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.

Monty opens one of the other two doors, and there is no 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!

To learn about another probability puzzler, read my post about answering the birthday problem in statistics!

Zoot says

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.

Jim Frost says

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.

John Puopolo says

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.

Ronald says

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.

Adnan says

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?

Klaus 74 says

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.

Trevor says

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

Ronald says

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.

CuriousGato says

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.

Trudy Mahoney says

I don’t get the optical illusion, but I can understand the Monty Hall scenario.

Ronald says

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).

Vienna Raglin says

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.

Mona says

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.

Mel says

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.

Jim Frost says

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%.

Ken says

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.

william rich says

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.

Raymond Taddeucci says

Yes I realized that too by increasing the number of doors. I have finally given in to the fact that I was wrong.

Jim Frost says

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!

Zach says

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.

Geoff says

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.

Jim Frost says

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.

John Anderson says

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.

Raymond Taddeucci says

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.

Jim Frost says

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.

Raymond Taddeucci says

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.

Jim Frost says

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!

Ronald says

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.

Vince says

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.

Jim Frost says

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!

Nubley says

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.

Jim Frost says

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.

Beavis says

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%

Keith Hawley says

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.

Rodrigo Mourão Nunes says

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

Jim Frost says

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.

Keith Hawley says

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

Jim Frost says

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!

Rick Evans says

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.

Jim Frost says

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.

Rich at Large says

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.

David says

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.

Jim Frost says

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!

Ronald says

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.

Daniel Diggs says

This is a wonderful response I have never heard before. An excellent way of viewing the problem

George says

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

Ronald says

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.

Jim Frost says

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.

Ronald says

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.

Dean says

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.

Dean says

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.

Jim Frost says

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).

Ronald says

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.

Jim Frost says

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.

Dean says

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.

Jim Frost says

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.

Russell says

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

Jim Frost says

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!

Ronald says

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.

Jim Frost says

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!

Harry Chu says

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?

Jim Frost says

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.

Stephen says

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.

Jim Frost says

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.

Colin Scrivener says

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

Jim Frost says

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!

Russell Kennerley says

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

Jim Frost says

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!

Russell Kennerley says

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.

Jim Frost says

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.

Russell Kennerley says

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.

Jim Frost says

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!

Russell Kennerley says

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

Brent says

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

Jim Frost says

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!

Frank Goring says

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. 🙂

Jim Frost says

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!

Zach Dorman-Jones says

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.

Zach Dorman-Jones says

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.

Jim Frost says

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.

Rachel Williams says

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.

Jim Frost says

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.

Matthew Allen says

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.

Jim Frost says

Hi Matthew,

That’s a great way to explain it!

Billy says

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.

Jim Frost says

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.

Dan Holgate says

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.

Joe McCollum says

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.

Jim Frost says

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.

Jack2 says

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

Jim Frost says

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.

Jack says

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.