and now you’re left with this choice: the door you chose (still a 1/3 chance), or the door you didn’t choose and that Monty didn’t open (2/3 chance). The thing to know is that there are still 3 doors in play – you can still choose the door Monty opened! Of course you never would, but the illusion has been created that you are choosing from 2 doors, when you are actually still choosing from 3, it’s just that you now know of at least one door for certain that has a goat. Since your door is still 1/3, and the goat door is now 0, the remaining door must be 2/3. 1/3 + 1/3 + 1/3 = 1.
I think this specific part of the problem is the hardest to wrap your mind around. Took me years, to be honest. I believed in the correct answer, and I intellectually understood why it must be correct (and why it tests out and verifies this as fact), but it just didn’t “feel” right for the longest time, even after I understood the problem. So don’t feel bad, they call it a “brain teaser” for a reason.
Hi Mike K, please explain why the 2/3 chance doesn’t become a 1/2 chance after the elimination of door 3 when it is revealed that door 3 is a dud. Surely only 2 doors are in play following this elimination?
Jim illustrated that there are essentially 9 combinations when you made your initial choice of door 1, after door 3 is revealed to not be the door with the prize why are there still 9 combinations being considered, do the combinations now not shift to only 4 since you are now aware that door 3 cannot be chosen? Appreciate your response.
Hi Nqobzin,
I agree with Mike K’s answer to you. I just wanted to add that it’s important to understand that Monty’s choice is a non-random process, which helps explain why the full two-thirds probability gets shifted to the remaining door.
You have two groups of doors that are defined by your original random choice. You have your one door and Monty has his group of two doors. Your group has a 1/3 chance and Monty’s has 2/3 chance.
Monty opens a door. However, this is not a random choice. The probability that the door he opens has the prize is ZERO. The probability for the entire group is 2/3 and doesn’t change because the location of the prize has not changed. Because the probability for the door he opens is zero, using simple algebra, you can calculate the probability for the unopened door in Monty’s group must be 2/3. 0 + 2/3 = 2/3.
If Monty made a random choice, the probabilities would be different, and he might well have revealed the prize. But, it wasn’t random and he will never reveal the prize. That affects the probabilities as I describe.
No. Each door has a 1/3 chance. When you choose door 1, your door has a 1/3 chance. The other 2 doors combined have a 2/3 chance. When door 3 is opened, that 2/3 chance is effectively transferred to door 2. Door 1 still has a 1/3 chance; nothing about opening door 3 changed that. You’re now left with this choice: door 1 with a 1/3 chance, or door 2 with a 2/3 chance.
As mentioned elsewhere, this is easier to understand if you imagine 100 doors instead of just 3. If you chose 1 door out of 100, and Monty opened 98 other doors to reveal goats (remember, he CAN’T reveal the prize), would you say that both remaining doors have an equal, 50/50 chance?
How is that different to sticking to your original door?
Choosing to stick to your current door is still choosing a door – that and the already open door means you still have the same chance.
If door 3 is opened and you picked door 1, you now effectively have the choice between door 1 and door 2. Whether you choose door 1 or 2 from there, you’re also effectively taking door 3 meaning that either way you now have a 2/3 chance no matter what you pick.
This means we can safely ignore door 3 as it becomes irrelevant and you have a 50% chance of picking the right door, no?
When the doors are originally chosen, the odds are 1 in 3. If you got to choose two doors at the beginning, your odds would be 2 in 3. Opening the second door accomplishes the same thing. Its as though you chose that door and the door you switch to. In that case your odds are 2 in 3. If you stayed with the original door you chose, then your odds would remain 1 in 3.
I’ve loved this problem since I read about it in Marilyn vos Savant’s column in 1990 (and I love the story of everyone thinking she was wrong, but she was not), and it amazes me that in 2021 people are still arguing about it. The fact is that you benefit from switching – if you think otherwise, you are wrong, period. What people need to realize is that the MHP is essentially a classic con that uses misdirection to fool you into thinking that something that isn’t relevant, is. If you chose a door, and then Monty DIDN’T open a door but merely said, “Great, now would you like to stay with your original door, or would you like to choose both of the other doors?”, everyone would switch without a moment’s hesitation, and a game show that wants to make a profit can’t have that. So, they introduce a “red herring” – something that isn’t relevant to the problem at hand, but which distracts and misdirects you to think that it is relevant and possibly alter your choice. It is DESIGNED to trick you and fool you, so I guess it’s not surprising that it tricks so many, even today.
When you choose your door, it has a 1/3 chance of winning. Nothing that happens in the rest of the scenario changes that; it’s literally meaningless that Monty opens a door at all, because you already knew that a goat was behind at least one other door, and revealing which door that is doesn’t change a thing. It is irrelevant. You are choosing between a set of one door and a set of two doors, so the odds are 1/3 and 2/3 respectively. The red herring throws you off this logical trail and makes you think you might be choosing 50/50, when in fact revealing a goat doesn’t change the odds at all – all it does is shift the 2/3 chance from two doors to one door – the door that you didn’t choose.
I think an easier way to look at it is: When you picked your door the odds were 1 in 3. When the second door is revealed, your odds for the door you picked are still 1 in 3. But you can trade for the third door. Since one door is revealed, trading for the third door is the same as if you picked it plus the door that was revealed. That’s 2 of 3.
Hi. I’m not Jim but let me answer your question.
The trick in this problem is that since the probability of having failed remains being the same 2/3, it looks like if we didn’t update the information, as if we didn’t eliminate possible scenarios. But that is not the reason. What occurs is that the scenario in which you could have get it right was also reduced by half (as I will explain soon) so the proportion does not change.
It’s like which occurs in the example: Imagine that there are two men called A and B. Person A has 50$ in his pocket and person B has 100$ in his, so B has twice as A. If A spends half of what he has (25$) and B also spends half of what he has (50$), the money remaining in their respective pockets will be 25$ and 50$, so B still has twice as A. The proportion did not change despite the money is not the same as in the beginning.
In Monty Hall problem, note that since according to the rules of the game the host knows the locations of the contents and must reveal one door that is not which the contestant chose and neither which has the car, that means that he has two possible doors to reveal when the player’s one is which has thar car, but he only has one possible to reveal when the player’s one has a goat.
So, supposing you pick door 1, the possible cases are:
1) Door 1 has the car (yours) => Probability 1/3, but it splits in two sub-cases:
1.1) The host then reveals door 2 => Probability 1/6
1.2) The host then reveals door 3 => Probability 1/6
2) Door 2 has the car => Probability 1/3. Here the host is forced to reveal door 3.
3) Door 3 has the car => Probability 1/3. Here the host is forced to reveal door 2.
That means that if for example door 2 is revealed, not only must we remove case 2). We must also remove case 1.32) because we cannot be in one of those times when the host reveals door 3 being the car in door 1. We could only be in case 1.1) or in case 3), that had 1/6 and 1/3 chances respectively. Since the probabilities must always sum 1, the original 1/6 of case 1.1) turns into 1/3, and the original 1/3 of case 3) turns into 2/3.
Hi Jim,
Why doesn’t this problem get recalculated once the first door is revealed. The first door maintains a value of 1/3 chance even when it is determined to have 0 value.
I meant that everything goes as usual with the Monty reveal. Then one of the two contestants has the choice of switching with the other contestant. I assume that the odds are then 33%/66%. But if both contestants choose to switch, is it then 50% for each? Or does that end up putting them both back to where they were before the switch, which would mean 33%? Or does it change to 50%?
Ok, dumb question. What if there are two contestants rather than one? One of the two has the prize door, the other does not. If they switch doors with each other, are their chances each 66% or 50%. Is it possible for the both to have a 66% chance, as this is greater than 100%, so I would think not. Does it matter whether only one chooses or both agree to switch?
Hi Gregory, if there are two doors and the prize is placed randomly, and Monty has no role, then it is truly 50/50 for those two doors. There’s no benefit to switching. The reason why it becomes 66% in the actual game is because of Monty’s non-random role in removing a non-prize door from consideration. But I don’t see that in your scenario.
That article isnt exactly wrong in its conclusion as it does not conclude that Jim would be wrong but that in reality its not as sure to yield the desired results as most mathematicians will have you think. After all nothing is perfect. Including math.
Hi Clay,
Probabilities are long run averages for outcomes. Consequently, you know that in the long run, you’ll win 67% of the time versus 33% of the time by switching. However, some contestants will, in fact, lose by switching. To that extant, I’d agree that nothing is perfect. However, to maximize your chances of winning, you want to switch.
An analogy is that rolling a pair of dice and getting snake eyes (two 1s) only happens ~2.8% of the time. However, despite that low percentage, you will eventually roll snake eyes!
Math and probability theory are exceptionally good at calculating probabilities for games of chance like these!
Your own comments omit an important piece of information: Monty knows which doors have goats behind them, and makes sure when he opens door 2 to reveal one with a goat. But this doesn’t give you any new relevant information — you already knew at least one of the remaining doors, 2 or 3, had a goat behind it when you guessed door 1. You also know there’s a 2/3 chance the prize will be behind either door 2 or door 3; those odds aren’t changed because Monty verifies there’s a goat behind one of them.
Had Monty opened one of the remaining doors at random the situation would be as you described, but the assumption here is that he’ll make sure to choose a door with a goat. This is a very different situation than if you, say, accidentally heard the goat make a noise behind door 2. When I first saw the Monty Hall problem I made the same mistake you’re making. But you have to look at not just what you know about what’s behind the doors, but how you know it.
Think of it this way: had Monty initially offered you a choice of picking door 1, or letting you have the prize if it’s behind either of the doors 2 or 3, you would choose the latter since you’d then have a 2/3 chance of winning as opposed to 1/3. If Monty’s shtick is to always reveal a goat behind one of the unguessed doors, it doesn’t matter whether it’s 2 or 3.
If you still don’t see it (and this is what worked for me), imagine there are a million doors and you choose door 1. You have a one in a million shot of winning. The prize is located behind one of the other doors with a probability of 999,999/1,000,000. Monty knows where it is, and verifies that 999,998 of the doors other than door 1 have goats behind them by opening them. But not at random — if the prize is behind one of the 999,999 doors other than door 1 (and it is with very high probability), that will be the door Monty will not open. I think you can see that the odds of the prize being behind the door Monty didn’t open are immense, while the odds it’s behind door 1 are still one in a million.
I was fully in agreement with you based on the table above with every outcome listed, until I realised your presumption is also wrong. 3 of the outcomes are not possible as the door is opened by an outside influence. The probability of picking correct first time is 33%, but that doesn’t statistically affect the actual choice. The only choice will be between the one with the prize and one without after Monty intervenes
Hi Craig, the table shows that by switching, you have a 67% chance of winning. You end up having a choice between two doors, but that doesn’t mean the probability is 50/50, as shown in this post. Given Monty’s nonrandom process, there’s a 67% chance the prize is behind the door he offers. Think of it it this way. If there’s only a 33% chance that your initial door selection is correct, then you must have a 67% chance of winning by switching to the only other door.
This solution omits an important piece of information. Assume three doors: 1, 2, and 3. You pick 1, Monty opens 2 revealing a goat. That leaves doors 1 and 3 unknown. Doors 2 and 3 have a .67 probability of having money. But, importantly, there is a second combination, doors 1 and 2, which also have a .67 probability of having money. At this point you have two combinations, each with a .67 probability. It becomes a choice between two equally probable combinations. Effectively this is the same as a probability of .5 There is no advantage to either staying or switching.
Hi Michael, Sorry that is incorrect. You have a 67% of winning if you switch or a 33% of winning if you stay. In your scenario, door 2 is revealed and has a 0% chance of having the prize.
Hi jim, on Quora i posted the following question: “How do I convince my friend that in the monty hall problem, switching only improves your chances when the host reveals one of the doors? He says switching will always improve your chances even if the host never gives you any hint or help.” And every one that answered agreed with me and each other that my assumption is correct and my friend is wrong. Except for one person who answered. He believes my friend is right. Or atleast thats what i am interpreting from his answer. Could you tell me who right? Also if you would like to see that person who’s answer was different i will provide the link it. It would be good if you could because he seems to be a qualified individual who holds a PhD and used math and formulas in his answer. Here is the link:
https://www.quora.com/How-do-I-convince-my-friend-that-in-the-monty-hall-problem-switching-only-improves-your-chances-when-the-host-reveals-one-of-the-doors-He-says-switching-will-always-improve-your-chances-even-if-the-host-never-gives/answer/Alfredo-Sep%C3%BAlveda-Jim%C3%A9nez?ch=10&share=1d08733f&srid=oyle
I think I see why you thought I was repeatedly asking the same question. You write “You asked how we determine the probability for the contestant.” That is not the last question I asked. I asked:
Q Contestant Probability: How is the probability that the car is behind a door from the contestant’s perspective determined?
This is not the same as:
Q2 Contestant Probability: How is the probability (for the contestant) of randomly selecting the door that has the car behind it if each door is equally likely to be selected determined?
I have always agreed the answer to Q2 Contestant Probability is 1/3.
Hi Kicab, at the beginning, there is a 1/3 chance of the car being behind any of the three doors. After Monty performs his non-random process, the probabilities become 1/3 behind the contestants door and 2/3 behind Monty’s unopened door. As we discussed a length, the probability changes for Monty’s two doors because he uses a non-random selection process. Hence, that last door he offers has 2/3 chances of having the car. Check your email for a message from me. –Jim
I’m sorry I caused you such frustration. I looked at my questions and they are different and variations from my view. But I can see why from your view they look like repeated questions and I’m regressing. Regrettably, I still do not understand your answer to my last question.
Hi Kicab,
You asked how we determine the probability for the contestant. That’s simple. Three doors, one prize, and the contestant makes a random choice of doors. Hence, 1/3. That’s basic probabilities. In the future, I will be adding many articles about probability. Until that I point, I can’t refer you to one. But, it’s one random choice from three doors. That should be straightforward enough.
You don’t need to use CIs to answer your question. I use CIs in another post to illustrate how hypothesis testing can exclude 50% as the probability. But, to understand that initial determination, it’s just simpler to understand it from a basic probability standpoint. I didn’t include your very long text about CIs because it’s not a rabbit hole we need to go down to answer your question.
I’d recommend beefing up your basic probability knowledge to be able to answer your own question if my explanation isn’t working for you.
Hi Jim.
Sorry, I didn’t add more explanation as to what I actually did. After I played a few games I realized that the simulation was exactly Monty’s perspective showing the key facts about two roles of probabilities. So, I switched to creating the simulation in Excel. I’ll use your words since you say it best regarding Monty’s perspective and shown playing the game and with simulation:
Fact 4: Many-games Probability: “Monty knows that overall contestants will win 67% of the time by switching.”
Fact 5: Single-game Probability: “he’ll know for each game whether contestant has a 100% or 0% chance of winning by switching.”
Monty’s perspective is clearly objective and matches the facts, especially
Fact 6: Once the car is placed behind a door (Step 1 is completed), the car is never behind any other door during the rest of the game.
This has been my perspective from the beginning and that is why I couldn’t understand why the correct probability of winning by switching is 66 2/3% for a single game. But when you explained that it is from the contestant’s perspective, I wondered what that meant—hence, questions Q4 and Q5 (from Jan 6 comment).
What puzzles me is that you say for the contestant both (a) the probability that the car is behind a door after Step 1 and (b) the probability of selecting the door with the car behind it where each door is equally likely to be selected are 1/3. From Monty’s perspective, the probability for (a) is 0 or 1, and for (b) it is 1/3.
Q Contestant Probability: How is the probability that the car is behind a door from the contestant’s perspective determined?
Hi Kicab,
Once and for all, the same probabilities apply. Contestant wins 1/3 of the time by not switching and 2/3 of the time by switching. Monty has perfect knowledge which allows him to know the outcome sooner but the outcomes follow those same probabilities. By perspective, I just meant that the contestants don’t know where the car is but Monty does. However, the probabilities are the same. I’m not going to keep answering the same questions over and over again.
It seems like you’re regressing in your knowledge. I don’t mean to sound harsh, but you should already know the answer to your last question. There are three doors and one car. The contestant doesn’t know where the car is and picks one door at random. Hence 1 out of 3 chance.
We’re getting to the point where you’re spinning wheels now. You were so close and somehow you seem to have lost ground. I don’t have time to answer essentially the same questions (or variations of them) over and over. Apparently, I’m not able to answer them in a manner that is clear to you. Consequently, I think you’ll need to find your answer elsewhere. Again, I don’t want to seem harsh, but I’ve answered many of your questions in quite some length!
Hi Jim,
I played the game 1000 times. I set it up in Excel as it is easy to do so that with one click 1000 runs are done. I have five columns: Step 1 Car Door, Step 2 Contestant Door, Step 3 Opened Door, Step 4 Switch Door, Step 5 Result When Switched (Win/Lose).
The results for first 1000 are: 680 wins by switching; 329 times car behind door 1, 340 behind door 2, 329 behind door 3. So, considerable support for both equal distribution of the car behind the three doors and that switching wins 2/3 of the time over many games. Note that I didn’t need to run the simulation to know that the frequency of wins by switching would be about 66.7%. I read Marilyn vos Savant’s response in Parade when it was published and accepted then and since that switching wins 2/3 of the time across many games. Recall that I had replied earlier that “I have always agreed (I realize I didn’t state this—sorry) that the simulation correctly shows that if the game is played many, many times and the car is randomly placed behind a door with each door equally likely to be selected for the car, then switching at step 4 wins 2/3 of the time.”
Yet, I still have questions about the difference between probabilities that apply to everyone and probabilities that apply to certain people and not others.
Q4: What do you mean by “for the contestant” the probability is 1/3?
Q5: Is this a subjective probability and if so, subjective to what?
Hi Kicab,
I’m wanting you to play the actual game as I describe in my other post about the Monty Hall Problem. Not a simulation. I’m glad you see the correct answer is you win 2/3 of the time by switching from the simulation. But, I want you to go through the process and see it from Monty’s perspective. The fact that you’re not understanding the answers to your questions means you’re not understanding the process. Playing the game yourself should help with that.
When I say 1/3 probability, I mean that the contestant’s initial door choice has a 1/3 chance of having the car. That’s true for all.
There’s nothing subjective about this problem. The only difference is that Monty has more knowledge, which both affects the outcome and allows him to know the outcome sooner, but it doesn’t change the probabilities.
Playing the game with a friend and three cards will help you see this in action. Maybe you don’t have to play it 100 times if you’re just understanding how it appears to both Monty and the contestant. I recommend 100 times because it provides a formal hypothesis test with 93% power for rejecting the notion that switching only has a 50% chance of winning.
Thanks, Jim, but I won’t celebrate yet. I still have some questions and suggestions. Let me first recap the two solutions.
Contestant Solution: when you state that the correct solution is that the probability is 2/3 of winning by switching to the unopened door in Group 2 doors regardless of whether the car is behind that door or not, you mean only from the contestant’s viewpoint.
Q3: Would it help readers understand more if this was explicitly stated in the explanation?
Monty Solution: from Monty’s perspective the probability of the contestant winning by switching is either 1 if the car is behind the unopened door in Group 2 doors or 0 if the car is not behind the unopened door in Group 2 doors.
Q4: Why not tell readers that there are two correct solutions, depending on whose perspective you take?
Q5: But if there is only one correct solution (the Contestant Solution), why is Monty’s Solution based on perfect knowledge the incorrect one?
Hi Kicab,
No, you’re very close. And your questions reiterate what I asked you to do earlier. Play the game as I describe. Going through the process will help you intuitively answer the questions that you’re asking. I’ll answer questions below, but I won’t answer more questions until you complete your next assignment of playing the game 100 times with a friend. Record the results for always switching. During the process, note the items I discuss below. I strongly feel that you would’ve been able to answer your own questions by going through that process.
No, there are NOT different solutions for Monty and the contestant. What is different is their knowledge. After the contestant picks their initial door, Monty KNOWS 100% for sure whether the contestant will win by staying or switching. Meanwhile, the contestant does not know for sure. However, a smart contestant can calculate that they have a 2/3 probability by switching. And Monty knows that overall contestants will win 67% of the time by switching.
From Monty’s point of view about switching, he’ll know for each game whether the contestant has a 100% or 0% chance of winning by switching thanks to his perfect knowledge. However, the probabilities still apply to him. If Monty plans to runs the game 100 times (whether with one contestant or different contestants), he’d expect over the long run that contestants will win about 67 times out of 100 by switching versus only 33 times if they stick with their original door. So, it’s the same answer and the same benefits for switching.
Here’s an analogy. Imagine we have a weighted coin where heads appears 67% of the time and tails 33% of the time. The coin is flipped and the result is hidden from the contestant, who has to pick heads or tails. Meanwhile Monty observed the outcome and knows which one the contestant should choose after each flip. However, when pondering future flips, Monty knows that contestants will win twice as often if they pick heads.
Same answer. Just different amounts of information along the way.
PLAY THE GAME! You seem to be resisting that but I think it would be informative. Play it both as Monty and the contestant and you’ll see that the percentages of wins for switching is the same. The only difference along the way is that as Monty, you’ll know the outcome before the contestant.
Happy New Year, Jim!
When I read “from the contestants stand point the probability of the car being behind any given door is 1/3” I wondered why would you include “from the contestants stand point”?
That’s when I had my “Aha!” moments for your correct solution.
Aha 1: The probability the car is behind each door is 1/3 for the contestant who has no knowledge and the probabilities are different for Monty who has perfect knowledge.
Aha 2: For the contestant, the probability the car is behind the Group 1 door is 1/3 from the moment the door was selected to the end of the game. While for Monty, who has perfect knowledge, from the beginning the probability is 1 if the car is behind the selected door and 0 if it is not.
Aha 3: While the probability that the car is behind the door in Group 2 that Monty opens stays 0 from the beginning for Monty, for the contestant the probability changes from 1/3 before the door is opened to 0 after it is opened.
Aha 4: For the contestant, the probability the car is behind the Group 2 door that is not opened is 2/3. While for Monty, who has perfect knowledge, the probability that the car is behind that door is 0 if the car is not behind that door or 1 if it is.
But then I remembered that when I asked if the probability for the unopened door from Group 2 doors was the same for everyone, you said it was. Was that a typo or,
Q1: Are the probabilities for each door different for the contestant and Monty (depending on knowledge or lack thereof)?
Q2: If the probabilities are not different than what is the one probability that the car is behind the Group 1 door that applies to both people who don’t know whether the car is behind that door and people who do know?
Hi Kicab,
Happy New Year! And, this is a great way to celebrate with you understanding the Monty Hall Problem! I’m glad we could work through this together.
You’re right, understanding it from Monty’s Perspective is crucial. That’s why I always pointed out how he was using a non-random process that affects the probabilities that the contestant faces. He has perfect knowledge and probabilities (other than 0 and 1) don’t apply to him.
I must’ve misunderstood your question about probabilities for everyone. I thought you were asking about all contestants who play. All contestants have the same probabilities. But, no, as I state throughout my comments to you and in the blog post, Monty has perfect knowledge. The probabilities do not apply to him. His intentional actions based on perfect knowledge alter the final probabilities for the contestant.
It’s as if you flipped a coin but you knew the outcome. For someone without perfect knowledge, the odds are 50/50. But for someone who knew the outcome, it’s 100% for one side of the coin and 0% for the other!
I don’t get it why some people make counter arguments for this article, even after it is proved in the article itself. And it is correct of course. I too tested this in real life for 52 times (Which is enough FOR ME, in addition to seeing that the 2/3 win change when choosing to change is true) and the result was 32 wins and 20 losses ~2/3 = 66,666%.
Hi, I think some people just can’t get past the fact that there are two doors and, in their minds, it has to be 50/50. However, after understanding the process, it becomes clear that’s not the case!
I think you can probably get the gist of the solution empirically with fewer than the 100 trials I recommend. That’s my recommendation because I’m thinking in terms of a hypothesis test where I want a 90% chance (actually 93.7% statistical power) of being able to reject the notion that the probability of winning by switching is 50%. To have a ~90% chance of being able to accomplish that with a formal hypothesis test, you need about 100 trials.
Each door has a probability of 0.33 for having a prize. If we stick not to switch then probability of winning is 0.33. The combined probability of two door having the prize is 0.66 and as per the problem Monty always opens empty door hence its switching the door make 66.666…percent chances
Yep! Exactly right! 🙂
Jim, thanks for your patience. I’m sorry it appears I am not focusing on the process or don’t understand it, because I think I do. But you can confirm or tell me where I am wrong. Because I think I understand the process is why I was asking for confirmation at different points of the process.
First, here is how I saw and still do see the Monty Hall process:
Step 1: A car (the prize) is placed behind one of three doors and a goat is placed behind each of the other two doors. After the placements, Monty knows which door has the car behind it and which doors have a goat behind them [“because he has perfect knowledge”].
Step 2: A contestant who does not know which door has the car behind it selects a door, hoping to select the door with the car behind it and win the car.
Step 3: Monty, knowing which two doors have a goat behind them and which has the car behind it, deliberately and knowingly opens a door that does not have the car behind it. If the contestant selected the door with the prize behind it, Monty has a choice of two doors to open; otherwise, only one choice when the contestant selects a door with a goat behind it.
Step 4: Monty offers the contestant the option of switching to the other unopened door not selected.
Problem question: What should the contestant do?
Second, I have always agreed (I realize I didn’t state this—sorry) that the simulation correctly shows that if the game is played many, many times and the car is randomly placed behind a door with each door equally likely to be selected for the car, then switching at step 4 wins 2/3 of the time.
Third, your table of the nine sequences of the process does show that 2/3 of the time switching wins.
Fourth, it is the two sentences I asked about in my last response that I still do not understand what you mean. It may be because I may not have been clear in what I was asking. So, let me try again. At Step 2 the contestant will select a door. Let’s assume that the pick is random with equal probability of picking each door.
Question 3: Do you see a difference between
(a) the probability a door has a car behind it is 1/3
(b) the probability of picking the door that contains the car is 1/3?
If yes, why? If no, why not? Thanks.
Hi Kicab,
Ok, that’s great, thanks for writing that out. I just wanted to be sure that we are working from the same page.
There’s just one thing I’d add to your process, which dovetails nicely with your question.
Part of the process is that there are two groups of doors in this game that are created when the contestant picks the initial door. Group one contains the contestant’s single door, which has a 1/3 chance of having the car. And Monty’s group of two doors, which has a 2/3 chance of having the car. We know that there is a zero probability that Monty will open a door with the car. Consequently, if his initial set of two doors contains the car (2/3), the door with car will 100% be Monty’s unopened door (2/3). Given this inevitability, the 2/3 probability for the set of two doors still applies to Monty’s single unopened door after he opens the other one. I think that’s just the final step of understanding you need to understand the solution.
So for your Question 3, yes, from the contestants stand point the probability of the car being behind any given door is 1/3. And their chances of picking the the correct door is also 1/3. And, as mentioned, that means the other two doors combined have a 2/3 probability. Given Monty’s non-random actions, that 2/3 probability is fully transferred to his final unopened door. Again, if either of his two doors have the car, that door WILL definitely, positively be the door that he let’s you switch to at the end. I think it’s just that last portion you’re having difficulty with.
Keep in mind that it’s Monty’s perfect knowledge combined with the rule of never opening a door with a car that confers the 2/3 probability to the his single, unopened door at the end.
If you want to understand this more intuitively, I highly recommend that you play this game with a friend using cards as I write about in my other post about the Monty Hall Problem. In that game, be Monty. I think that you’ll find that buy playing the game and seeing it from Monty’s view that if his doors contain the car that his knowledge and rules constrain him to opening a particular door and forcing the car to be behind his final door. So, if you don’t fully understand it from what I’ve written, that’s what I want you to do next. Set aside an hour or so, get a friend, play it out 100 times, and record the results for always switching. I did this with my daughter and she caught on!
I think by seeing it from Monty’s point of view, it’ll become clear!
Happy New Year!
Jim, thanks again for the details. I understood your explanation but am confused by these two statements:
1. “Any door that you pick has a 1/3 chance of containing the prize.”
2. “the door he opens by definition has zero probability from the beginning [of having the car behind it].”
I will state my confusion as a deductive argument so you can tell me which premises are false and which inferences are invalid, and why.
Deduction
Step 1 (Premise 1): “the door he [Monty] opens by definition has zero probability from the beginning” of having the car behind it.
Step 2 (Inference 1): Therefore, from the beginning the probability of winning by selecting that door was 0.
Step 3 (Inference 2): Therefore, from the beginning not “Any door that you pick has a 1/3 chance of containing the prize [car behind the door].” In particular, the door Monty opens did not have a 1/3 chance of containing the prize from the beginning.
Where does the logic fail and why?
Hi Kicab,
Your sequence starts at the wrong point. The game begins with the random placement of the car behind a door. There are three doors and the contestant can pick one at random. Consequently, the contestant has a 1/3 chance of picking the correct door at the outset because he has absolutely no information about where the car is located.
The next step is when Monty opens a door. I’ll answer you question with a question. Assume you’re Monty. You have perfect knowledge about where the car is located. You also strictly adhere to your rule of not opening a door that has the car. Given those two assumptions, what are the chances that you’ll open the door with the car?
So, yes, both are true. The contestant has a 1/3 chance of picking the door that contains the car at the start of the game. And then Monty has a zero probability of opening the door with the car. Monty is really operating outside the realm of probability because there’s no chance involved because he has perfect knowledge.
Now, it seems like you’ve been spinning your wheels thinking about in your own way. And, I also want to be sure that you are taking in my comments. So, before I answer any more questions from you, I’ve got an assignment for you! I want you to write in your own words a summary of Monty’s process. You can refer back to my previous replies to you. But, I want you to focus on Monty’s process, how it ensures that if one of his two doors has the car, that door becomes the last unopened door. And, how switching flips the outcome compared to staying with your original door. Focus on the process and then work in the probabilities as needed. This is all stuff I’ve covered in my previous replies to you.
So write that up so I can see that you’re absorbing and at least understanding my comments to you. You haven’t touched on my comments at all to show that you’re taking it onboard and I want to be sure before spending more time answering endless questions from you. And, again, you’re spinning your wheels trying to come at it from that single angle you’re fixed on. I think working on it from a process standpoint first (maybe putting yourself in Monty’s shoes would be helpful) and then adding in the probabilities will help you understand it from another standpoint. Your most recent questions show me that you don’t understand the process, which is making it hard for you to understand the probabilities. So, focus on the process first and then I think the probabilities will come to you more easily.
After you do that, I’d be willing to answer more questions if you have any.
Hi Jim,
I just posted a comment and then read through the other comments and I feel like you might have been really close to explaining that plants prefer water over Brawndo (because you can talk to them). I hope you get that reference from Idiocracy, if not go watch it, I have a feeling you will truly appreciate that movie!
I enjoyed reading this article, I was looking for a good explanation to share with my family and friends who want to be “Monty Hall Problem Deniers”. I understand why someone without a math or science background might be confused, but I really don’t understand how educated people have a hard time understanding this. It’s not that hard. It didn’t make immediate sense to me, but I understood that the difference between a 50%/50% chance ( which is what most people believe the answer to be at first glance), and a 33%/66% chance are large enough to prove in a relatively small simulation. I ran 100 tests in Excel and sure enough, 33%/66% +/- 1%. Then I did 1,000 tests in Excel and the results were nearing perfect in accordance with the law of large numbers. My point is, if such a simpleton like me can figure it out, why would it stump a statistician? Surely they would see that they can just run a simulation to prove the answer. Whether you call it the law of large numbers or the scientific method, we are taught methods to test our theories. I was appalled when I found an article from a Math Institute (written in 2015) that was arguing for the wrong side of the problem.
https://ima.org.uk/4552/dont-switch-mathematicians-answer-monty-hall-problem-wrong/
My aha moment came when constructing the simple test in excel. I knew what the results were going to be before I ran the tests. I was thinking of how many columns I will need when it dawned on me, only two. The first door you pick and the winning door. The door Monty opens is irrelevant when making the simulation. Either the first door you picked is the winner, or you would win by switching doors. When you understand that statement and what it means, it is easy to see that your odds are 33%/66%.
Hi Joe,
Thanks so much for you nice comment, and for recommending Idiocracy. I haven’t seen but I’ll check it out!
Well that’s one thing I don’t understand from the deniers. It’s not hard to simulate the process and, run it many times and observe the results, never mind the probability calculations! In my other post about the Monty Hall Problem, I run both a computer simulation and a physical simulation using cards and both confirm the results. Although, I was convinced by the math. I also agree that understanding that switching doors causes you to flip the outcome is a key point. And, you have a 2/3 chance of picking the wrong door (losing) at the outset, which means you have a 2/3 chance of winning by switching!
I haven’t read the article you link too. I’ll have to check it out but I’m shocked that a Math Institute would write something like that!
Thanks again and Happy New Year!
I wrote to you last night about being able to explain the solution to another using your section above. This morning I did so with my wife. She had heard of the problem before but had forgotten the solution. She said it was the best explanation she had ever heard. I laughed and said that that’s I wrote the guy last night.
Hi, that so makes my day! Thanks for letting me know! Now you just need Monty Hall to offer you a door to switch to! 🙂
Your “Here’s how it works” section is, by far, the best simple explanation of this problem I’ve seen. I think I have a grasp enough of it to explain to another person. Thanks!
Thanks for the detailed explanation, Jim.
I understand what you wrote so I must be misunderstanding something else. There are two items where that might have occurred.
Item 1. I have understood that after a door is opened revealing no car behind it, “the contestant has a 2/3 chance of winning by switching” as meaning that the probability the car is behind the door the contestant can switch to, is 2/3.
Question 1a. If this is not correct, what is the probability the car is behind the door the contestant can switch to?
Question 1b. Is this probability from 1a the same for everyone, including for example those who think the probability is 1/2?
Item 2. The probability of winning by selecting the door that was opened after it was opened is 0, of course. You state that before the door is opened the probability of winning by selecting that door is 1/3.
Question 2. Why does the probability change from 1/3 to 0 by opening the door? I see a couple of possible reasons but want to make sure I understand. For example, it could be because of additional information or it could be because in fact the car was not behind the door or some other reason.
Hi Kicab,
1a. You are correct. The probability that the car is behind the door that the contestant can switch to is 2/3.
1b. This probability is the same for everyone. Believing that the probability is 1/2 does not change the underlying reality. This illustrates how knowledge is power! As you gain more information and understand how a process/system works, you can better predict the outcomes using accurate probabilities.
2. When the game starts, we have no information other than that the car was placed randomly behind one of the doors. Any door that you pick has a 1/3 chance of containing the prize. This is the only truly random part of the process.
There are two sets of doors in this game. The door you choose, which has a 1/3 chance. And Monty’s two doors, which collectively has a 2/3 chance of having the car. At this point, Monty makes a non-random choice to open a door that does not have the car. There’s zero probability that the door he opens has the prize because he has perfect knowledge and yet his set of two doors still has a 2/3 chance because that was the random part of the process at the beginning. Consequently, we know that there is a 2/3 chance the car is behind Monty’s unopened door.
So, the probability really isn’t changing from 1/3 to 0. Monty’s knowledge is perfect so the door he opens by definition has zero probability from the beginning.
First, I apologize for my tone and attitude. That was not my intention. But there is no reason to be that way. I’m sorry.
Second, I don’t believe my first comment had that attitude or tone but it’s not always easy to see things from others’ perspectives.
Third, my comments are not opinions but facts.
Let’s start the discussion and I will ask questions so I can better understand where my errors are. I think these are three facts that determine the solution.
Fact 1: There are nine possible combinations of door for the car and door for the contestant before the game starts.
Fact 2: There are 3 combinations are possible after the car is placed behind a door. If that is not a fact, why not?
Fact 3: There is one combination possible after the car has been placed behind a door and the contestant selects a door. If that is not a fact, why not?
Hi Kicab,
I’m glad we can move forward more constructively. What you need to realize is that the Monty Hall Problem is solved mathematically and there is one correct solution. The mathematically correct solution is that you win 2/3 of the time by switching. Additionally, computer simulations have played this game hundreds of thousands of times and have confirmed that mathematical solution. You say that your comments are “facts.” However, you need to realize that you’re essentially arguing that 2 + 2 = 5 because you’re disagreeing with the mathematically correct solution. Instead of digging in on the wrong answer, I’d recommend opening your mind and trying to understand how the correct solutions works.
I’ll get to your questions in just a minute. But, I’d highly recommend trying to understand the two explanations I provided in my previous reply to you. You haven’t discussed them at all and I’m pretty sure that you have not tried working through them. Take a moment to consider them. It shows you the HOW and WHY you win more often by switching. Right now you’re stuck spinning your wheels by coming at it from one direction. Try understanding the solution using the other explanations I mention.
You seem to be misunderstanding the table. For your first “fact,” you’re a bit off. The table does NOT represent 9 possible combinations of “the car and door before the game starts.” Instead, the table lists out all possible sequences and possible outcomes. All of these outcomes can happen. There are 9 possible sequences the game can take. These incorporate the initial placement of the prize, your initial choice, the door that Monty opens, and you’re decision to stay or switch. All of those factors boil down to the 9 possible sequences and outcomes in that table.
Given the placement of the car and your initial door choice, the last two columns indicate the outcomes when you stay or switch. As I mentioned, notice how switching ALWAYS flips the outcome. If you win by staying, then you lose by switching. If you lose by staying, then you win by switching. The table makes it clear that you win twice as often by switching doors (6 times vs 3).
For your fact 2 and 3, the car can be placed behind one of the three doors. And, yes, after the contestant picks a door, the contestant either has picked the correct door or not–although the contestant does not know which condition is correct at this point. However, using simple probability, the chances of picking the correct door initially is only 1/3. Right there you know that the probability of winning by sticking with that initial door can only be 1/3. However, it doesn’t end there because Monty offers you the option of switching doors.
Now, you need to take the next step and understand the how and why it works. Upon the initial selection, the contestant only has a 1/3 chance of winning if they stay with their original door. However, switching will flip the outcome, which means the contestant has a 2/3 chance of winning by switching. That’s what I mean by trying to understand what I wrote about how switching flips the outcome. And you need to understand Monty’s process whereby if the prize is behind one of his two doors (which happens 2/3 of the time), then that door ultimately becomes Monty’s unopened door to which he lets you switch.
The real trick to this problem is understanding that Monty has full knowledge and bases his actions on that knowledge. When you understand that his behavior is not random but based on his insider information and the rule that he will never reveal the prize, you can use that understanding to improve your chances of winning. You seem to be missing that entire angle.
Finally, if you aren’t connecting with the math and logic of how it works, I recommend you try it empirically. Read my other post on Revisiting the Monty Hall Problem. I show a computer simulation of the outcome. Additionally, I perform an empirical experiment that confirms the results. If nothing else convinces you, grab a friend and play the game 100 times as I describe in that post. One of you is Monty and the other is the contestant who always switches. Record the win/loss outcomes. You’ll see how the winning percentage converges on 66.6% when you always switch.
I have tried posting four responses but they fail to show. Here’s number five. I believe it’s because they show why the solution of 2/3 probability of winning by switching is wrong. The explanations have several fatal mistakes. Here is one.
Mistake 1. Calculating the probabilities using all possible combinations. There are nine possible combinations of door for the car and door for the contestant. However, there are only three possible combinations after the car door has been selected and there is only one possible combination after the contestant selects door. Hence, calculation of the probability of winning by switching using all nine combinations when only one combination exists is a mistake. This is easily seen by looking at the column of Win/Lose for each combination. There are no 1/3s or 2/3s or any number. Win or Lose means with probability 1.
Hello Kicab,
I’ve seen your comments, and no, I haven’t approved them because of your attitude and tone. I’m okay with differences of opinion. However, this is a mathematical problem with a correct solution. There’s no room for your opinion on it! You come in here claiming there many “fatal mistakes.” Next time, try coming here with some humility and a mindset for trying to understand. Yes, this problem is counter-intuitive. That’s why I call it a statistical illusion.
Did you really think that you found something that one else did over the past several decades? Statisticians have looked at this, and we’re unanimous in that switching doubles your odds. But, apparently, you found something we all missed?! And think about it a bit. Computers can simulate this game and play it fairly. And guess what. Simulations that play the game hundreds of thousands of times always show that you double your chances of winning by switching. Explain that!
So, even if you don’t believe the math (which is conclusive), the computer simulations also come to the same conclusion. In another of my posts about the Monty Hall Problem, I show one such computer simulation. I also conduct an empirical experiment and play the game 100 times. You can try them yourself.
About your comment relating to the 9 rows in the table. It’s straightforward. There are 9 possible combinations of choices and outcomes. As you can see, of these 9 outcomes, you win 3 times when you don’t switch and 6 times when you do switch. Using elementary math (6 / 3 = 2), you find that you double your chances of winning by switching.
I’m going to provide the two simplest explanations I can think of to explain it to you.
1) As you can see, when you switch, it causes the opposite outcome. For example, if you pick door 1 and the prize is behind door 1, you win if you stay but lose if you switch. At the beginning of the game, you have a 2/3 chance of picking the incorrect door. That initial door you choose probably does not have the prize. Consequently, switching doors reverses that outcome. In this case, having an initial 2/3 chance of losing means that switching gives you a 2/3 chance of winning. That’s about as simple as I can put it.
2) For a slightly more complex reason, consider that Monty uses a non-random process. That process isn’t obvious with only 3 doors but becomes apparent with more doors. Imagine that there are 101 doors and 1 prize. You pick one door, and that leaves 100 doors for Monty. It’s almost guaranteed that the prize is behind one of Monty’s doors (100/101). Next, Monty systematically opens 99 of his 100 doors. During this process, he carefully does not reveal the prize. He gets to his final door, which he does not open, and offers you the chance to switch. Do you? If you do switch, you’ll increase your chances of winning by 100 times!
The trick is understanding that Monty intentionally avoids opening the door with the prize. With his 100 doors, the prize is almost certainly behind one of them. Using his rules, he cannot open that door. Hence, it becomes the last, unopened door at the end of his process.
The same thing applies to 3 doors, but Monty’s intentional selection, non-random process is less obvious because he only opens one door.
If you have genuine questions and an open mindset for learning, I’ll discuss them with you. But if you come here claiming everyone else is wrong and that there are fatal mistakes, I will not discuss it with that type of tone.
Hi, Keegan Hall
Sorry, but I must tell you that you are wrong. If the other person (the host) randomly reveals one door from the other two and just by chance it results to have a goat, then the chance to have the car is 1/2 for each remaining door, no longer 1/3 vs 2/3.
Firstly, you can see it in an intuitive way: If the other person is doing it without knowledge, then it makes no difference if you (the player) are who also makes the revelation, because both are doing it randomly. For example, you could pick door 1 and then decide to reveal door 3. But in this way what you are doing is basically selecting which two doors will remain closed, I mean, it is the same as if you said: “I want to keep closed doors 1 and 2”, and then you revealed door 3. Now, if door number 3 results to have a goat, which one do you think is more likely to have the car, door 1 or door 2? According to your reasoning, since you declared door 1 first, it must be only 1/3 likely because it is the equivalent to the staying door. But what about if you had said it in the opposite order: “I want to keep closed doors 2 and 1”? Does it seem logical that door 1 has 2/3 probabilities now? So does the order in which you declare them determine which is more likely to have the car? The doors don’t know that order.
You can also visualize it in the frequentist way. If you played the game 900 times, in the first selection you would get the door that hides each content in about 1/3 of them, so about 300 times goat1, about 300 goat2, and about 300 the car. Then, since the host will randomly open a door from the rest, he has 1/2 chance to reveal each of the other two remaining contents. The possible cases are:
1) In 300 games you pick the door that hides the car.
1.1) In 150 of them the host reveals goat1.
1.2) In 150 of them the host reveals goat2.
2) In 300 games you pick the door that hides goat1.
2.1) In 150 of them the host reveals goat2.
2.2) In 150 of them the host reveals the car.
3) In 300 games you pick the door that hides goat2.
3.1) In 150 of them the host reveals goat1.
3.2) In 150 of them the host reveals the car.
So, if a goat results to be revealed, you know you are not in case 2.2) and neither in 3.2). That restricts you to a subset of 600 games. You win by staying in 300 (cases 1.1) and 1.2) and by switching in 300 (cases 2.1) and 3.1), so each strategy wins 1/2 of the time of this subset of 600 games.
This is different as if the host had known the positions and followed the rule of always revealing a goat. In all the 300 games of case 2) he would have revealed the other goat, and the same in case 3), so you would win by switching in 600 games (case 2) and 3) that are 2/3 of the total 900, and by staying in 300 games (cases 1.1) and 1.2), that are 1/3 of the total 900.
Thx for your reply. Yeah, the thing about being suspicious makes a good point. The MHP is interesting because it is an exercise on how the ‘correct’ answer disgrees with our intuition.
However, our intuition is based on real-life experiences of being offered alternatives. Why are am being offered a change?
I guess what I am saying is that the MHP is disappointing to the extent that part of the trick of the problem is that the assumption for the problem (host behaviour) is different than real world behaviour that forms our intuition.
Of course the MHP is interesting b/c of the math.
I just find the problem flawed.
Hi Peter,
I guess I’d have a different take on it. I absolutely love the Monty Hall Problem for several reasons.
One reason is that it symbolizes life for me. This problem contains a mixture of luck, process knowledge, and information. By understanding the process and deducing what Monty knows, you can use that information to make an informed choice that increases your chances of winning. Yet, there is still luck involved. So you can make the correct choices that maximize your chances of winning but you can still lose. It’s life in a nutshell. Those who understand the system and use the available information can improve their expected outcomes. I also appreciate the psychological factors that make this solution feel intuitively wrong.
For me, I don’t see the assumptions as flawed. Probability questions always have assumptions that are necessary to calculate an answer. These are just the assumptions for this given problem. Of course, if you change the assumptions, you’ll change the correct answers. Given the assumptions as stated, the best approach is to switch doors.
What has always bugged me about the Monty Hall problem is that it makes certain assumptions that make the math work.
Some of the assumptions we can agree on. For example Monty will never reveal the prize. (which incidentally implies that Monty knows what is behind the doors)
However, it is also assumed that Monty _must_ reveal a door. Why is this important?
Well, Monty can decide which he reveals a door, he may do so conditionally based on his knowledge of what is behind the door.
So what?
Well, Monty’s behaviour can range from ‘evil monty’ (only offers a switch if you originally chose correctly), ‘nice monty’ (only offers a switch if you originally chose wrong), game theory equilabrium monty (offers choice randomly 2/3 of the time you chose correctly, 1/3 of the time you chose wrongly) etc.
Based on Monty’s behaviour, the correct action is to not switch, switch, switch/not (equally) respectively.
Monty Hall himself has said that a choice was not always offered.
To make my point in another way.
We bet $1000 dollars on the colour of a randomly chosen card. You say red. I look at the card.
_Then_ I say. I will let you switch if you want.
What do you do?
Hi Peter,
I agree that Monty can behave in different ways. But the assumptions for the classic Monty Hall problem is that he’ll always open a door, he’ll never open a door showing the prize, and he’ll always offer the chance to switch. Those are the assumptions for this puzzle. Of course, if you change the assumptions, the answers changes.
I remember as a kid that I’d occasionally watch the game show. It was usually when I was home sick from school. Whenever Monty offered the chance to switch, I did think he was trying to trick them into unwittingly giving up the prize! I was suspicious! However, at least for this puzzle, we’re assuming he always offers the chance to switch.
He does not “remove” a door from the equation, figuratively he does by eliminating something you would logically choose but he doesn’t take a door out of the problem. I think that may be where you’re getting confused
Intent has absolutely NOTHING to do with the Monty hall problem if he still opens a door you have not selected that has a goat behind it. Okay if you pick a door yeah it has a 1/3 chance of being right and the other two doors have a collective 2/3 chance of being right. Now, he opens one of the 2/3 doors and there is a goat behind it; now right here I want you to imagine that you can still choose the open door okay… so you know that the two doors that you have not chosen have a collective 2/3 chance of having a car, but you don’t want to choose the open door because you know it’s a loser. BOTH doors still have a 2/3 chance of being right (if you ignore the fact that you can clearly see a goat) you just don’t want to choose the open one.
The problem works in real life, as long as the 2/3 goat door is opened it doesn’t matter if it’s a video game; a host opening it accidentally; or your friend running up on stage, the problem always works!
Because you as the person who picked the door would NEVER choose the one that you originally selected. Your friend on the other hand would IF he doesn’t know who one you picked; if he saw you pick a door I would still be 2/3 vs 1/3.
The fact that Monty knows the location of the car is precisely what leads to the results not being 50/50.
When he chooses door one, the contestant knows the odds are 1/3 that the car is behind door one, and 2/3 that the car is behind either door two or door three. The contestant already knows that at least one of the doors two or three has a goat behind it, so Monty revealing that to be the case doesn’t tell the contestant anything that changes the probabilities.
A good way to more intuitively see this is to imagine the problem with, say, a million doors. If the contestant picks door one, he has only a one in a million chance of being right. Out of the remaining doors at least 999,998 have goats behind them — and Monty knows which those may be, so can always open those to show the goats. But if you’re the contestant, and you know Monty knows where things are, this doesn’t really tell you anything you didn’t already know. Door one still has only a one in a million chance of having the car, and the odds are 999,999 in a million it’s elsewhere. That Monty knows which doors don’t have the car behind them doesn’t change that.
Another way to think about this is not to confuse it with the case of the contestant accidentally discovering that a goat is behind door number three. Suppose the contestant accidentally heard the goat make a noise behind the door — a random event rather than one Monty planned. Then you’d be right to think the car has an equal chance of being behind door one or two. But Monty doesn’t open door three as a random event that might reveal a goat or reveal a car — if the car had been behind door three, he would have opened door two instead. All he’s doing is affirming what was already known to the contestant when he picked door number one and initially computed the odds: that at least one of the remaining doors has a goat behind it.
Here is the problem I have with the Monty Hall problem. It is flawed. Yes the probability rules show an increase in chance of winning if you change doors, however if this is a real world situation Monty Hall knows where the car is.
Basically at the start you have a 33.3% chance of selecting the correct door. Monty Hall opens a wrong door. Meaning 1 door will have the car, and the other door will have nothing. So in the real world situation it is 50/50 if the car is behind your door or not. The door Monty Hall opens should not be factored into the probability as he knows it contains nothing.
This question is an example of the model not being a true reflection of observable results.
I think that’s a very smart question. The answer is that the knowledge of which door you picked is essential to getting the best odds. Like the Monty Hall problem itself, it becomes more intuitive when you try it with more doors. With a little analysis, it is obvious that with the rules you outlined, Ted has a 50/50 chance of choosing the correct door, no matter how many original doors there were. Whether we started with 3 doors or 1000, for Ted it comes down in the end to a random choice between two doors.
It’s not a paradox, and it’s a little easier to demonstrate why if we take Ted out of the equation, and use 1000 doors. So you pick a door, and then Monty reveals 998 doors without a car. Regardless of which one you picked, one of the two doors that Monty did not reveal has the car. So now we have a new rule that you must determine which of the remaining two closed doors to open by coin flip (this is exactly equivalent to the role of Ted). Monty gave you critical knowledge by opening those 998 doors: either you picked the correct door initially–very unlikely–or the car is behind the other door. The coin flip is forcing you to disregard this knowledge. It’s not a paradox that you have a different probability of success now, because we have significantly altered the game, so we can expect the probability calculations to be different as well.
I like this question because it really gets to the heart of why the Monty Hall problem is so counterintuitive: most people who have trouble with the solution do so because it’s hard to grasp how the scenario is different from a simple 50/50 choice. Once you overcome that hurdle, you really understand the problem, and have gained an insight into probability theory.
There’s no paradox. You have better information than Ted. Thus, you are more likely to select the correct box. The probability that any given box contains the car doesn’t vary, it’s either 1 or 0. What does vary is the probability that you or Ted will select the correct box.
I think the issue here is that your adding a piece to the Monty problem that isn’t stated. The Monty problem only factors from Ted’s friend, the original “contestant”, perspective of 1 in 3 doors. The variables can and will likely change when adding or subtracting to the original stated position.
I have three questions:
1. What is the difference between a non-random process and a random process?
2. How do non-random processes get probabilities assigned to the outcomes?
3. All processes have steps. In the Monty Hall game, what are the specific steps of the non-random process (to which you reference)?
I helped people understand this problem by changing the problem to 10 doors. You pick 1 door, Monty then shows you 8 empty doors, should you switch? Pretty obvious now. 🙂
Hi Randy,
I agree with that. Somewhere way back in these comments another reader and I discuss this approach. It seems to work better with more doors because it emphasis that there is a non-random process at work, which affects the probabilities. That process is still present with 3 doors but its implications are not as obvious.
I’m not sure if it’s clear, but my comments posted on 10/3 were a response to the “monkey wrench” brought up by Mike in his post of 8/29.
The probabilities are still one third and two thirds, but Ted doesn’t have the same amount of information that you do and therefore can’t correctly compute the probabilities. In particular, Ted doesn’t know what door you picked. Let’s suppose you picked door one, after which Monty opened door three to show it had no prize. If Ted knew you picked door one, and knows Monty’s choices in opening a non-prize door had to be either door two or door three, his analysis would be the same as yours. But for all Ted knows you picked door two. In that case, door one would have the two thirds chance of having the prize. Without knowing which door you picked, he can only guess which door, one or two, has the advantage. Seeing the problem through Ted’s eyes, it’s a toss-up. But we know more than Ted since we know you picked door one, and that the real probabilities are one third for door one, and two thirds for door two.
oops, should be “and that 1 minute elapses in 1/2 mile”…
It’s 50-50 once we know that Monty is going to open a door with a goat (that trashes randomness). As someone replied (paraphrasing) since we know from the get-go that Monty will always open a dud door, then we should know that the 2 doors we didn’t choose are ‘linked’ doors, and in effect act as a single door…
This reminds me another classic teaser where: if a car goes 30mph for 1/2 mile, how fast does the car have to go to average 60mph for 1 mile? The convoluted answer is that 30mph=2 miles/minute and that 1 minute elapses in 1 mile, while 60mph=1 mile/minute, whereby no more time is available to ‘average’ 60mph across entire 1 mile, thus goofy answer is infinite. Hmmm, every cop I’ve run this past said tell that to the judge…
Hi Matt, I’m pretty sure you meant it’s 33/67 because of the linked doors. It’s because it’s not random that you’re twice as likely to win by switching.
I don’t see here problem at all. It is only to options here. 1 option the door you chose- 33.3333% chance. 2 option the two other doors 66.6666% chance, the two doors works like one because one is open. Is nothing to even thing about. Just switch and take the 66.666% chance
Yes, exactly!
Here is my issue with the question that tripped me up, while initially got that Monty would use his knowledge to not pick the door with the car behind, i actually never thought it would get to that point and here is why. No where in the problem does it state that Monty will always offer to switch doors, my brain actually said that he will only offer to switch when i have choosen the right door and he knows it.
So then he would wait for me to choose, if i choose correctly he will then make the offer for me to switch. If i do i lose, if i don’t i win.
If on my first choice i don’t pick the car he stays silent. And so in this situation i lose because he never gives me the option to switch.
I can see the logic, thought not stated that if he choosing it is logical he would not choose the door with the car behind it. What i cannot see(agree with) is him always offering the option to switch because it is not stated he always offers it, and thus my logical assumption was he would only offer it when i have choosen correctly and he wants another chance at making me choose incorrect.
Is there a fault in my logic i am missing somewhere?
Whether Monte knows or not doesn’t change the odds, as long as he opens a door that doesn’t contain the prize. The odds are still 2/3 that it is behind the one he didn’t open.
Hi, Mike.
You must distinguish two different things: one is the frequency with which each option will be the correct one, and another is the frequency with which a player selects the correct one. The switching door is the correct option 2/3 of the time, and the staying one is only 1/3. So, if you are the original player who knows which option is which and you always decide to stick with your first option, you will win 1/3 of the time.
Instead, a person that does not know which door is which cannot force his selection to always be the switching door. That person has 1/2 probability to select each. If played several times, about half of them would select the switching door, and the staying door in the other half, so that person ends winning 50% of the time but not because each door wins 50% of the time, but because the advantage that gets with one is compensated with the disadvantage that gets with the other. They average:
1/2 * 2/3 + 1/2 * 1/3
= 1/2 * (2/3 + 1/3)
= 1/2.
To put another more clear example: imagine a psychopath raped lots of persons and forced them to play this macabre game to survive. Each person will have two bottles, one green and the other red, where the green one always contains a deadly poison and the red one always contains a natural juice. Each person is forced to drink one of of those bottles, but has no clue about which has the poison, so the person has 50% chance to select either and so 50% chance to survive. No one can see what other person selected, so each selection will be completely random and only about half of the total persons will survive. But that does not mean that the green bottle is killing 50% of the persons that drink it. It kills 100% of the persons that drink it, only that about half of the total population selects it:
1/2 * 100% + 1/2 * 0%
= 1/2
Okay, thank you for the great explanation. I just have one more monkey wrench to throw in. Let’s say you go through the first round with Monty. You’ve picked your door and he’s taken one away. You decide 5o stick with the door you got. Great. Now, they bring your friend Ted in from offstage, and also ask him to pick the door with the car. Ted has no idea what door you picked or what door has been left over by Monty. There are two doors, one with the car and one without. Does Ted have a 50/50 chance of picking the right door? Won’t he pick your door half the time and be right half the time? Can one door still represent a 33% chance of being right and the other a 66% of being right, bur because Ted has no knowledge of what transpired, he we still win the car half the time? Hopefully you can see the paradox I am trying to draw out. How can Ted be a winner half the time but you only be a winner a third of the time if he’s just as likely to pick your door as the one with the car behind it?
i agree
The colour on square A and B is the same, but by having them adjacent to in case A a lighter colour it looks darker and in case B it’s adjacent to a darker colour and that makes it look lighter.
Or in short, don’t believe you’re lying eyes, trust what you can measure or calculate.
I see it this way…
Let’s say:
Door 1 = Donkey
Door 2 = Donkey
Door 3 = Car
If you choose door 1 he will be FORCED to open door 2. He can’t open 1 since you picked it and can’t open 3 since it has the prize.
You switch = you win
You don’t = you loose
If you choose door 2 he will be FORCED to open door 1. He can’t open 2 since you picked it and can’t open 3 since it has the prize.
You switch = you win
You don’t = you loose
If you choose door 3 he won’t be forced, he will have a choice between door 1 and 2…
You switch = you loose
You don’t = you win.
If you picked a donkey with your first pick and then switched, you 100% won the prize. Since there is 2 donkeys and 1 prize, the chance of you picking a donkey is 2/3. Meaning that you obviously have a higher chance of winning the price when switching.
This was super hard for my little brain to understand. Lol
Hi Dylan, that’s a great explanation of how it works!
The problem never states Monty Hall’s reasoning as to which door he opens. Since the problem does not state that Monty Hall opens a door that he knows does not have the prize, a person could assume Monty Hall was just opening any random door. In that case, there is a one out of three chance the door he opens has the prize behind it. If Monty Hall opens a random door and it happens to be one of the doors with no prize behind it, the contestant can not increase his chances to win the prize by switching choices. For the problem to work, Monty Hall has to know which door has the prize behind it and the contestant has to know Monty would not open the door with the prize behind it.
Hi Stephen,
It’s a reasonable assumption that Monty knows the location of the prize and won’t randomly open a door that could contain the prize. Yes, that’s part of the trickiness of the solution!
You’re correct, if Monty did open the doors randomly, the contestant doesn’t gain by switching. However, 1/3 of the time Monty would open the door with the prize. That just didn’t happen on the show and it would nullify the whole contest.
For your last part, for the problem to work, only Monty needs to know which door contains the prize. The contestant doesn’t have to know that Monty knows. It’s Monty’s knowledge and non-random choice that drives the probabilities in this scenario. The contestant’s knowledge only determine whether the contestant can take advantage of the process to increase their chances of winning. If the contestant doesn’t know, there’s still an advantage to switching, but the contestant won’t be able to take advantage of it knowingly. It would be like flipping a coin that has a 2/3 chance of heads and 1/3 chance of tails. If you knew, you’d pick heads. However, if you didn’t know, it’s still advantageous to pick heads, you just wouldn’t know that was the better choice!
The difficulty in understanding of this problem is that you end up focusing on the final choice, not the choice that brought you there (some life lesson etc… etc…)
i.e. Whatever I choose initially, I have a 50:50 shot of having it right by the time I get to choose to either hold or switch.
The crux is that you are more likely to choose a goat with your first choice (2/3) (the other goat always removed after) and therefore more likely to need to switch to win, due to that first choice disadvantage.
This used to bake my noodle, then I got real drunk and it made perfect sense, then I signed up to some weird website to post a comment. You people are weird, I like it.
This is correct. I have the same conclusion too.
I’m confused. Brain hurts, but I love it.
Scratch the 100 door scenario. It’ll help me visualize better if it’s three doors. Everything else is the same.
If it’s a three card game and the second contestant is brought out after a goat is revealed and then the contestant is told to pick a door are his chances at that point 50/50? At that point I see the contestant basically playing a new game independent of the “switch or stay game” which he knows nothing about. Now it’s just pick one of two cards.
Hi Eric,
I’m glad you love it even though your brain hurts! I think those are some of the best puzzles!
In this case, you have to remember that what the contestant thinks is irrelevant to the outcome. The important part is understanding Monty’s process–and that hasn’t changed. The contestant might have a misunderstanding of that process, which doesn’t affect the true probabilities, but it will affect the contestants decision making. So, yes, he is left with two cards and he might think the odds are 50/50, but in actuality they are still 33% and 67%. He might not think it’s to his advantage to switch, but it still is. You have to understand the true underlying process to have the best chances.
That really ties into the larger picture of using accurate data to make decisions–which is why I love statistics. You use the best methodologies to collect the best data available, apply the correct analyses, and that gives you the best chances of making the correct decision. If you have incorrect data, that can lead you to making bad decisions. And that’s the situation the new contestant is in if he thinks it’s 50/50. He might make a poor decision because of bad data.
The universe doesn’t behave as you think it ought to behave. It behaves as it actually behaves and it’s up to you understand it correctly and make decisions accordingly.
I really like your explanation – and there are plenty out there. Plus there are few enough outcomes that you can easily build a table, which you did, illustrating all possible choice and results. You can’t refute that. Similar to your last poster, I read you could perform the same exercise with a deck of cards. You could take the deck of 52 and ask somebody to pick a card. What are the odds they picked the Ace of Diamonds? 1 in 52. What are the odds you hold the Ace in the remaining stack? 51 in 52. I could examine my stack of cards and start flipping over all cards that are not the Ace of Diamonds. No matter how many cards I flip over and show you, your 1 in 52 odds of picking it do not change….even if I flipped over all but the last one in the deck.
Hi Bob,
I think using a deck of cards is a great way to illustrate that principle! And, I do think the illusion falls apart when you add more doors, or more cards.
Hi Jim
Could you comment on the following:
Imagine a variation on the Monty Hall Game where everything is the same except that the game is played with 100 doors and Monty opens 98 doors revealing all goats (the statistical illusion collapses with larger numbers) and right after the contestant is asked to “stay or switch” doors he leaves the studio and a new contestant walks out from an isolation booth. The new Second contestant is completely ignorant of everything that has happened between Monty and the first contestant. The 2nd contestant sees two closed doors and 98 open doors with goats. Monty tells him one thing: he has a 50/50 chance of opening one of the two closed doors and winning the prize.
Is Monty correct about the SECOND contestant’s 50/50 odds?
If contestant one stays with his door are his odds 1 in a 100?
If contestant 1 switches are his odds 99 in 100?
Hi Eric,
I agree, the illusion falls apart when you consider more doors. It becomes more obvious that Monty is selectively opening doors that don’t have the prize.
For your scenario, it’s important to remember that it’s Monty’s process that counts rather than what the contestant thinks happens. Knowledge about the process helps you make the correct decision. Inaccurate knowledge won’t change the actual probabilities but will change the contestants decision making process and, in this case, lead him/her astray.
So, to answer your questions, Monty is incorrect when he says it’s 50/50. And yes, if the contestant stays with his door, there is 1/100 chance of winning. And, there is a 99/100 chance of winning by switching.
Of course, the contestant won’t know what Monty’s process was and would likely take Monty at his word. That could lead the contestant astray. But, the underlying probabilities are still 1/100 and 99/100. Knowledge is key!
Imagine being in this game. If the rules of the game were told to you with complete openness then it will be obvious how to play it. You MUST be told that if you pick incorrectly then the door the game show host opens will not be the prize door, because he KNOWS that the prize is behind the other door and always does this. But read the way the problem is set up above and find out where you were told this. Nowhere. The problem is neither probabilistic or statistical. It’s one of incomplete information, trust, communication etc. A game theory problem of determining the best strategy with poor knowledge. Not the kind of game I enjoy. I already have serious lack of trust of people in our society and don’t need to have it reinforced in a game.
Hi Zoot,
That is the basis of the puzzle. That information isn’t presented to you, but using logic you can figure it out. People know that the host won’t open the door with the prize. However, many people don’t take it the next step further and realize that the host uses his insider knowledge to selectively (not randomly) choose a door that doesn’t contain that prize. When you realize that, it affects how you calculate the probabilities.
I guess as with any puzzle, whether one enjoys it or not, it’s a subjective matter. I like it because you can figure it out. You just need to look into it more deeply to obtain all the information you need.
I think it’s important to realize that for this to make sense, you need to understand that Monty cannot open the door you originally chose. That door is “off limits” to Monty.
Hi, Adnan.
In your 2 doors scenario, notice that the host can only reveal a goat when you have chosen the car, so he cannot fulfill the condition of always revealing a goat from the rest. Instead, in the 3 doors scenario, there is always at least one goat available to be revealed in the other two doors regardless of what you caught, so the fact that he purposely shows it, knowing where to find it, does not say anything about if your door is more likely to have the car or not.
But you can think about it with another perspective. Suppose you play a lot of times, like 900. The car should appear in each door about 1/3 of the time (about 300). For simplicity, suppose that you always pick door 1.
1) In 300 games the car is behind door 1 (yours). Here the host can reveal any of the other two doors, because both have goats, and since we don’t know if he has preferences, we can only assume that he reveals each with 1/2 probability.
1.1) In 150 of them he reveals door 2.
1.2) In 150 of them he reveals door 3.
2) In 300 games the car is behind door 2. The host is forced to reveal door 3.
3) In 300 games the car is behind door 3. The host is forced to reveal door 2.
So, if for example, door 2 is revealed, we must discard case 2), which means that the cases in which you could have failed were reduced by half —> they were 600 in total and they are only 300 now (case 3). But the problem is that the cases in which you could have been right were reduced by half too —> they were 300 and now they are only 150 (case 1.1), because not everytime that the car is behind your door the host will reveal the same other door. He sometimes reveals door 2 and sometimes door 3.
As you see, the proportions 1/3 vs 2/3 remain the same but not because we are considering the original cases, but because both cases (when you have chosen wrong and when you have chosen right) were reduced by half, and to reduce both by half makes their proportions remain the same.
If door 2 is revealed, you can only be in case 1.1) or in case 3), which is a subset of 450 games. You win by switching in 300 of them (case 3) and by staying in 150 of them (case 1.1). The same occurs if door 3 is revealed.
Ok Jim, lemme try to come at it from a different angle to demonstrate what’s bugging me. Couldn’t read all the comments TLDR; so my apologies if you’ve already answered it.
Let’s say there are only 2 doors. Now the probability that the car will be behind any of the doors is 50/50, right. So when I choose, let’s say door 1, since I’m choosing randomly the initial probability is 50/50. Now, when Monty opens door 2 revealing the goat, as per your explanation this ‘knowing and non-random’ act of Monty cannot change my original probability of being right, i.e. 50/50. Whereas we undoubtedly know that now the probability of me being right has also changed to 100.
Hence what’s bugging me is you saying that when I chose randomly my probability was 1/3 and it will remain so even after Monty eliminating one door. If so, then why can’t I reproduce it in the 2 door case I just presented?
Trevor, you are incorrect in that the host cannot lower his probability of having the car from 2/3 to 1/2 no matter what he does. If you pick Door 1 as in your example, and the host opens Door 2 there is still a 2/3 chance that the car is behind Door 3 because if the car was behind Door 3 he would have opened Door 2. There are only three combinations of two doors that the host inherits after the contestant’s selection. They are goat/goat, goat/car, and car/goat, each equally likely. Opening a door with a goat in each combination leaves the car twice of the three.
Hi everyone. I am really enjoying the conversation. I understand the theory behind the statistical probability and why it is believed that switching creates the highest probability of winning.
I want to throw something out for debate. There is something that is discussed but I am not sure that is accounted for in the probability. The fact that one door not chosen is automatically eliminated after we choose our door. It does not matter what door number it is, but we know it is an incorrect door (a goat). I will propose that our initial odds could actually be 50-50 because of that.
If I choose door 1, either door 2 or door 3 is automatically revealed. It does not matter which door it is. That only leaves one of the non-chosen doors and the one we chose. And because a door with a goat is always revealed in this argument, then that door is statistically irrelevant to the probability arguement.
So it will always come down to the door we picked, and the door that wasn’t opened. And they both have the same probability of containing a car?
I am not sure if I even fully support my hypothesis, but I feel like it is worth discussing? Here is what I think the breakdown of the odds based on choice.
I will use other door as a term because it does not matter what number it is because there is only one other door remaining at this point when you have to make the choice to stay or switch.
Car behind door 1
Pick door 1 and stay – win
Pick door 1 and switch – lose
Pick other door and stay – lose
Pick other door and switch – win
Car behind Door 2
Pick door 2 and stay – win
Pick door 2 and switch – lose
Pick other door and stay – lose
Pick other door and switch – win
Door 3 would be the same.
Based on an assumption that the initial odds are 33% to pick the correct door, then the arguement to switch makes mathematical sense, however, because the door which is revealed is not random, the initial choice is out of three, but your odds are 1 out of 2??
I am curious to hear some opinions on this.
Thanks
Trevor
In my previous comment, the number 900 is the total amount of times you play. You cannot say that the total games are not 900 but 1200 if you are only playing 900. The contents are suppossed to be placed randomly at first, so each should appear about in 1/3 of the time in each door (300 games). I also only considered only if you always selected door 1 to make the reasoning easier. The locations of the contents are completely independent of the door you choose, so they are not going to put the car behind door 1 in 450 games (1/2 of 900) only because you are selecting it.
In the 300 games of case 1, when the car is behind door 1 (yours), you said that there is the possibility that host reveals door 2 in all those 300, so let’s suppose that it occurs. That means that if he reveals door 2, you could be only in the 300 games when the car is behind number 1 or in the 300 when the car is behind number 3, so switching and staying win 50% each.
But that would also mean that there is no game of those 900 you are playing in which the car is behind number 1 and the host reveals number 3. You’ve already sold out the only 300 games when the car was behind door 1. So if he revealed door number 3, you could only be in the 300 games when the car is behind number 2, and therefore you would have 0% chance by staying and 100% chance for switching.
Anyway, staying would still win only 300 games (1/3 of the total 900), only that the proportion would not be symmetrical according to which door is revealed. The 1/3 is the average of the two cases. The host would reveal door 2 in 600 games (2/3) and door 3 in 300 (1/3). You win by staying in 1/2 * 2/3 + 0 * 1/3 = 1/3 in total.
========================
The reasoning is the same as if you got a job in which you have to go the two days of the weekend. On Saturdays they send you to work in place A, but on Sundays they can send you to work in place B or in place C, sometimes one and sometimes the other (it is completely random). That means that after you have worked several weekends, in total you would have gone more to place A than to place B or to place C, because there is the same number of Saturdays as of Sundays, and on all Saturdays you go to place A, but only in some Sundays you go to place B.
In the total days you work, you do not go to each place 1/3 of the time, but 1/2 to place A, 1/4 to place B and 1/4 to place C.
The list you made:
1. Choose door 1, Door 1 has prize, Door 2 revealed, (Stay: Win, Switch: Lose)
2. Choose door 1, Door 1 has prize, Door 3 is revealed. (Stay: Win, Switch: Lose)
3. Choose door 1, Door 2 has prize, Monty has to choose 3 (Stay: Lose, Switch: Win)
4. Choose door 1, Door 3 has prize, Monty has to choose 2 (Stay: Lose, Switch: Win)
is the same as if I made this list:
1. It is Saturday and you go to place A.
2. It is Sunday and you go to place B.
3. It is Sunday and you go to place C.
and I concluded that you go to each place 1/3 of the time, because there are three cases.
Only because there are two places you can go on Sundays there are not going to be two Sundays and only one Saturday in each weekend. In the same way, when you select the correct door (number 1 in our example), the host has two possible choices, while when you choose a losing door, he only has one possible choice, but that does not make the correct door being yours twice as frequent as the others.
Ronald says “1) In 300 games the car appears in Door1 (yours). In this case, the host is free to reveal each of the other doors, and since we don’t know if he has preferences for one over the other, we can only assume he reveals each with 1/2 probability.
__1.1) In 150 of them he reveals Door2.
__1.2) In 150 of them he reveals Door3.
2) In 300 games the car appears in Door2. The host can only reveal Door3.
3) In 300 games the car appears in Door3. The host can only reveal Door2.”
The problem with this is that this does not indicate the possibilities properly. In scenarios 1.1 and 1.2 Monty could potentially open up door 2 300 times in 1.1 or potentially open door 3 300 times in 1.2. However, in scenario 2 and 3 Monty has no option but to open the door without the car. So when you are looking at the potential scenarios there are not 900 there are 1200. You are hiding 300 potential scenarios by “splitting” them in half and just saying he’ll choose door 2 half the time and door 3 half the time. Thus we are once again brought back to a 50/50 probability. Increasing the numbers doesn’t change the error.
The problem is that you know the door Monty is going to open will not have a car behind it so it does nothing to increase your knowledge about what is behind the other door as compared to your own.
Computer simulations will pump out a 66% chance if the input in is failing to capture the true nature of the situation.
This same problem that I pointed out with what Ronald said can be applied to the 9 row chart of the author produced. There is a missing row of input and that is the door that Monty opens. If we just reduce it to choosing door 1 the chart should produce 4 rows not 3.
1. Choose door 1, Door 1 has prize, Door 2 revealed, (Stay: Win, Switch: Lose)
2. Choose door 1, Door 1 has prize, Door 3 is revealed. (Stay: Win, Switch: Lose)
3. Choose door 1, Door 2 has prize, Monty has to choose 3 (Stay: Lose, Switch: Win)
4. Choose door 1, Door 3 has prize, Monty has to choose 2 (Stay: Lose, Switch: Win)
If someone has a counter to this point I’d like to see it because I haven’t seen anything compelling that answers this objection.
I don’t get the optical illusion, but I can understand the Monty Hall scenario.
Hi, Vienna
In the second decision, when you have to pick from two doors, you already have more information about one than about the other. Two options don’t imply 50% chance for each, that is only when we have exactly the same information about them, which is not the case here.
What makes the difference is the fact that each door was left by a different person: one was left by the contestant and the other by the host. While the contestant chose randomly, the host already knew the positions and the rules forced him to not reveal the contestant’s door and neither which has the car. The contestant chooses a goat door with a frequency of 2 out of 3 times, so the host is who purposely avoids to reveal the car from the other two doors in those same 2 out of 3 times. In this way, the switching door is which results to have the car 2 out of 3 times, not 1 out of 2.
It’s like if the host was a guide that, given the fact that he knows the results, he tried to help you to find the car when you failed, but was trying to trick you when you chose correctly. It was easier for you to fail, so it is easier that he is telling you the truth.
You can see it better supposing you played a lot of times, like 900. In about 1/3 you should pick each content, so 300 times the goat1, 300 times the goat2 and 300 times the car. So 600 times a goat and 300 times the car. After your selection the host is forced to reveal a goat from the other two doors, and it occurs in all the 900 cases.
So, once the goat is revealed, if you always keep your original door, you will still find the car in about 300 cases (1/3 of 900), because the revelation didn’t shuffle the contents. Instead, in the 600 cases that your door already had a goat, the revealed goat must be the second one, and so the car must be in the other door the host avoided to reveal (the switching one). So, if you always switch, you win about 600 times (2/3 of 900).
So I don’t see why the table doesn’t account for the eliminated door. Once Monty opens a door, there is no longer an option to chose that door, thus it needs to be taken out of the stats for the final choice to switch over. The win-loss analysis doesn’t make sense to me, if you get to keep the stats of that additional option in then the “switch option” needs to be adjusted to where you have to choose which door you will select. How in 3 options is there only 2 options (to win if you switch)? There’s only two doors left, why keep the 3rd option in after it has already been eliminated by Monty? Further, if it’s kept in, why is there not more options to chose where you switch to?
So it sounds like in reality, the Monty Hall problem is really less of a statistical illusion and more of a psychological theory because it wouldn’t work without Monty. Then again, I do not understand variable change.
I agree with this explanation, if in scenario 2 the doors are opened and there are only 2 choices remaining then the probability is 50-50.
But if the doors are not opened and we always switch then I agree with the explanation that probability of winning is increased by switching.
Why does the chance pass to the other door and not your door? That is where I’m mathematically hung up…I don’t see anything that would require the already opened door 1/3 to go the door that could be switched to as opposed to the already chosen door.
Hi Mel,
One thing thing to remember is that your initial choice is the only random part of this process and it sets the initial probabilities.
Right after your initial choice, you have two sets of doors. One set contains the single door you chose. Because you chose it randomly, there’s 33% chance it contains the prize. The other set contains two doors and because the process is random up to this point, we know there’s a 67% chance that this set of two doors contains the prize.
When Monty opens one door in the set of two doors, it’s important to note two points. First, you selected your door randomly and nothing Monty does will change your door’s probability. You picked your door, one out of three, hence it will always have a 33% chance. Additionally, the set of two doors will always have a 67% chance because they were established by your random choice.
The other thing to realize is that Monty’s choice is not random. There’s a zero percent chance that the door he opens contains the prize. Consequently, when he opens a door in the set of two doors, the one unopened door retains the full 67% probability of that group.
So, think about it in terms of the portions that are random and how that fixes probabilities for each set of doors. Then, the second part of the process is non-random, which allows the final door in the other group to retain the full 67%.
There are three possibilities:
Scenario 1, the prize is behind Door 1
Scenario 2, the prize is behind Door 2
Scenario 3, the prize is behind Door 3
Each scenario has a 1/3 chance of happening and so you have a 1/3 chance of being correct in your initial pick.
If you pick Door 1, half the time Monty will reveal the prize is not behind Door 2, leaving scenario 1 & scenario 3 as possibilities. The other half of the time he will reveal it is not behind Door 3, leaving scenario 1 & 2 as possibilities. At first glance, this would seem to leave 2 equally likely scenarios regardless of which door Monty opens. However, if you think about it, this would mean that in your initial pick, you selected the prize door (correct scenario) half the time. That does not make sense because we know the initial odds were 1 in 3. Obviously what Monty does after you make your initial choice can’t affect the odds of that choice, only subsequent choices.
Then again, after the first player choice, we have two sets: on with the selected door, and one with the the other two doors; obviously the two door set has the better chance of containing the car. The gamemaster simply eliminates a non-car door from the two door set. The proof is the greater number of cars won by the flippers.
Yes I realized that too by increasing the number of doors. I have finally given in to the fact that I was wrong.
Hi Raymond and Zach,
Humans in general aren’t great at probabilities. And, this problem even tripped up the experts! So, you’re in good company!
I used to be a Monty Hall Problem skeptic. I had thought about it, and it was so obvious that the solution is 50/50. But then I found out that empirical data from simulations clearly and consistently show that it is better to switch. So instead of spinning my wheels arguing for the 50/50 position, I decided to make myself understand the problem, the solution, and why my intuition was fooled. This effort was was very rewarding.
The point is, that if you think it doesn’t matter whether or not you switch, you are wrong. Provably. Empirically. Full-stop. There is no shame in being wrong. You may not care enough to understand the solution, and that is perfectly fine; we all have a finite lifetime. But if you do care, the best course is to start from the realization that it is in fact better to switch, and work from there, rather than trying to justify an incorrect position.
It is hard to convince people especially non-mathematicians. The way that has worked for me is to expand the game till there are 999 goats and one car. After you pick, Monty opens 998 doors with goats (hopefully well behaved and not too smelly). Now you get the option to stick or change. Most people now realise the chances of them being right initially were vanishingly small.
Hi Geoff, I agree. In fact, somewhere in these comments, I mention that line of reasoning. It’s easier to illustrate that it’s a non-random process when you expand that process to include more doors.
Your initial door has a 1/3 chance. The other doors together have a 2/3 chance. But Monte removes one of those doors from consideration. So now the 2/3 chance passes to that other door.
All the probabilities have to add up to 1, and the door you initially picked still has a 1/3 chance.
Okay, I understand your reasoning, but I think you are wrongly disregarding 2 of the scenarios. Using your table, there should be a second prize door 1, with stay/win in the second column and switch/lose in the third column. You are treating doors 2 & 3 as a single possibility of selection, but each has a its own possibility of being selected. Again, please use my original example and point out which scenario I listed would not happen in game show.
Hi Raymond,
There’s really only 6 scenarios. It’s a given that you’re going to choose door 1. The prize can be behind one of three doors (3 possibilities). You can stay or switch (2 possibilities). 3 X 2 = 6. Hence, there are six possible scenarios. It’s really that simple.
I’m counting each door as one equally weighted possibility. There are three options for doors. Therefore, you wouldn’t double count door 1.
I am on the same page with Harry. You misinterpreted his answer. I’ll explain my reasoning, which is similar to Harry’s:
For any door picked, there are 8 possible scenarios. In this example, I will ALWAYS pick door 1. Here are the scenarios:
1) prize is behind door 1, door 2 is opened, I stay with door 1- WIN
2) prize is behind door 1, door 3 is opened, I stay with door 1- WIN
3) prize is behind door 1, door 2 is opened, I switch to door 3- LOSE
4) prize is behind door 1, door 3 is opened, I switch to door 2- LOSE
5) prize is behind door 2, door 3 is opened, I stay with door 1- LOSE
6) prize is behind door 2, door 3 is opened, I switch to door 2- WIN
7) prize is behind door 3, door 2 is opened, I stay with door 1- LOSE
8) prize is behind door 3, door 2 is opened, I switch to door 3- WIN
50/50 chance
When the prize is not behind the door I initially picked, such as scenarios 5-8, Monte is only able to open one door so as not to reveal the prize and not open the one I picked.
Please point out my error in logic. Please use my example.
Hi Raymond,
For one thing, there are six possible scenarios–not 8. There are three scenarios for Staying and three scenarios for Switching. In the table below, I show the six scenarios and their outcomes. The assumption is that you always choose door 1. For example, in scenario 1, you choose door one and the prize is behind door one. Therefore, if you stay, you win. And, you lose if you switch.
As you can see in the table, there is one winning scenario out of three possible scenarios in the Stay column (1/3). However, there are two winning scenarios out of three possible scenarios in the Switch column (2/3). Consequently, by always switching you’ll double your chances of winning from 1/3 to 2/3.
I hope this helps!
Vince,
Suppose you pick Door1 and the host reveals Door2, so only doors 1 and 3 are still closed. Take into account that the host knows the positions and rules imply that he must always reveal a door that isn’t the contestant’s selection and neither the one that hides the prize (the car). That means that if the car was in Door3, he was forced to reveal Door2. Instead, if the car was in Door1, we cannot be sure that he would have revealed Door2, because both Door2 and Door3 would have goats and either could be opened. So it is easier that the reason why he is opening Door2 is because the car is behind the 3 than because it is behind the 1.
You can see it better supposing you played a big number of times, a large enough number so the proportions are not very different to the actual probability. For example, imagine you played 900 games. The car should appear in about 300 in each door (1/3 of 900). Suppose for simplicity that you always pick Door1 in all the 900 games.
1) In 300 games the car appears in Door1 (yours). In this case, the host is free to reveal each of the other doors, and since we don’t know if he has preferences for one over the other, we can only assume he reveals each with 1/2 probability.
__1.1) In 150 of them he reveals Door2.
__1.2) In 150 of them he reveals Door3.
2) In 300 games the car appears in Door2. The host can only reveal Door3.
3) In 300 games the car appears in Door3. The host can only reveal Door2.
So, if Door2 is revealed, you could only be in case 1.1) or in case 3), which is a subset of 450 games. You win by staying in 150 of them (case 1.1) and by switching in 300 of them (case 3). So, you win by switching twice the times as by staying.
So, don’t get confused. The reason why the probabilities of your door are still 1/3 is not because we are counting the original cases, but because both the cases in which you could have failed and in which you could have succeeded were reduced by half (in the example, success cases were reduced from 300 to 150 and failure cases were reduced from 600 to 300) and to reduce both by half results in their respective probabilities being maintained.
But that’s not what happened. You are fudging a statistical edge based on choosing one door out of three. Once door three is opened and you are invited to switch, any case involving door three is moot and you are left with a 50/50 chance of guessing.
Of course, the chance that it is behind door one is either 100% or 0% as that does not change.
Like most supposed paradoxes, the poser insists on telling a story that is not represented by the actual facts.
Hi Vince,
The first thing you need to realize is that this is not a “supposed paradox” and there’s no “fudging” go on. There was confusion about the correct answer back in the 1980s, but the standard, accepted solution since then is that you double your chances of winning by switching. This solution has been proven both by mathematics and by computer simulations.
Ronald has some great points. And, I’ll tackle several of your other misconceptions.
First, you seem think that the just because there are two choices that it must be 50/50. That’s true under only very specific conditions, which the problem does not satisfy. Namely, the event must be an independent, random event.
The only part that is truly independent and random is the choice of the initial door. There are three doors and you’re choosing one. Therefore, your initial door has a 33% chance of winning. So, your door isn’t even starting at 50%!
There’s a 33% chance the prize is behind the door you choose, which means there’s 66% chance it’s behind the other two doors. Monty’s actions are NOT random. He will systematically open a door that does not have the prize. Now we’re getting into the realm of conditional probabilities. The end result is that the door that is unopened (the one you’d switch to), now has the full 2/3 chance of winning because of Monty’s non-random opening of a non-winning door. I show the calculations for this elsewhere in the comments section if you’re interested.
The two mistakes are thinking that your door starts at 50% when it’s only 33%, and then thinking it’s a random process after that when it isn’t. Monty isn’t random and it affects the outcome.
I hope that helps!
Don’t get why door 3 appears in the table of outcomes. I though Monty Hall had already opened that door and there was nothing behind it. Therefore you would not be switching from or to door 3.
Hi,
The table does not reflect a scenario where the prize is only behind door 3. Instead, the table contains all possible combinations of prize doors, your choices, and whether you switch or not.
The text above and below the table describes how the table works. In a nutshell, by showing all possible combinations, you can see how the chances of winning are 66% when you switch. A row in the table represents one combination of the sequence of events and the outcome for whether you switch or stay. Look in the Prize Door column to see which door contains the prize for each possibility.
There is an option for 50/50. If you flip a coin to decide when given the choice to switch or not to switch; this gives you the probability of winning 50%. To summarize:
1. Never switch – probability of winning is 33%
2. Randomly switch – probability of winning is 50%
3. Always switch – probability of winning is 66%
Hi Rodrigo
I am not completely sure, but what I think you are saying is that, if you have initially chosen the door with the car behind it, Monty has two doors he can open, either of which will leave a goat behind the remaining door, and so in both cases you get a goat if you switch. So, TWO ways to lose. Is this what you meant?
In fact although Monty has a choice of two doors, in any one game he can only pick and then open ONE. Whichever he chooses will have the same outcome and it is the outcome that matters, rather than exactly how you got there. So there is only ONE way you can get a goat if you switch and two ways you can get a car; so double the chance.
Hope this makes sense now.
I am happy to admit that when I first came across the Monty Hall problem the answer seemed unbelievable. I kept an open mind however and, once I had followed the reasoning given, it all made sense. This is what still makes it such a great puzzle years later.
The problem with this solution is because when we pick the door that cointains the winning prize, Mounty Hall has 2 options.
Imagine I choose door1 and the prize is in door 1, then 2 things can occur, MH can open door 3 and if you switch, you lose, or can open door 2 and if you switch, you also lose! therefore, you chances of winnig the prize increases to 50% every time because instead of a 1 out of 3 choise between doors, you now have a 1 out of 2 coise for either switch or don’t swicth
Hi Rodgrigo,
This is the correct solution. I know it can be hard to wrap your mind around. You’re right that if your initial choice has the prize behind it, if you switch, you’ll lose. However, that scenario only happens 1/3 of the time.
The other 2/3 of the time, you’re initial door choice doesn’t have the prize behind it. In that case, when you switch you win. Hence, your chance of winning by switching is 2/3.
I tried to send this comment but don’t think it went, so trying again.
Apologies if you get it twice!
Hi Jim
I came across your fascinating website by chance. It seems that people do become remarkably entrenched in their views and I admire the patient, polite and logical way you try and explain to them why the solution is what it is.
The simplest way to prove the answer, I think, is as follows. I have used the version where a car and two goats are behind the doors, partly because that is the version I first read about, and partly, for me at least, because it makes the reasoning easier to follow than using win/lose or prize/no prize options.
So, the contestant makes a guess which door hides the car with a 1/3 chance of success.
There are then just three possibilities for what are behind the other two doors:
1 Goat/goat
2 Car/goat
3 Goat/car
In case 1, it doesn’t matter which door Monty opens, the remaining door will hide a GOAT.
In case 2, Monty has to pick the door hiding the goat and so the remaining door will hide the CAR.
In case 3, again Monty has to pick the door hiding the goat and so the remaining door will hide the CAR.
So, if the contestant chooses to swap doors the chance of picking the car is 2 out of 3.
Regards
Keith
Hi Keith,
I only got your comment once. Maybe the system ate it the first time?! I’m glad you resent it.
Of all my posts on my website, the Monty Hall Problem by far elicits the strongest emotional responses! And, I understand it. The answer just feels wrong when you first hear it.
Thanks for sharing your explanation. I really like it because it simplifies the problem into something that’s easier to understand by focus on the two remaining doors. I’ve tried to explain it using different approaches figuring different people will respond to different explanations. Your explanation is a great one to add to the mix!
Hi Jim.
Does not the 2/3 probability argument depend on the entire process being treated as a single event? Most people perceive the process as two events.
Event 1 – contestent picks door 1 with 1/3 probability of winning.
Event 2:- 1/2 probability of winning. car
Most of us interpret the Vos Savant question as asking about the second event.
If I combine the events into one and ask what is the probability of NOT winning car after picking two doors then probability against winning in Event 1 is 2/3. and in Event 2 is 1/2. Probability against winning for two independent events is 2/3 x 1/2 = 1/3. Thus your overall probability of winning in two tries is 1-1/3 = 2/3.
Seems Vos Savant is right about the whole process but her phrasing of the question is misleading.
Please explain where my logic fails.
Hi Rick,
The question for the Monty Hall question is, when Monty offers you the option to switch doors, do you switch doors or stay with your original choice?
The answer to that question depends on the probability of winning for switching compared to the probability of winning by staying. The question itself is not asking about first or second events, or anything like that. It’s asking about what the probability of outcomes at that final stage. The difficulty does not lie in correctly interpreting the question but rather in correctly understanding the process (particularly Monty’s non-random role) and then basing the probabilities on that underlying process. The scenario does intentionally leave Monty’s role vague and you have to deduce that he non-randomly reveals a non-prize door using his inside knowledge, which throws the probability off from what most people expect. We’re not dealing with random, independent events in the 2nd step.
The correct answer is that you have a 67% of winning by switching and a 33% of winning by staying. Therefore, the smart move is to switch. That’s the answer to the question.
To get to the correct answer, you need to understand how the process works. Yes, there are two parts of the process. However, the contestant does not get “two tries.” At the end, the contestant has one door and winning depends on that one door.
Here’s how to break the steps down.
Step 1: The contestant’s random choice divides the set of 3 doors into two group. One group with one door (33% chance of having the prize) and the other group has two doors (67% chance of having the prize). This is the random part of the process. The probability for the one door group (the contestant’s initial choice) does not change from 1/3 here on out because nothing affects this group.
Step 2: Monty removes a non-prize door from the second group. This is a non-random process that relies on Monty’s insider knowledge about the location of the prize. If the prize is in the second group, it is 100% guaranteed to be the final unopened door of that group because Monty will open the door that does not have the prize. Consequently, the chances of the prize being behind the door you can switch to is (2/3 * 1) = 2/3. This seems counter-intuitive. Remember, it’s a conditional probability of a non-random process. If the prize is in the two door group (2/3), then Monty’s non-random process guarantees it to be behind the unopened door (1/1) that Monty allows you to switch to. Hence, 2/3 * 1/1 = 2/3.
Note that there is never a point where it’s a 50/50 choice (which you state is the probability at your Event 2). The confusion is not related to a misinterpretation of the question. It’s related to a misuse of probabilities by not correctly breaking the process down into random and non-random parts and then using a conditional probability. You don’t multiply the probability of event 1 by event 2 because they are not independent, random events.
I found this article on the internet after reading about this “problem” in a book. I perfectly understood the 2/3 chance of winning by switching, but then my devious mind went to the “Monty Hell ” scenario that you mention. If the host only sometimes reveals what’s behind a door, and other times not, then the best bet is to stay put with your initial pick, since the host is probably only going to reveal a door when he wants you to switch away from the door you selected, because the big prize is behind it. If you picked a door with nothing good behind it, then the host would not open any door at all, would not offer you the chance to switch, and you’d be stuck.
If you know you are going to switch doors, then you are first choosing one of the doors you hope is a goat. That choice is a 2/3 chance you picked the goat. Then Monty shows you the other goat and it becomes obvious which door is more likely to have the prize, so you switch to it. You already determined to switch doors, so there is no 50/50 choice. Of course, what you are thinking in your brain does not change the actual probability, so the odds are still better if you switch no matter how you look at it.
Hi David,
That’s a great way to look at the problem. Your initial choice is the only random part of the problem. Consequently, you have a 2/3 probability that your initial choice is incorrect and a 1/3 that it is correct. Switching doors flips your outcome compared to staying. If you are going to lose by staying, then you’d win by switching. And, vice versa. So, you have a 2/3 chance of losing if you stay because we know your initial choice is probably wrong. Consequently, you have a two-thirds chance of winning if you switch. There’s never a point in this problem where it’s 50/50!
Jim, thank you for your appreciation.
The code you posted is obviously correct, since choosing a goat at the beginning is necessarily a victory for switching, and choosing the car is necessarily a victory for staying. However, it has already been seen that there are people who are not satisfied with this, since they want to see explicitly the part of the revelation and the part when the player chooses the other remaining door when switching.
Here I share a link to a code I made in Python in which the contestant always decides to switch. Since it includes explicitly the other steps, it is more complicated, but that way nobody can make any excuse.
https://drive.google.com/open?id=1vxvGAeJfpc8MsimKI50TWfn6akiMSk2p
By the way, with Monty “Fall”, I was referring to the variant of the game in which the host randomly chooses a door. I have seen that they give it that other name to make it clear that they are not the same problem and have different results.
I’ve also seen a third case they usually call Monty “Hell”. In there, the host knows the positions but only reveals a goat and gives the opportunity to switch if the player chose the car, so it is impossible to get the prize with the switching strategy. The revelation is a hoax that he makes on purpose trying to make the player wrong. However, looking for that name on Internet you may also find another problem that has nothing to do with this, so that name is ambiguous.
This is a wonderful response I have never heard before. An excellent way of viewing the problem
There are only 4 paths the game can take and here’s the probability of each path delivering a win:
1)Choose right Prob 0.33 followed by Don’t Switch, Win Prob 1: Total this path 0.33 x 1 = 0.33
2)Choose wrong Prob 0.67 followed by Don’t Switch, Win Prob 0: Total this path 0 x 0.67 = 0
Therefore Total Win Prob for Don’t Switch is 0.33
3)Choose right Prob 0.33 followed by Switch, Win Prob 0: Total this path 0
4)Choose wrong Prob 0.67 followed by Switch, Win Prob 1: Total this path 0.67
Therefore Total Win Prob for Switch is 0.67
Dean, I forgot to say: In Monty Hall rules, the revealed door is not random. That is the precisely the reason why the 1/3 vs 2/3 works.
If the contents of the doors are:
Door1 Door2 Door3
Goat Goat Car
and the player chooses Door1, it is sure that the host will reveal Door2. So all the games are valid.
If the host does it by random and just by chance it results to be a goat, it is true that its result will be 1/2. That is what they call “Monty Fall Problem”. The “Fall” is because it alludes that the host fell and by accident revealed a door, which just by coincidence it resulted to have a goat.
Hi Ronald, I’m not sure Dean will be back. But, you raise good points. Monty does not follow a random process and simulations must accurately portray his systematic removal of a non-prize door. He’ll never pick your door and he’ll never open the door with the prize. Thanks to this non-random process, if the prize is behind one of the doors in his initial group of doors, it is guaranteed to be behind the one he doesn’t open. Because there’s a two-thirds chance it’s behind one of his initial doors, there’s a two-thirds chance it’s behind that one final door he doesn’t open.
And, one point of clarification, it’s the Monty *Hall* problem. It’s named after the host of the original TV game show.
It seems you didn’t take the time to think about the comment.
First, the code Jim showed is not the only one made. There are already others that make that the host always reveals a goat from the remaining because he knows where to find it, that is, the host never fails to follow the rules of the game, and they show that the probabilities are 1/3 vs. 2/3.
Second, the point is not whether the desired amount of valid games is completed or not (10,000 in your case). Completing the total does not make the proportion real again. Of the first 10,000 iterations that the code performed, some were discarded, remaining with the same number of times choosing goat at the beginning than times choosing the car, which is an error. The discard was unfair, because it eliminated a larger proportion of the type that could have won by switching, than from those that could have won by staying. Then, the code has to do some extra iterations to be able to complete the 10,000 valid games, but to those extra iterations the unfair discard is applied again: it eliminates some and remains with the same amount of one type as the other. So there are still missing iterations to complete the 10000. It does extra repetitions again, to which the unfair discard is applied again, and so on.
To make it easier for a moment, suppose the code only discarded games in which the contestant had chosen goat (which are precisely games in which he would have subsequently won by switching). So the code has to repeat each of those discarded games in order to complete the 10,000, but in doing so, the new game may not have the same result as the previous one, because the contestant may now choose the car at the beginning instead of a goat. On the contrary, the times when he originally chose the car remained intact. In total, the replacement resulted in an increase of the times in which the player chooses the car at the beginning and in a reduction of the times in which he chooses goat, which translates into an increase and reduction of winnings by staying and switching respectively.
Now, in your case, both the times you can win by switching or staying can be replaced by its opposite result, but from those games that were sent to replace, double were victories for switching, so there will be more replacements of victories for switching to victories for staying, than replacements in the opposite direction.
To fix your code, every time the host fails to fulfill the condition, instead of starting the game from the beginning make it only the host who has to repeat his choice of the revealed door until he reveals an appropiate one. That is, do not make the contestant to choose again; his choice remains. The only one who has to repeat is the host if he fails.
Ronald, my code only introduces the revealed door as a variable, the same as the the premise of the Monty Hall Problem does, then creates the rules set out in the Monty Hall Problem. If an attempt does not follow the rules rather then discard the attempt, the code resets and tries again. This is only done so we don’t lose attempts and only take into consideration options that actually follow the rules set out in the Monty Hall Problem. The resets are only there so we still have 10000 attempts, they change nothing on the probability of the end result. Remember you have to tell code exactly what you want it to do or it will throw out something you didn’t ask for.
Jims code tells the program to pick a door and a prize door and give you the odds they’re the same.
My code tells the program to pick a door and a prize door and also reveal a door that does not match the picked door or the prize door and give you the odds on which of the two remaining doors has is the prize door.
My code does what they Monty Hall Problem is asking, Jims does not.
You say I should keep the answers that were discarded by the rules set out, but why should I keep impossibilities, why should I keep the times when either the host reveals the players door or the prize door? They’re physical impossibilities due to the rules of The Monty Hall Problem. The rules are only there so the host cannot pick the winning door or the contestants door. Which in turn makes the code reflect The Monty Hall Problems rules.
As I said, as a programmer I know for a fact your code doesn’t work, it does not account for the revealed door. It only tells you if your initial choice is right or not, which is not what you’re looking for with the code.
In the 100 doors scenario, yes on your initial choice your chances of being right are 1/100, but Monty has removed 98 from the denominator, they’re no longer yours and neither are they Montys. Monty doesn’t have any doors, he’s simply altering the odds by providing impossibilities where there were once possibilities.
Dean, your comment was rude and I’ve edited it to remove the rudeness. This is a place for polite discussion. Rudeness will not be tolerated on my site. That’s your one and only warning.
First, the code is not my code. It’s sample code that comes with the simulation software. And, it was written by the author of the software. However, I can validate that it is correct.
Here’s how it works. The software picks a prize door randomly. Then it chooses a door for the contestant. At that point, it determines the result if the contestant stays. It then switches the results for if the contestant switches.
In a previous comment, I’ve explained how switches causes the opposite outcome based on what we know about the game. For example, if staying with the original pick causes the contestant to lose, then switching will cause the contestant to win. That’s how the software keeps track of wins by staying versus wins by switching. It’s simple logic.
What you’re not understanding is that Monty’s process systematically removes non-winning doors. This systematic removal of non-winning doors affects the probabilities. Specifically, there is a conditional probability at play. If the door is in Monty’s original set of doors (regardless of the number), there’s a 100% chance that it will be his final unopened door. We know that based on his process of removing only non-winning doors. So, when Monty’s two doors have a 2/3 chance of having the prize. We know that the final door also has a two-thirds chance of containing the prize (2/3 * 1 = 2/3).
Dean,
the code you showed alters the results. The host must always reveal a goat from the two doors that the contestant did not choose, and he can do it because he knows the positions. In your code, he does not know the positions; he makes a random selection and can fail. In case he fails, you discard the game and try it again until the condition is fulfilled. At first glance it may seem that this yields the same results, but it doesn’t, and the reason is that the proportion of games that you discard from those that would win by switching is greater than the proportion you discard from those that you would win with your original choice.
As you are posing, there are 9 cases that can occur, which come from the three possible choices of the contestant, and from the three possible doors that the host can reveal. (Instead of placing the cases according to the numbers of the doors, I will place them according to their contents, in order to more easily illustrate which cases should be discarded, but it is the same).
Contestant’s selection Revealed door
—————————————————————–
1) Goat 1 Goat 1 –> Discarded
2) Goat 1 Goat 2
3) Goat 1 Car –> Discarded
4) Goat 2 Goat 1
5) Goat 2 Goat 2 –> Discarded
6) Goat 2 Car –> Discarded
7) Car Goat 1
8) Car Goat 2
9) Car Car –> Discarded
So, for the times you choose goat 1, which are three cases, you are discarding two and only one survives. The same with the goat 2. On the other hand, when you choose the car, you only discard a case and the other two survive.
To see it better, suppose you play 900 times. Since you are 1/3 likely to pick each content, in about 300 games you should pick the car, in about 300 the goat 1 and in about 300 the goat 2. If we apply the Monty Hall rules (the host always reveals a goat because he knows the positions, therefore it is not necessary to discard any games) this is what on average should happen:
1) In 300 games your door has the car. If this occurs, the host can reveal either goat 1 or goat 2.
1.1) In 150 of them the host reveals the goat1.
1.2) In 150 of them the host reveals the goat2.
2) In 300 games your door has the goat1. In all those 300 the host reveals the goat2.
3) In 300 games your door has the goat2. In all those 300 the host reveals the goat1.
You win with your original choice in cases 1.1) and 1.2), which are in total 300 games (1/3 of 900). You win by switching in cases 2) and 3), which are in total 600 games (2/3 of 900).
Instead, the following occurs with your code:
1) In 300 games your door has the car.
1.1) In 100 of them the host reveals the goat1.
1.2) In 100 of them the host reveals the goat2.
1.3) In 100 of them the host reveals the car. ————–> Discarded.
2) In 300 games your door has the goat1.
2.1) In 100 of them the host reveals the goat1. ———-> Discarded.
2.2) In 100 of them the host reveals the goat2.
2.3) In 100 of them the host reveals the car. ————-> Discarded.
3) In 300 games your door has the goat2.
3.1) In 100 of them the host reveals the goat1.
3.2) In 100 of them the host reveals the goat2. ———> Discarded.
3.3) In 100 of them the host reveals the car. ————> Discarded.
So of the 900 runs, you discard 500 games; you only consider 400 as valid. Of those 400, you win by switching in 200 (1/2 of 400) and by staying on 200 (1/2 of 400). That change in the proportion happened because of the times you got a goat (600), you discarded 400 (2/3 of 600), while of the times you got the car (300), you discarded 100 (1/3 of 300). So at the end you save the same amount of games of each type.
Even if you keep iterating until the desired number of games is completed, the same thing will continue to happen with the rest: you will discard some, keeping approximately the same amount of each type, and so on.
You must fix your code in order that the host can reveal a goat because he knows where to find it, and also knows which the player’s selection is to avoid it, instead of repeating games until the condition is fulfilled.
Ronald, you’re correct and you raise great points. Monty knows all about this contest. His actions are not random. He won’t pick the contestants door. He also won’t open the prize door. He’ll only open a door that the contestant did not pick and that does not contain the prize. It’s the non-randomness of his process that throws people off. Thanks for clearly showing how the code in this comment thread does not work correctly.
Why is this the accepted solution when it’s easily debunked. The problem is you’re treating the whole set up as a single premise when there is actually three separate premises where one premise sets up the next. Premise 1 is followed by premise 2 and that is followed by premise 3.
Premise 1 is your initial choice, I think everyone here agrees the chance of you picking correctly is 1/3.
Premise 2 follows that, the Host must pick a door and it must be wrong, how many options does he leave behind, 2, he either left the wrong choice or the right choice based on what you picked. So he has a 1/2 chance of leaving behind a wrong choice as one wrong choice has been eliminated by his action. Tense is important. When we’re working out what has been left behind (Future), we must first know what has been removed (Past) to get our probability (Present).
Premise 3 is your finally, switching doors is simply a rephrasing of the question, this or that. Again past is hugely important as one door is no longer pick-able, it has been removed from the equation and no other door has gained over another. To the person choosing, we’re back at square one. Premise 3 is a re-write of Premise 1, but with two doors to choose from instead of three. So our final answer is 1/2.
The Monty Hall Problem sets up Premise 1 and 2, then asks you to solve premise 3. When solving premise 3 you have to have Premise 1 and 2 in the past and past only gives answers that are certain for the present to factor into future choices.
When people tackle this problem they treat it all as one premise, present only, and that screws with your working out.
Also for those saying just increase the number of doors, it changes nothing. You’re adding choices to premise 1, then removing them in premise 2 so premise 3 is unaltered as adding a number to a problem and then removing that same number puts you back to where you were before you added those numbers.
To put it really simply, you and your friend and everyone else, who has a lottery ticket, have a lottery ticket. You’ve already chosen your ticket but don’t know if you’ve won or not. Assuming what you friend tells you is true, they tell you one of the two of you have a winning lottery ticket. This immediately eliminates all the other lottery tickets. What is the probability that switching with your friend will give you the winning ticket? It’s 50:50, either you have it or they do, the scenario has already run through every other premise and you’re just left with premise 3, to switch or not.
Using the accepted solution for The Monty Hall Problem, your ticket is 1/possiblelotterywinchance and there’s is possiblelotterywinchance-1/possiblelotterywinchance your friend has it. But your friend already told you only one of you has it, you know that for a fact (as long as your friend is telling the truth which we’re assuming they are for this scenario). So the answer The Monty Hall Problem gives you, is fundamentally flawed as we have only two denominators and one numerator. 1/2 in fraction form.
The lottery scenario is the same as the The Monty Hall Problem, but I replaced the doors with lottery tickets and the host with your friend, that is all I did and The Monty Hall Problem fails dramatically.
This is why I don’t understand why the solution presented to The Monty Hall Problem is the consensus, it fails on a basic mathematical, statistical and presentation basis. Never in Maths and Statistics do you present three separate premises as one premise.
If I ran this scenario through a computer, there’d be 6 possible outcomes, a win for switching, a win for sticking, a loss for switching, a loss for sticking and two eliminated outcomes. This leaves us with 2 possible wins to 2 possible losses out of 4 probable outcomes. this gives us 2/4 no matter what we choose which a computer logically defines as 1/2.
You next question is why did I eliminate 2 outcomes? Well one outcome is picking the same door as the host, since we know the host cannot pick our door that outcome is eliminated. The other option is switching to the door the host revealed, we’re never presented with the option to pick that same door again, we cannot pick it according to the game show scenario and to the fact that we’re not stupid enough to pick it if we were given the option, so that outcome is eliminated. That is why I eliminated them, they are logically impossible outcomes.
Next I’ll use your table to show the flaw in your argument.
Line 1, works perfectly. Host gets to pick between wrongs.
Line 2, the host doesn’t get to pick.
Line 3, this is just line 2 the host still doesn’t get a choice.
Because of this, your table is showing an eliminated outcome along side an actual outcome. As long as the host cannot choose the eliminated door these outcomes do not have the same value, either they split one value in half between them or one takes all the value making the other impossible.
This works for every three lines of your table, one of the three must be eliminated.
Line 4, the host cannot pick anything but the remaining wrong door.
Line 5, works perfectly. host gets to pick.
Line 6, this is line 4 again.
Line 7, the host can only choose a specific door.
Line 8, same as line 7.
Line 9, works perfectly. Host gets to pick.
If we eliminate each repeated line we get 3 wins to the stay and 3 wins to the switch. 3:3 in ratio speak, which is equivalent to 50:50.
Your table is flawed as it assumes the host is making a choice in every outcome. When they’re not making a choice, the other scenario when they’re not making a choice are the same scenario, it’s not a different outcome, it’s a repeated outcome. The simplest way to show this is to replace door numbers with the values the table assigns.
When you pick a winning door, replace the door number with win. When you pick the wrong door, replace the door number with lose. See any repeats? If you do that’s a problem, because it means that you’ve set up your table with a 1/3 chance on the winning door, a 2/3 chance on a losing door and a 2/3 chance on another losing door before you even started, giving you a total of 5/3 which is a statistical impossibility. The denominator must never be bigger then the numerator or you have more then a certainty which statistics never allows. Statistics always starts with a certainty which is, you will choose one, it then divides that certainty by the number of choices, which of these three will you choose. And finally it multiplies by how many corrects there are, there is one favourable result. You will choose one = 1, which of these three = 3, there is one favourable result = 1.
1 / 3 * 1 = 1/3
If you remove a door, we have choose 1, of these 2, 1 favour.
1 / 2 * 1 = 1/2
Switch effectively means re-choose in this scenario. You’re deciding over and you have two choices in your new decision regardless of what you decided previously.
Long post, really got into this. Again I find it weird that The Monty Hall Solution is so widely accepted when it’s so flawed.
Reading your revisited article now and as a programmer I can already see a flaw in your code. The software isn’t recording the result for staying and switching, your software is recording the result from your initial choice, your choice being right and one of the two others being right. Basically what are the chances of you being right when you pick a door. 1/3, there’s three doors. ELSE isn’t doing what you want it to, it’s not deciding switches you’ve just named it that.
NAME doorOne doortwo doorThree
COPY (doorOne doorTwo doorThree) doors
COPY 10000 rptCount
REPEAT rptCount
SAMPLE 1 doors prizeDoor
SAMPLE 1 doors guessDoor
IF guessDoor = prizeDoor
SCORE 1 stayingWinsScore
ELSE
SCORE 1 switchingWinsScore
END
END
SUM switchingWinsScore switchingWinsCount
SUM stayingWinsScore stayingWinsCount
DIVIDE switchingWinsCount rptCount switchingWinProbability
DIVIDE stayingWinsCount rptCount stayingWinProbability
PRINT stayingWinProbability switchingWinProbability
You give that code to a competent programmer as a solution to the problem and they’ll tell you to try again and stop wasting their time.
For your code to work, your code needs to incorporate the reveal of a door before recording if you win or lose, since that’s how the problem is set up.
A simple fix is:
NAME doorOne doortwo doorThree
COPY (doorOne doorTwo doorThree) doors
COPY 10000 rptCount
REPEAT rptCount
SAMPLE 1 doors prizeDoor
SAMPLE 1 doors guessDoor
SAMPLE 1 doors revealDoor
WHILE (true)
IF prizeDoor != revealDoor
IF guessDoor != revealDoor
IF guessDoor = prizeDoor
SCORE 1 stayingWinsScore
BREAK
ELSE
SCORE 1 switchingWinsScore
BREAK
END
ELSE
‘guessDoor cannot equal revealDoor try again
END
ELSE
‘prizeDoor cannot equal revealDoor try again
END
END
SUM switchingWinsScore switchingWinsCount
SUM stayingWinsScore stayingWinsCount
DIVIDE switchingWinsCount rptCount switchingWinProbability
DIVIDE stayingWinsCount rptCount stayingWinProbability
PRINT stayingWinProbability switchingWinProbability
With this your computer should return a 0.5 (Or similar) chance for each staying and switching results, or something close.
Your initial code missed key variables such as the reveal door not being equal to the guess door and the reveal door not being equal to the guess door. In fact you didn’t even include the reveal door at all, which is a major factor of the result as it impacts the other two variables.
If your code is the code being given out to people as a 101 of statistics, I fear for future statisticians as they’re being fed incomplete code and being told it’s complete.
Code is touchy, miss something or misunderstand it and it’ll do something different to what you want without telling you that’s what it’s doing. It doesn’t tell you this because you told it that the thing you didn’t want it to do is the thing you want it to do, it doesn’t know any different.
So what did I do to the code, I added in premise 2. The host removes a door. This door shall be known as revealDoor. Let’s add that variable in.
We know that the prizeDoor cannot be equal to the revealDoor, so that’s our first addition after adding the variable. We want to know if the reveal door is equal to the prizeDoor and if it isn’t we continue, otherwise we restart. Ah, I thought we missed something, before starting our arguments (IF), we need to add a restart for when when a the revealDoor equals either the prizeDoor. WHILE will tell us to keep going as long as we don’t BREAK the cycle. ELSE doesn’t BREAK it so it try’s again without reaching a result and without using up an attempt. We want it to do this so that our prizeDoor fulfills the condition of not being the revealDoor like in our scenario.
Next, same thing for comparing the guessDoor with the revealDoor, continue when they’re not the same, restart if they are.
And that’s it. I revised this multiple code multiple times and rewrote it whenever I found an issue or mistake, the repeat was actually added at the end when I realized we’d lose attempts if we didn’t repeat on null attempts. BREAK simply refers to end of a loop caused by WHEN.
I’m not sure if my additions match up to your code language as your code doesn’t appear to match with C# or C++ and you didn’t state the programming language in your example. I used C++ when adding WHEN and BREAK for the repeat, please use your programming languages equivalent when coding it in. (Assuming they’re not the same that is.)
Ok, so first I revealed the flaws in your table, then I revealed flaws in your code. What’s next.
How about your trial with your daughter? Everything else before this part relies on your flawed code so we can skip it as debunking the code debunks the things that use the code.
First, well done for beating the odds, but it proves nothing apart from you having poor instincts before switching. Theory and Practice can have differing results because of thing known as True Random. Even if a 50% chance of being right, you can be right 100% of the time in your sample. It’s a gamblers fallacy to think your previous results affect your current results, you’re not guaranteed to get it right 50% of the time or 30% of the time or even 1% of the time even if your chance is being right is 50%, alternately you’re not guaranteed to get it wrong 50% of the time or 30% of the time or even 1% of the time even if your chance is being right is 50%.
And that’s where your other one ends…
Geez, I added another length to the post. But I managed to debunk every example you had. The number of broken methods used to enforce the falsity is insane.
It’s always interesting to see how strongly some people cling to the incorrect answer. I’m not going rebut your points here. I’ve rebutted similar points throughout the posts and other comments.
I do want to point out that the code I use in the other post is correct. The way it works is quite simple. If you stay and win, that means you’d lose by switching. And, if you’d lose by staying, that means that you’d win by switching. In other words, switching causes the opposite result of staying. And, that’s how the code works. Simple logic.
Think of it from the standpoint of Monty’s process. And, it’s easier to understand when you have more doors. You have one door and Monty has a set of doors. Let’s say there are 100 doors. You pick one and Monty has the other 99. Monty’s process is to open all doors except one while taking care to not open the prize. Given that process, if the prize is behind one of Monty’s original set of doors, it’ll be behind his one unopened door. So, you pick your door. Monty then opens 98 doors from his group, which leaves one opened door. We know that if the prize was originally in Monty’s group of 99 doors, it is now behind his one unopened door.
Do you switch? Of course! There’s a 99% chance that the prize is behind Monty’s unopened door because he systematically opened 98 non-prize doors. The only way you win by staying is if your 1/100 initial choice is miraculously correct. It’s possible but unlikely.
The problem with 3 doors follows a similar logic–just fewer doors. If the prize is behind one of Monty’s two doors, then we know it’ll be behind his one unopened door at the end. Two-thirds of the time it’ll be behind his set of doors. Therefore, two-thirds of the time it is behind his one unopened door. That’s why you win two-thirds of the time by switching.
Thank you Ronald,
I am convinced that Jim is right.
It is confusing because as Colin very clearly explained there is actually two separate parts to the game.
Because the first choice does not result in any opened doors it therefore has no relevance to what follows. Any door of the three you choose will not win you the car. You had a choice of 1 out of 3. But SO WHAT? you won no car; But Monty steps in and reduces your choice now to 1 out of 2 and this time we are going to open the door. So for the second part of the game your chances are 50/50.
Now for the switching problem.
Harry has carefully listed out all the options, but this is not entirely accurate. Note on the left side he has listed four options after picking door #1, but only two for each when choosing either of the other two doors.
This is the error.
Note that if #1 has been chosen, we leave Monty a choice; the two remaining doors are now equivalent. Option #1 and #2 together is identical to option #3 and #4 together. We can’t distinguish these so we should rule out one pair. This leaves a total of six options.
If door #2 (or #3) is chosen we constrain Monty to open only one door because he knows where the car is but from our point of view we still don’t know where the car is. But we do know that of the two remaining choices that Harry has listed we are better to switch. Of three choices we had originally, in two cases Monty was constrained, unbeknownst to us, to actually convey the information of where the car is. In only 1 out of our original 3 choices did Monty have a choice, but again we do not know that. Hence our 2 out of 3 chances of winning, but only if we switch. If we decide to stay we will only win 1 out of 3 times.
So with just the two choices we have available we will win if we switch 67% of the time; if we stay with our first choice we will win only 33% of the time. It’s all in how you frame the question.
Regards to all
Russell
Hi Russell,
Hey, that’s great that we’re all on the same page now! I have just a couple minor quibbles with what you write, but in the main we’re on the same page.
The first choice does affect things. It defines the outcome for both whether you stay or switch. For example, if you by chance pick the door with the prize and stay, you do in fact win the car.
Also, in this case, it’s not how you frame the question because the Monty Hall Problem is based on a specific set of rules that are clearly stated. Instead, I’d say that it is easy to overlook the implications of those rules. Specifically, it’s easy to overlook the fact that Monty uses his knowledge to affect the outcomes in a non-random manner. It’s also easy to overlook the conditional probability that if your initial choice is incorrect, then the probability that prize is behind the one remaining door is 100%. In other words, when your initial choice is incorrect, Monty’s intentional process winnows the other doors down to the one with the prize. Hence, because there’s a two-thirds chance that prize is behind the other two doors that you did not choose at the beginning, there’s still a two-thirds chance that is behind the single other door at the end.
As other readers have mentioned, this process is easier to understand when you have more doors. Suppose we follow the same rules except that we have 100 doors. You pick one door and Monty has the other 99 doors. There’s a 99% probability that the prize is behind one of Monty’s doors. Next, Monty opens 98 of his doors one-by-one while taking care not to reveal the prize. In the 99% of the cases where your initial choice is incorrect, this process systematically winnows Monty’s set doors down to the one that has the prize. Consequently, in this scenario, you have a 99% chance of winning by switching. You only lose by switching when your initial choice of one door out of 100 doors is miraculously the correct choice!
I’m not sure if you’ve read my follow post to this one. If not, you should check it out!
Hello, Russell.
What tends to confuse in this problem is the presumption that each of the options must be equally likely in any case, as if it was a rule, but note that it is not always true. For example, when two people are competing in something, the odds of winning do not have to be the same for each one, right? We could put someone random from the street to run in the 100 meters against Usaint Bolt, and in that case it would be incorrect to say that both have a 50% chance of being the winner just because they are two options. There is a clear advantage in the case of Usaint Bolt, for having been the world champion, while the other person may not even be a runner.
In Monty Hall case, it occurs the same reasoning. You will always end with two options, but they were left by two different persons, one with more chances to leave the correct one than the other person. The contestant chose one door randomly from the three, meaning that in 2 out of 3 times on average he would fail. On the other hand, the host knew the positions and couldn’t reveal the contestant’s selection and neither the prize one, which means that everytime the contestant fails (2 out of 3 games), the other door the host leaves closed is which has the prize. So, we will always end with two doors, but the switching one will have the prize in 2 out of 3 times on average, not in 1 out of 2.
To make an analogy, if the 50% chance was right, then you could win the jackpot of the lottery with 50% chance too, which is absurd. You would only need to follow this strategy: Suppose you buy a ticket and its number is 456432. You don’t see the results on the day of the contest but you tell a friend to do it for you. You tell him that if your number was not the winner, he must write yours and the winner together on a piece of paper. For example, if the winner is 989341, he would have to write:
/////456432,,,989341/////.
On the other hand, if by chance of life yours was the winner, then he would have to write yours and any other that he could think of. For example, he writes:
/////456432,,,278226/////
He gives you the paper but you still don’t know if you won or not. Note that with these conditions you have managed to be in the same situation as in Monty Hall problem: despite what your first selection is, there will be always two possible options remaining on the paper, one of them is necessarily the winner and your option is also one of them. All the rest is discarded. But do you think yours is 50% likely to be right and so you will start winning the lottery 1/2 of the time applying this?
Of course this is incorrect. There are two options but the prize was not distributed on them with a 50% random process. It was very difficult for you to buy the correct number, and since the other one that your friend writes must be the correct if you failed, once you see the paper you know it is almost a certainty that the other number is the correct, not yours.
Thanks for the great explanation, Ronald! That’s a great point. Just because there are two options it does not mean that the probabilities are equally split 50/50!
Let’s break it down
1/2/3
Let’s say car is number 1 but the contestant didn’t know that.
Scenarios
1. #1 is picked, you opened #2, he stayed, he won.
2. #1 is picked, you opened #2, he switched, he lost.
3. #1 is picked, you opened #3, he stayed, he won.
4. #1 is picked, you opened #3, he switched, he lost.
5. #2 is picked, you opened #3, he stayed, he lost.
6. #2 is picked, you opened #3, he switched, he won.
7. #3 is picked, you opened #2, he stayed, he lost.
8. #3 is picked, you opened #2, he switched, he won.
If you only switch, 2 out of 4 you’ll lose, and if you only stay, 2 out of 4 you’ll won
4 out of 8 scenarios that the contestant won, it’s a 50/50 situation. See?
Hi Harry,
There are multiple problems with your list.
Your list doesn’t factor in which door the prize is behind. Look at #2 in your list. In that scenario, the prize is behind door #2, so Monty would not open #2 because that would reveal the price, but would instead open #3. Hence, when the contestant switches, they switch to door #2 and win.
Additionally, your list doesn’t correctly list all of the scenarios. For one thing, there are only 8 scenarios in your list. There are 9 scenarios, which are based on the three possible doors you can choose initially and the three possible doors that the prize is behind (3 X 3). You have 4 scenarios for picking door #1. Only two scenarios for picking door #2 and door #3.
Nice try, but the solution is still 66/33. Read the post I just published (Revisiting Monty Hall) where I solve this problem using both a computer simulation and an empirical experiment.
Hi Jim this is the clearest explanation I’ve read, thanks so much for it. To everyone who doubts what Jim is saying I can guarantee you that you are mistaken.
Thank you, Stephen. I really appreciate that!
Very soon I have a follow up post coming out that looks at this problem from several different angles.
Hi Jim,
I believe there is two questions in this problem:
1) poor definition of the original question. Does the game player increase his probability of success;
a) ‘from the original game question’ i.e 1 out of 3 doors or
b) does he increase his probability of success in a ‘new game’, 1 out of 2.
If the correct question (game definition) is ‘a’, then your answer is correct – by changing, the probability is increased to 2/3
However, if ‘b’ is the now correct answer i.e. you have started a new game – then changing has no effect and the probability is 50/50.
I suggest the problem is poor definition of the game – not the statistics or illusion.
Thanks – really great stuff and thought
Colin Scrivener
5/7/19
Hi Colin,
You make a great point. The exact definition of a problem is critical for determining the correct answer.
For the Monty Hall Problem, there is one standard definition for it as a puzzle.
It’s all one game. A game consists of the following:
1) Three doors with a prize randomly behind one.
2) The contestant chooses one door, but it stays closed for now.
3) The host knows where the prize is located and will always open one of the other two doors that does not contain the prize.
4) The host offers the contestant the option to switch to the other closed door.
5) The contestant decides to stay or switch and their door is opened to reveal the outcome.
That’s all one game that involves the same three doors and the prize does not move. Another assumption is that there is no cheating or switching of any kind. It’s a fair game.
I’ve defined all in the post but wanted to reiterate it here.
Thanks for the great comment!
Sorry Jim for being a bit blunt. You are an expert and I respect that. But there is clearly an anomaly in your argument.
Going back to your opening remarks in the introduction to this discussion there is a section there headed ‘Here’s how it works:’
1. You pick the incorrect door by random chance. The prize is behind one of the other two doors.
BUT how do you know? You still only know that the prize is behind one of all three doors.
Say you actually picked the correct door by random choice. Do you say now, as before, that the prize is behind one of the other two doors? Monty opens his door. So you switch and there is no prize.
This outlines the two cases: you either pick the door with a prize or, more likely pick one of the no prize doors in your first move. You get another choice but this time we will OPEN the door you choose. How many can you choose from? You know for certain that the prize is behind one of those two doors, one of which you merely pointed to earlier. You choose one by, tossing a coin, or by gut feeling, and making a decision to stick with your first choice, or switching. You have a 50% chance of winning whichever you choose.
Surely it is clear that your very first choice is irrelevant.
In your introduction you say:
3. By process of elimination, the prize must be behind the door he does not open.
There is no process of elimination. Your choice has eliminated nothing. Monty has helpfully eliminated a false choice for you.
Can you see that in your description you eliminated in your mind the first door you chose because you assumed that your choice was wrong; and therefore concluded that the remaining door of the three was the right one. If you picked the wrong door the first time and then switch you will win every time; if you picked the right door the first time and you stick with your first choice you will win every time.
You just have two options or 50% chance.
Best regards, Russell
Hi Russell,
This isn’t “my argument.” This is the accepted solution by the mathematical/statistical community. This puzzle tripped up a bunch of people in the 1980s. Now, it’s a solved problem that has been proven mathematically, by simulation, and even empirically by the Mythbusters (CONFIRMED!) and James May’s Man Lab.
Do you really think you’re seeing something new in the problem that no professional mathematician has understood?
The point is that you don’t know which door the prize is behind. That’s why it’s a probability and not a certainty.
There is a 1/3 chance the prize is behind the door you pick originally. Simple probability. So, you have a 1/3 chance of winning if you stay with the original choice.
Consequently, there is a 2/3 chance the prize is behind the group doors you didn’t pick. In that scenario, Monty opens the door that the prize is not behind, which means that it must be behind the other door (that’s where the process of elimination comes in). Ergo, there’s a 2/3 chance that the prize is behind the other unopened door. (The probability for that door remains at 2/3 because, in this scenario, there is a 100% chance that the unopened door in this group has the prize–and 1 * 2/3 = 2/3. It’s a conditional probability.) Consequently, you have a 2/3 chance of winning by switching to that door.
In a nutshell, you only had a 1/3 chance of your first choice being correct, so there’s a 2/3 chance with the other two doors. If you remove the non-winning door from the other two doors, it’s just one door, which takes the full 2/3 chance by itself.
Or, said another way, switching always reverses the outcome. You have an initial 2/3 chance of losing. Consequently, switching reverses that and makes it a 2/3 chance of winning.
We’ll have to leave it at that because we’re going around in circles. My recommendation would be to read more about this problem. Maybe someone elsewhere explains it better than I do. I’d also be careful about assuming you know better than the entire mathematical community!
Thank you Jim for your reply.
How can you assert that Monty doesn’t ignore my first choice?
It’s good you rehearsed exactly the procedure; I began to wonder if we were talking about the same thing.
You say my first choice has a 1/3 chance of being correct. If I did choose the correct one I will not get the prize. That’s why I say that regardless of my first choice I will not know what was behind that door. I could choose all the doors but I still won’t win the prize because that door is never opened when I choose it.
If I choose a door AND OPEN IT then I have the 1/3 chance of winning. If I didn’t win the first time then I have another choice, now between the two remaining doors and now for the overall chance of winning the game, yes, I do have a 2/3 chance of winning. As you rightly say, Monty helps you by discarding your choice (i.e. not opening the door), even if it was correct. He now eliminates a no-win door and basically starts the game all over with only two doors to choose and THIS TIME he will open the one you choose. NOW your chance of winning, for this choice only, is 1/2. Your suggestion of now switching doors still gives the same answer, i.e. 1/2.
I ran a series of trials as you suggested and found the answer homing in to 50%. This was what made me look a little closer.
Please stop counting my first choice as part of the game. You say in your reply that my first choice has a 33% chance of winning. This is simply NOT true. Your first choice and my first choice have a zero – 0% – chance of winning BECAUSE Monty has intended to ignore EVERY TIME. You then state that there is a 2/3 chance the prize is behind one of the other two doors. This is wrong. All the while you have three identical doors closed in front of you the chance of winning is still 1/3. And you still see the prize as equally likely to be behind any of them. Monty knows and to make it easier for you to win with only one opportunity to choose he eliminates one of the options by opening a no win door.
Look at it as if you had two choices and three doors and Monty stands back watching. Thus you OPEN each door you choose. Then on the first choice you have 1/3 chance; so then you take you second choice and your chance will be 2/3 over a large number of runs. There is no increase of probability by swapping doors. The answer is exactly 66.7%
In your reply you say “Hence the prize is behind the other closed door”. But the prize may be actually behind the door you first chose. Think hard about that and you will see how you have jumped to an unfounded conclusion that the prize is not behind the door you first chose.
I don’t know how to explain it clearer.
Look forward to your comments.
Russell, I’ve noticed in your comments a tendency for snide comments and borderline rudeness while discussing this issue. I’m happy to discuss this with you, but only if you keep it polite. It’s all supposed to be a friendly, positive discussion.
I don’t understand why you think Monty ignores your choice? The rules by which this puzzle works is that Monty does not cheat. He doesn’t move the prize or ignore your choice. The assumption is that it’s all fair play.
Your first choice is a crucial part of the game. It defines the initial probability of your first choice being correct at 1/3 and the other two doors having a combined 2/3 probability. Of those two doors, Monty systematically removes a non-winning door. Because it’s systematic, non-random decision to remove a non-winning door, the remaining door still has a 2/3 chance.
The prize is randomly placed behind one of the three doors. Consequently, when you pick your original door, it has 1/3 chance of being the door with the prize. If you stay with the original door, you, therefore, have a 1/3 chance of winning. Your original choice matters.
And, if you have a one-third chance of winning by staying, logically you have a two-thirds chance of winning by switching because you can only switch to one door.
As for your series of trials, I’ve done some simulations. What we’re trying to do is distinguish a difference between a 50% of winning and a 66% of winning. The simulations I’ve run suggest you need to run it a good 50 times to really be sure. Probably more. I’ll have more on that later.
I left a note on here yesterday Jim. However, on thinking this over I am wrong.
Let’s take Monty right out of it.
You are confronted with three doors, one of which conceals the prize. You have one choice. Your chances of winning are 33.3%
If you are then given another choice, the remaining two doors are equally likely to conceal the prize. You choose either one and your chance of winning in this choice are now 66.7%.
HOWEVER
Monty steps into the picture.
He lets you make a choice of one of three doors while he knows what is behind each one.
NOW
He now proceeds, regardless of which door you chose, and it could have been the winning one, he proceeds to DISCARD your choice, and makes an informed choice of his own.
Notice that your choice was never fulfilled; the door remained closed.
What he has done is to up your chances by removing one door and he now asks you to choose one of the remaining ones, one of which conceals the prize.
NOW you have a 50% chance of winning. This is the only real choice you make because your first choice was disregarded and discarded.
If you have 100 people who want to play this game you will need fifty, yes 50, prizes to surrender.
You make two faulty assumptions in your ill thought out table in this article.
1. You count the door you pick first as though it is a win or loose. BUT that door remains shut. So that scenario can be ruled out.
2. Then, as some of your readers pointed out, you make it look like you get two choices in the next step whereas you either have one or the other.
I’m not very impressed with your mathematical logic or statistics.
Please reply to my email.
Hi Russell,
I’m not quite sure I understand what you’re saying. Monty doesn’t discard your choice at all. And, I assure you, the proven answer is that if you switch, you will win twice as often.
Think of it this way. Your original choice has a 33% chance of being correct and 67% of being incorrect. Consequently, 2/3 of the time the prize is behind one of the doors you didn’t choose. In that case, Monty opens the door without the prize. Hence, the prize is behind the other closed door. Monty doesn’t negate your choice. He just eliminates one door that the prize is not behind. That informed, non-random action on Monty’s is what increases your chances.
I’m always amazed at how many people doubt the proven answer. One thing I’d suggest is to play the game with a friend. One of you can be Monty and the other the contestant. Follow these rules. There are three doors. Place a prize randomly behind one door. The contestant selects one door. “Monty” then opens one of the other two doors that contestant didn’t pick and doesn’t contain the prize. The contestant is allowed to stick with the original door or switch to the other unopened door. For each round, record the result for if you stay with the original and switch to the other door. Repeat this multiple times and you’ll quickly find that you win more often by switch.
In the near future, I’m going to add the results of computer simulation to this blog post. But, I think the best way to dispel any doubts is to simply try it yourself!
The table you have drawn up Jim is correct; that is a complete representation of all outcomes. However you have overlooked the fact that the person doing the choosing has only two choices, not the three you have listed. If you now go through your table and rule out one of the mutually exclusive outcomes and then add up the columns you will get the right answer. Which is ??.
From pure logic: Say the problem is presented to you. You make your choice; Monty makes his. Now you go away and have a coffee or go away for a week. Now the prize is exactly where it was before and when you come back you can’t remember which door you chose (or even if you did) what is the probability of the prize being more likely behind one door than the other? How possibly could the choice you made a week before have any influence on the location of the prize?
You are obscuring the truth by baffling people about information provided by Monty in opening one of two wrong doors. He gave you no useful information – you already knew there were two false doors; he simply reduced your choice to two equally likely doors but you still have no clue as to which it is.
I don’t believe there have been computer runs to prove your theory. If the thinking is biased to a given result it is simply ‘garbage in – garbage out’
Now go back and rearrange your table as a real situation giving each scenario with making only two choices. Add the numbers and the answer is the same for both columns: 0.33
Is it not, then, equally likely that you choose doors “2 or 3” when the prize is behind door one? Once more, the outcome would be the same if you switch no matter which door you chose, and if we use the incorrect doors as a unit in one instance, we’re forced to do so in any instance. How can it be that listing two separate outcomes is only artificially inflating the outcome when it benefits your hypothesis to call it so? Citing established belief as the only “correct” solution is a poor precedent to be setting when that belief relies on biased calculation
Hi Brent,
If choose door 2 or door 3 and the prize is behind door 1, that’s covered by different rows in the table.
The table lists multiple doors under the Monty Opens column only when your initial choice is the door that prize is behind. From Monty’s perspective, he will reveal one of the two choices. It doesn’t matter which one because it won’t change the outcomes or probabilities.
The logic behind this problem is that your initial choice is most likely to be wrong. There’s two-thirds chance that you’ll pick the wrong door. That’s why it’s beneficial to switch because your initial choice is probably wrong. Monty helps by removing one of the incorrect doors from the set of doors you didn’t select.
The reason this hurts your brain so much, and the point behind the blog post, is that most people go into this problem with the wrong assumptions. We’re thinking independent, random probabilities. In reality, it’s a mixture of random probabilities when you first choose but then Monty acts with intention to modify the probabilities going forward.
No one is saying that this is true because it’s “established belief.” It’s true because it’s been proven mathematically and it’s been proven using computer simulations that run through it thousands of times. Further, this is an experiment you and a friend can do on your own where one of you is the contestant and the other is Monty. Just run through this scenario a number of times, record the results, and see how you fare when you switch and don’t switch. The Mythbusters did this and verified it empirically.
Yeah, I know, it hurts the brain, but it’s true!
Hi Jim,
Another way to look at it:
1. The probability the contestant chooses the door with the car behind it is 1/3
2.The probability Monty chooses the door with the car behind it is 0
3. The probability the unchosen door hides the car is therefore 2/3
And the contestant can deduce this before he even makes his choice. 🙂
Hi Frank,
That’s a great way to look at it too! It’s interesting how there are common sense ways of seeing the problem like that, yet the solution caused such a commotion originally!
Jim,
I do understand the difficulty of presenting the information in an intuitive way, considering how subjective intuitiveness is to begin with. I guess what I would do is to accompany the table with a probability tree, which is a good way to visually illustrate the conditional probabilities.
By the way, I love the optical illusion in the beginning. I needed to use a color picker to convince myself that the colors of squares A and B are (nearly) identical. Even knowing that, I still “see” the difference between them, and always will. Mind-blowing!
To the folks asserting that Jim has cheated by clumping outcomes together, consider this somewhat simpler scenario. I plan to flip a fair coin. If I get heads, I will drink coffee and eat chocolate cake. If I roll heads, then I will flip the coin again. If I gets heads that time, I will eat ice cream. Tails, and I will eat pecan pie.
Here is a table of possible outcomes:
| First action | Second action | Third action
—————————————————————–
| Roll heads | Drink coffee | Eat chocolate cake
| Roll tails | Roll heads | Eat ice cream
| Roll tails | Roll tails | Eat pecan pie
The way I have presented this makes it look like these are three equally likely outcomes. They are not! The probability that I end up eating chocolate cake is 50%; ice cream and pecan pie are 25% each.
Presenting the table shown earlier with separate rows for each of Monty’s possible choices introduces a similar catch. Those rows would not be equally probable as the ones where Monty has only one choice. Individually, they would be half as probable, for the same reason that eating ice cream in my example is half as probable as eating chocolate cake: conditional probability.
That is a cogent observation, and it seems you are on the right track to understanding the solution with a little more consideration. No offense to Jim Frost, but I think that the table is a little bit misleading, and somewhat obscures the correct solution. The key thing to understand about the table is that the “Monty Opens” column is extraneous; it’s just a function of the first two columns. I think the example would be clearer with the “Monty Opens” column removed entirely.
Hi Zach,
I sort of see what you’re saying. The “Monty opens” column might add a bit of confusion. And winning is a function of the first two columns. However, I’m trying to clarify the process–specifically Monty’s non-random decision about which door to open. You pick one door. Monty opens one of the remaining doors based on his knowledge, which connects to what is ultimately behind the door you switch to.
I’ll have to think if there is a clearer way to present that process information.
that is still incorrect. You are deliberately ignoring different possibilities. Your percentages represent of percentages of total possibilities and you are deliberately two separate outcomes as one. Your very first column of your chart, for example, has “2 or 3” as the door that Monty opens. If you choose the correct door you cannot lump together whether or not he chooses one incorrect door or another incorrect door as a single possibility. They are two separate outcomes and when calculating a percentage of total outcomes, you have to be included for the sake of correct mathematics. I do not know how to express my thoughts in terms of probabilities but based on your explanation I would expect that the probability calculations are doing something similar and do not account for different outcomes when choosing the correct door, instead and correctly lumping them together because it is “irrelevant” which it most certainly is not.
Hi Rachel, I know the answer doesn’t appear to make sense at first glance, but it’s been the recognized solution for decades now.
As for the table, yes, Monty can do one or the other but not both. And, both actions produce the same outcome. So, the table is correct and consistent with the recognized solution.
I think the key is that Monty’s turn disregards the door you have already chosen. This means that the probability of your choice being incorrect remains as 2/3. Monty’s turn doesn’t change this because he knows where one of the goats are and your chosen door wasn’t part of his selection.
To put it another way, if you lump together the two unselected doors (before Monty opens one), they have a combined probability of 2/3. The chances of of these two cards containing the prize are 2/3 regardless of Montys selection. This doesnt change because your selected door isnt included in his selection so there is still a 2/3 chance that the remaining door contains the prize.
Hi Matthew,
That’s a great way to explain it!
That is just wrong. You are listing one outcome for 2 different outcomes. If he opens 2, that is not the same as opening 3. You are grouping them together, they are separate outcomes. The chance when having 2 doors is 50/50, and you cannot count the odds of the door already eliminated. This isn’t very hard.
Hi Billy,
Back in the 1980s, there was some debate about the correct answer. However, over the intervening decades, the consensus has converged on 67/33 split as the correct answer. If you search around, you’ll see that there is no longer any debate about the correct answer. People have run computers simulations and have gotten this result. So, there’s no doubt that you have twice the chance of winning if you switch doors.
As I explain in the post, it’s easy to get caught up in the illusion that it’s 50/50 because you have two doors. However, there are underlying assumptions behind that outlook which are not being met. Namely, you’re assuming random, constant probabilities. Instead, Monty acts based on inside knowledge which changes the probabilities in a non-random manner. And, that’s why the results aren’t what you expect.
As for the cells in the table that have two outcomes. As I explain in the post, Monty can open either one and it does NOT change the outcome.
I will update this post at some point to include a computer simulation where I can run this experiment many thousands of times.
One way to grasp the concept of how Monty’s knowledge has affected the final choice is to multiply the number of doors. If there were 100 doors, and then Monty eliminated 98, you can see that unless your initial choice was the 1 out of 100 right door, then Monty has shown you where the prize is located.
I would say this problem is a mischaracterization of what Monty actually did. For the final deal, he always showed the least valuable prize first even if a contestant picked that door. Then he would show the middle prize, and finally the most valuable prize, even if nobody picked it.
For the regular part of the game, he would offer what the contestant turned down to other contestants. Then he would reveal the zonk – so the contestant could wind up with a zonk on the first selection.
Hi Joe, that may very well be the case. I did catch a few episodes of the show way back when. However, when people talk about the Monty Hall Problem, they’re historically referring to the scenario as described in this post. This scenario might be slightly different than the various ways he presented the information in the game show. It’s apparent you know how he worked better than I! Thanks for the information.
Your table is missing lines. the “Pick door 1” section should read
1 1 2
1 1 3
1 2 3
1 3 2
There are 2 doors Monty can open if you happen to choose the right door the first time. Don’t know why you lump them together.
He can only choose 1 door if you choose the wrong door the first time
Hi Jack, I explain the reason for this in the post. When you pick door 1 and the prize is behind door 1, Monty can pick either door 2 or 3. However, the outcome is the same. For example, if you don’t switch, you win. However, if I list both options on separate lines, that artificially inflates that outcome because it is listed twice. It’s the same one outcome for one scenario, so it’s listed once. You can see this for yourself. If you fill in the table as you show it, the probabilities don’t work out correctly.
Monty knows which door to open. So the choice of which door he opens is not probabilistic. That changes the probabilities. If he were to open a door at random, then your analysis would hold true, but he doesn’t, he removes a known empty door from the game.