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! Learn more about the Fundamentals of Probabilities.

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

saeed says

I think there is an illusion in 33-66 calculation. The 33% chance of winning if you stay with your choice is correct, but chance of 66% winning if you switch is not correct. It is, in fact, chance of your pick not winning which by illusion considered as switching your pick.

Gregory A. Cheatham AIA says

This problem has been modeled many times by computer, and the percentage always works out to be 66%. It’s a settled question.

Carolyn B says

Yes – the penny finally dropped when I overcame my assumption that a new game/rules/statistics had been created by his door opening.

It’s so interesting – we run 2 parallel concepts in our mind simultaneously. First (and the strongest and most compelling and hard to dismiss) is the FACT that ‘the car is now behind 1 of 2 doors’. But the second concept is the STATISTIC that ‘the car is 2/3 more likely to be behind the door we should switch to’. I’ve reflected on this, and decided that when FACTS conflict with STATISTICS, facts generally win our cognitive beliefs hands down.

Mike K says

That’s exactly why Monty opens a door, to introduce a red herring to make you think everything has changed and been reset, when in fact revealing a goat behind one door (a fact that you already knew) changes absolutely nothing, except that the remaining unopened and unchosen door has now adopted the 1/3 chance of the door that was opened, and now has a 2/3 chance of containing the prize. Your brain will do everything it can to reject this because “it doesn’t make common sense”, but as I saw elsewhere, you confirmed this for yourself by actually running the experiment. 🙂 The truth is, it’s one game from start to finish, and your initially chosen door was chosen from 3 doors, so nothing that happens afterwards can alter the odds from 1/3 for that door.

Carolyn B says

This explanation makes the most sense. And YET my brain has STILL taken so long to accept it!! I had to do 20 game runs to really accept it, and the result was so clear. Staying = I won 7/20 times. Switching = I won 16/20 times. Using a random generator, exactly replicating the scenario.

Goddamn brain.

Carolyn B says

Yes Frederick! That is my view too. By the host effectively removing one option, it has created a new ‘game’. A new game (1 of 2 doors has the car) with new statistical assumptions (the car has 50% chance of being behind either remaining door). The statistics of the first 3-door scenario CANNOT be applied to the new statistic of the new 2-door scenario. New game = new rules = new statistics. I would love to be convinced to switch, truly, but I just can’t see it!!

Frederick Christopher Holder says

Okay; thanks for taking the time to respond.

Having rethought this, I reimagined it in a way that made more sense to me.

Specifically, if we look at this from Hall’s perspective, should I pick the correct door (1/3 probability), Hall has two choices of door to open; if I pick an incorrect one (2/3 probability), Hall only has one choice.

Ergo, there is about a 67% chance Hall only has one choice, and it makes more sense to select the other door.

Or I have managed to come to the right conclusion for flawed reasons.

Ronald says

Hi, Frederick.

This is easy to understand making an analogy with the lottery. I don’t think you believe you have 50% chance to guess the number that will win the jackpot of the lottery. To make the experiment, you don’t actually have to play the lottery; all you have to do is to guess the number. You tell me the number you thought and then the day when they give the results you don’t see them but you let me do it for you. If I see that your number was not the winner, I will say the winning number out loud to you, but if luckily yours was the winner, then I will say any other incorrect that I can think of.

In this way, the winning number will always be one of the two that one of us said: the one that you thought of first (at random), or the one that I said later (already knowing the results). It is like in Monty Hall when there are only two doors remaining. But do you think that each of them has 1/2 chance of being the winner? Or do you think that it was easier for me to be the one who said the right one?

This is basically what happens at Monty Hall. Since the host knows the positions and never discards the door with the prize, then other door he leaves closed is like the number I’d say, and the revealed doors are like the ones neither of us said.

The difference is that in Monty Hall you have the opportunity to switch your guess to the other option that I would say.

Greg CHeatham says

Because if you pick one its 1/3. If you get to pick a second one, that’s 2/3.

Frederick Holder says

Okay; I have little doubt that I have managed to totally screw up the issues around this, but one begin with a 1/3 chance and, as presented, one selects Door 1. Hall obviously can only open either Door 2 or 3; selecting 1 would remove the whole point of the show. The absolute only thing to be sure of then is that the prize is not behind the door he opened and must be behind the two untouched doors.

At that point, the contestant effectively begins with a completely different and new scenario and faces just two choices, at least as I interpret it. Again, this definitely could be wrong.

Please explain to me why it is wrong to see this as an entirely new choice, or perhaps, in a slightly different situation, how it differs from the scenario where the contestant approaches the three doors at the beginning, but, before having to chose, Hall reveals that Door 3 is definitely not the door. Thus the contestant faces just a 50/50 choice.

Vinay Mimani says

You are a great teacher Jim! Thank you

Does this line of thought make sense –

Anytime you made your choice, there’s a 66% chance that you picked the goat.

Monty comes along and tells you that you can count out at-least this one door.

Now you are still left with a 66% chance that the door you had picked has a goat behind it. Why? Because that chance was locked in when you made the choice.

Monty pointing out a door with a goat behind it can’t affect that %

So basically, now you have a selection with a 66% chance of it being goat with just one other choice. Which means this other choice has (1-0.66) = 33% chance of it not being goat (as a choice can only be a goat or a car, this means 33% chance of not being a goat is the same as 66% chance of being a car)

Thus, if you switch, you go from

**66%(33%) chance of selecting a goat(car) to 33%(66%) chance of not selecting a goat(car)**

Did this make sense?

Thanks!

Kicab Castaneda-Mendez says

Before the game starts there are nine combinations. After the car is placed behind a door, there are three combinations. After the contestant selects a door, there is one combination.

Mike K says

Hi Nqobzin! It’s because the choice is really between 3 doors, not 2, even after Monty opens a door to reveal a goat. Like I said elsewhere, the opening of a door by Monty is a “red herring” – it’s meant to make you think that things have changed, but in reality it doesn’t change a single thing. Think of it this way: you choose your door from 3 doors = 1/3 chance, right? That means the other two doors combined have a 2/3 chance. When Monty opens a door to reveal a goat, he hasn’t given you any new information whatsoever – you always knew there was at least one goat behind at least one of the 2 doors you didn’t choose. By showing you a goat, Monty effectively transfers that door’s 1/3 chance to the remaining door that you didn’t choose, and now you’re left with this choice: the door you chose (still a 1/3 chance), or the door you didn’t choose and that Monty didn’t open (2/3 chance). The thing to know is that there are still 3 doors in play – you can still choose the door Monty opened! Of course you never would, but the illusion has been created that you are choosing from 2 doors, when you are actually still choosing from 3, it’s just that you now know of at least one door for certain that has a goat. Since your door is still 1/3, and the goat door is now 0, the remaining door must be 2/3. 1/3 + 1/3 + 1/3 = 1.

I think this specific part of the problem is the hardest to wrap your mind around. Took me years, to be honest. I believed in the correct answer, and I intellectually understood why it must be correct (and why it tests out and verifies this as fact), but it just didn’t “feel” right for the longest time, even after I understood the problem. So don’t feel bad, they call it a “brain teaser” for a reason.

Nqobzin says

Hi Mike K, please explain why the 2/3 chance doesn’t become a 1/2 chance after the elimination of door 3 when it is revealed that door 3 is a dud. Surely only 2 doors are in play following this elimination?

Jim illustrated that there are essentially 9 combinations when you made your initial choice of door 1, after door 3 is revealed to not be the door with the prize why are there still 9 combinations being considered, do the combinations now not shift to only 4 since you are now aware that door 3 cannot be chosen? Appreciate your response.

Jim Frost says

Hi Nqobzin,

I agree with Mike K’s answer to you. I just wanted to add that it’s important to understand that Monty’s choice is a non-random process, which helps explain why the full two-thirds probability gets shifted to the remaining door.

You have two groups of doors that are defined by your original random choice. You have your one door and Monty has his group of two doors. Your group has a 1/3 chance and Monty’s has 2/3 chance.

Monty opens a door. However, this is not a random choice. The probability that the door he opens has the prize is ZERO. The probability for the entire group is 2/3 and doesn’t change because the location of the prize has not changed. Because the probability for the door he opens is zero, using simple algebra, you can calculate the probability for the unopened door in Monty’s group must be 2/3. 0 + 2/3 = 2/3.

If Monty made a random choice, the probabilities would be different, and he might well have revealed the prize. But, it wasn’t random and he will never reveal the prize. That affects the probabilities as I describe.

Mike K says

No. Each door has a 1/3 chance. When you choose door 1, your door has a 1/3 chance. The other 2 doors combined have a 2/3 chance. When door 3 is opened, that 2/3 chance is effectively transferred to door 2. Door 1 still has a 1/3 chance; nothing about opening door 3 changed that. You’re now left with this choice: door 1 with a 1/3 chance, or door 2 with a 2/3 chance.

As mentioned elsewhere, this is easier to understand if you imagine 100 doors instead of just 3. If you chose 1 door out of 100, and Monty opened 98 other doors to reveal goats (remember, he CAN’T reveal the prize), would you say that both remaining doors have an equal, 50/50 chance?

Broken Brain says

How is that different to sticking to your original door?

Choosing to stick to your current door is still choosing a door – that and the already open door means you still have the same chance.

If door 3 is opened and you picked door 1, you now effectively have the choice between door 1 and door 2. Whether you choose door 1 or 2 from there, you’re also effectively taking door 3 meaning that either way you now have a 2/3 chance no matter what you pick.

This means we can safely ignore door 3 as it becomes irrelevant and you have a 50% chance of picking the right door, no?

Greg Cheatham says

When the doors are originally chosen, the odds are 1 in 3. If you got to choose two doors at the beginning, your odds would be 2 in 3. Opening the second door accomplishes the same thing. Its as though you chose that door and the door you switch to. In that case your odds are 2 in 3. If you stayed with the original door you chose, then your odds would remain 1 in 3.

Mike K says

I’ve loved this problem since I read about it in Marilyn vos Savant’s column in 1990 (and I love the story of everyone thinking she was wrong, but she was not), and it amazes me that in 2021 people are still arguing about it. The fact is that you benefit from switching – if you think otherwise, you are wrong, period. What people need to realize is that the MHP is essentially a classic con that uses misdirection to fool you into thinking that something that isn’t relevant, is. If you chose a door, and then Monty DIDN’T open a door but merely said, “Great, now would you like to stay with your original door, or would you like to choose both of the other doors?”, everyone would switch without a moment’s hesitation, and a game show that wants to make a profit can’t have that. So, they introduce a “red herring” – something that isn’t relevant to the problem at hand, but which distracts and misdirects you to think that it is relevant and possibly alter your choice. It is DESIGNED to trick you and fool you, so I guess it’s not surprising that it tricks so many, even today.

When you choose your door, it has a 1/3 chance of winning. Nothing that happens in the rest of the scenario changes that; it’s literally meaningless that Monty opens a door at all, because you already knew that a goat was behind at least one other door, and revealing which door that is doesn’t change a thing. It is irrelevant. You are choosing between a set of one door and a set of two doors, so the odds are 1/3 and 2/3 respectively. The red herring throws you off this logical trail and makes you think you might be choosing 50/50, when in fact revealing a goat doesn’t change the odds at all – all it does is shift the 2/3 chance from two doors to one door – the door that you didn’t choose.

Greg says

I think an easier way to look at it is: When you picked your door the odds were 1 in 3. When the second door is revealed, your odds for the door you picked are still 1 in 3. But you can trade for the third door. Since one door is revealed, trading for the third door is the same as if you picked it plus the door that was revealed. That’s 2 of 3.

Ronald says

Hi. I’m not Jim but let me answer your question.

The trick in this problem is that since the probability of having failed remains being the same 2/3, it looks like if we didn’t update the information, as if we didn’t eliminate possible scenarios. But that is not the reason. What occurs is that the scenario in which you could have get it right was also reduced by half (as I will explain soon) so the proportion does not change.

It’s like which occurs in the example: Imagine that there are two men called A and B. Person A has 50$ in his pocket and person B has 100$ in his, so B has twice as A. If A spends half of what he has (25$) and B also spends half of what he has (50$), the money remaining in their respective pockets will be 25$ and 50$, so B still has twice as A. The proportion did not change despite the money is not the same as in the beginning.

In Monty Hall problem, note that since according to the rules of the game the host knows the locations of the contents and must reveal one door that is not which the contestant chose and neither which has the car, that means that he has two possible doors to reveal when the player’s one is which has thar car, but he only has one possible to reveal when the player’s one has a goat.

So, supposing you pick door 1, the possible cases are:

1) Door 1 has the car (yours) => Probability 1/3, but it splits in two sub-cases:

1.1) The host then reveals door 2 => Probability 1/6

1.2) The host then reveals door 3 => Probability 1/6

2) Door 2 has the car => Probability 1/3. Here the host is forced to reveal door 3.

3) Door 3 has the car => Probability 1/3. Here the host is forced to reveal door 2.

That means that if for example door 2 is revealed, not only must we remove case 2). We must also remove case 1.32) because we cannot be in one of those times when the host reveals door 3 being the car in door 1. We could only be in case 1.1) or in case 3), that had 1/6 and 1/3 chances respectively. Since the probabilities must always sum 1, the original 1/6 of case 1.1) turns into 1/3, and the original 1/3 of case 3) turns into 2/3.

Erik says

Hi Jim,

Why doesn’t this problem get recalculated once the first door is revealed. The first door maintains a value of 1/3 chance even when it is determined to have 0 value.

Gregory Cheatham says

I meant that everything goes as usual with the Monty reveal. Then one of the two contestants has the choice of switching with the other contestant. I assume that the odds are then 33%/66%. But if both contestants choose to switch, is it then 50% for each? Or does that end up putting them both back to where they were before the switch, which would mean 33%? Or does it change to 50%?

Gregory Cheatham says

Ok, dumb question. What if there are two contestants rather than one? One of the two has the prize door, the other does not. If they switch doors with each other, are their chances each 66% or 50%. Is it possible for the both to have a 66% chance, as this is greater than 100%, so I would think not. Does it matter whether only one chooses or both agree to switch?

Jim Frost says

Hi Gregory, if there are two doors and the prize is placed randomly, and Monty has no role, then it is truly 50/50 for those two doors. There’s no benefit to switching. The reason why it becomes 66% in the actual game is because of Monty’s non-random role in removing a non-prize door from consideration. But I don’t see that in your scenario.

Clay Goeke says

That article isnt exactly wrong in its conclusion as it does not conclude that Jim would be wrong but that in reality its not as sure to yield the desired results as most mathematicians will have you think. After all nothing is perfect. Including math.

Jim Frost says

Hi Clay,

Probabilities are long run averages for outcomes. Consequently, you know that in the long run, you’ll win 67% of the time versus 33% of the time by switching. However, some contestants will, in fact, lose by switching. To that extant, I’d agree that nothing is perfect. However, to maximize your chances of winning, you want to switch.

An analogy is that rolling a pair of dice and getting snake eyes (two 1s) only happens ~2.8% of the time. However, despite that low percentage, you will eventually roll snake eyes!

Math and probability theory are exceptionally good at calculating probabilities for games of chance like these!

Joe Johnson says

Your own comments omit an important piece of information: Monty knows which doors have goats behind them, and makes sure when he opens door 2 to reveal one with a goat. But this doesn’t give you any new relevant information — you already knew at least one of the remaining doors, 2 or 3, had a goat behind it when you guessed door 1. You also know there’s a 2/3 chance the prize will be behind either door 2 or door 3; those odds aren’t changed because Monty verifies there’s a goat behind one of them.

Had Monty opened one of the remaining doors at random the situation would be as you described, but the assumption here is that he’ll make sure to choose a door with a goat. This is a very different situation than if you, say, accidentally heard the goat make a noise behind door 2. When I first saw the Monty Hall problem I made the same mistake you’re making. But you have to look at not just what you know about what’s behind the doors, but how you know it.

Think of it this way: had Monty initially offered you a choice of picking door 1, or letting you have the prize if it’s behind either of the doors 2 or 3, you would choose the latter since you’d then have a 2/3 chance of winning as opposed to 1/3. If Monty’s shtick is to always reveal a goat behind one of the unguessed doors, it doesn’t matter whether it’s 2 or 3.

If you still don’t see it (and this is what worked for me), imagine there are a million doors and you choose door 1. You have a one in a million shot of winning. The prize is located behind one of the other doors with a probability of 999,999/1,000,000. Monty knows where it is, and verifies that 999,998 of the doors other than door 1 have goats behind them by opening them. But not at random — if the prize is behind one of the 999,999 doors other than door 1 (and it is with very high probability), that will be the door Monty will not open. I think you can see that the odds of the prize being behind the door Monty didn’t open are immense, while the odds it’s behind door 1 are still one in a million.

Craig Thompson says

I was fully in agreement with you based on the table above with every outcome listed, until I realised your presumption is also wrong. 3 of the outcomes are not possible as the door is opened by an outside influence. The probability of picking correct first time is 33%, but that doesn’t statistically affect the actual choice. The only choice will be between the one with the prize and one without after Monty intervenes

Jim Frost says

Hi Craig, the table shows that by switching, you have a 67% chance of winning. You end up having a choice between two doors, but that doesn’t mean the probability is 50/50, as shown in this post. Given Monty’s nonrandom process, there’s a 67% chance the prize is behind the door he offers. Think of it it this way. If there’s only a 33% chance that your initial door selection is correct, then you must have a 67% chance of winning by switching to the only other door.

Michael Thomas says

This solution omits an important piece of information. Assume three doors: 1, 2, and 3. You pick 1, Monty opens 2 revealing a goat. That leaves doors 1 and 3 unknown. Doors 2 and 3 have a .67 probability of having money. But, importantly, there is a second combination, doors 1 and 2, which also have a .67 probability of having money. At this point you have two combinations, each with a .67 probability. It becomes a choice between two equally probable combinations. Effectively this is the same as a probability of .5 There is no advantage to either staying or switching.

Jim Frost says

Hi Michael, Sorry that is incorrect. You have a 67% of winning if you switch or a 33% of winning if you stay. In your scenario, door 2 is revealed and has a 0% chance of having the prize.

Javier Gonzalez says

Hi jim, on Quora i posted the following question: “How do I convince my friend that in the monty hall problem, switching only improves your chances when the host reveals one of the doors? He says switching will always improve your chances even if the host never gives you any hint or help.” And every one that answered agreed with me and each other that my assumption is correct and my friend is wrong. Except for one person who answered. He believes my friend is right. Or atleast thats what i am interpreting from his answer. Could you tell me who right? Also if you would like to see that person who’s answer was different i will provide the link it. It would be good if you could because he seems to be a qualified individual who holds a PhD and used math and formulas in his answer. Here is the link:

https://www.quora.com/How-do-I-convince-my-friend-that-in-the-monty-hall-problem-switching-only-improves-your-chances-when-the-host-reveals-one-of-the-doors-He-says-switching-will-always-improve-your-chances-even-if-the-host-never-gives/answer/Alfredo-Sep%C3%BAlveda-Jim%C3%A9nez?ch=10&share=1d08733f&srid=oyle

Kicab Castaneda-Mendez says

I think I see why you thought I was repeatedly asking the same question. You write “You asked how we determine the probability for the contestant.” That is not the last question I asked. I asked:

Q Contestant Probability: How is the probability that the car is behind a door from the contestant’s perspective determined?

This is not the same as:

Q2 Contestant Probability: How is the probability (for the contestant) of randomly selecting the door that has the car behind it if each door is equally likely to be selected determined?

I have always agreed the answer to Q2 Contestant Probability is 1/3.

Jim Frost says

Hi Kicab, at the beginning, there is a 1/3 chance of the car being behind any of the three doors. After Monty performs his non-random process, the probabilities become 1/3 behind the contestants door and 2/3 behind Monty’s unopened door. As we discussed a length, the probability changes for Monty’s two doors because he uses a non-random selection process. Hence, that last door he offers has 2/3 chances of having the car. Check your email for a message from me. –Jim

Kicab Castaneda-Mendez says

I’m sorry I caused you such frustration. I looked at my questions and they are different and variations from my view. But I can see why from your view they look like repeated questions and I’m regressing. Regrettably, I still do not understand your answer to my last question.

Jim Frost says

Hi Kicab,

You asked how we determine the probability for the contestant. That’s simple. Three doors, one prize, and the contestant makes a random choice of doors. Hence, 1/3. That’s basic probabilities. In the future, I will be adding many articles about probability. Until that I point, I can’t refer you to one. But, it’s one random choice from three doors. That should be straightforward enough.

You don’t need to use CIs to answer your question. I use CIs in another post to illustrate how hypothesis testing can exclude 50% as the probability. But, to understand that initial determination, it’s just simpler to understand it from a basic probability standpoint. I didn’t include your very long text about CIs because it’s not a rabbit hole we need to go down to answer your question.

I’d recommend beefing up your basic probability knowledge to be able to answer your own question if my explanation isn’t working for you.

Kicab Castaneda-Mendez says

Hi Jim.

Sorry, I didn’t add more explanation as to what I actually did. After I played a few games I realized that the simulation was exactly Monty’s perspective showing the key facts about two roles of probabilities. So, I switched to creating the simulation in Excel. I’ll use your words since you say it best regarding Monty’s perspective and shown playing the game and with simulation:

Fact 4: Many-games Probability: “Monty knows that overall contestants will win 67% of the time by switching.”

Fact 5: Single-game Probability: “he’ll know for each game whether contestant has a 100% or 0% chance of winning by switching.”

Monty’s perspective is clearly objective and matches the facts, especially

Fact 6: Once the car is placed behind a door (Step 1 is completed), the car is never behind any other door during the rest of the game.

This has been my perspective from the beginning and that is why I couldn’t understand why the correct probability of winning by switching is 66 2/3% for a single game. But when you explained that it is from the contestant’s perspective, I wondered what that meant—hence, questions Q4 and Q5 (from Jan 6 comment).

What puzzles me is that you say for the contestant both (a) the probability that the car is behind a door after Step 1 and (b) the probability of selecting the door with the car behind it where each door is equally likely to be selected are 1/3. From Monty’s perspective, the probability for (a) is 0 or 1, and for (b) it is 1/3.

Q Contestant Probability: How is the probability that the car is behind a door from the contestant’s perspective determined?

Jim Frost says

Hi Kicab,

Once and for all, the same probabilities apply. Contestant wins 1/3 of the time by not switching and 2/3 of the time by switching. Monty has perfect knowledge which allows him to know the outcome sooner but the outcomes follow those same probabilities. By perspective, I just meant that the contestants don’t know where the car is but Monty does. However, the probabilities are the same. I’m not going to keep answering the same questions over and over again.

It seems like you’re regressing in your knowledge. I don’t mean to sound harsh, but you should already know the answer to your last question. There are three doors and one car. The contestant doesn’t know where the car is and picks one door at random. Hence 1 out of 3 chance.

We’re getting to the point where you’re spinning wheels now. You were so close and somehow you seem to have lost ground. I don’t have time to answer essentially the same questions (or variations of them) over and over. Apparently, I’m not able to answer them in a manner that is clear to you. Consequently, I think you’ll need to find your answer elsewhere. Again, I don’t want to seem harsh, but I’ve answered many of your questions in quite some length!

Kicab Castaneda-Mendez says

Hi Jim,

I played the game 1000 times. I set it up in Excel as it is easy to do so that with one click 1000 runs are done. I have five columns: Step 1 Car Door, Step 2 Contestant Door, Step 3 Opened Door, Step 4 Switch Door, Step 5 Result When Switched (Win/Lose).

The results for first 1000 are: 680 wins by switching; 329 times car behind door 1, 340 behind door 2, 329 behind door 3. So, considerable support for both equal distribution of the car behind the three doors and that switching wins 2/3 of the time over many games. Note that I didn’t need to run the simulation to know that the frequency of wins by switching would be about 66.7%. I read Marilyn vos Savant’s response in Parade when it was published and accepted then and since that switching wins 2/3 of the time across many games. Recall that I had replied earlier that “I have always agreed (I realize I didn’t state this—sorry) that the simulation correctly shows that if the game is played many, many times and the car is randomly placed behind a door with each door equally likely to be selected for the car, then switching at step 4 wins 2/3 of the time.”

Yet, I still have questions about the difference between probabilities that apply to everyone and probabilities that apply to certain people and not others.

Q4: What do you mean by “for the contestant” the probability is 1/3?

Q5: Is this a subjective probability and if so, subjective to what?

Jim Frost says

Hi Kicab,

I’m wanting you to play the actual game as I describe in my other post about the Monty Hall Problem. Not a simulation. I’m glad you see the correct answer is you win 2/3 of the time by switching from the simulation. But, I want you to go through the process and see it from Monty’s perspective. The fact that you’re not understanding the answers to your questions means you’re not understanding the process. Playing the game yourself should help with that.

When I say 1/3 probability, I mean that the contestant’s initial door choice has a 1/3 chance of having the car. That’s true for all.

There’s nothing subjective about this problem. The only difference is that Monty has more knowledge, which both affects the outcome and allows him to know the outcome sooner, but it doesn’t change the probabilities.

Playing the game with a friend and three cards will help you see this in action. Maybe you don’t have to play it 100 times if you’re just understanding how it appears to both Monty and the contestant. I recommend 100 times because it provides a formal hypothesis test with 93% power for rejecting the notion that switching only has a 50% chance of winning.

Kicab Castaneda-Mendez says

Thanks, Jim, but I won’t celebrate yet. I still have some questions and suggestions. Let me first recap the two solutions.

Contestant Solution: when you state that the correct solution is that the probability is 2/3 of winning by switching to the unopened door in Group 2 doors regardless of whether the car is behind that door or not, you mean only from the contestant’s viewpoint.

Q3: Would it help readers understand more if this was explicitly stated in the explanation?

Monty Solution: from Monty’s perspective the probability of the contestant winning by switching is either 1 if the car is behind the unopened door in Group 2 doors or 0 if the car is not behind the unopened door in Group 2 doors.

Q4: Why not tell readers that there are two correct solutions, depending on whose perspective you take?

Q5: But if there is only one correct solution (the Contestant Solution), why is Monty’s Solution based on perfect knowledge the incorrect one?

Jim Frost says

Hi Kicab,

No, you’re very close. And your questions reiterate what I asked you to do earlier. Play the game as I describe. Going through the process will help you intuitively answer the questions that you’re asking. I’ll answer questions below, but I won’t answer more questions until you complete your next assignment of playing the game 100 times with a friend. Record the results for always switching. During the process, note the items I discuss below. I strongly feel that you would’ve been able to answer your own questions by going through that process.

No, there are NOT different solutions for Monty and the contestant. What is different is their knowledge. After the contestant picks their initial door, Monty KNOWS 100% for sure whether the contestant will win by staying or switching. Meanwhile, the contestant does not know for sure. However, a smart contestant can calculate that they have a 2/3 probability by switching. And Monty knows that overall contestants will win 67% of the time by switching.

From Monty’s point of view about switching, he’ll know for each game whether the contestant has a 100% or 0% chance of winning by switching thanks to his perfect knowledge. However, the probabilities still apply to him. If Monty plans to runs the game 100 times (whether with one contestant or different contestants), he’d expect over the long run that contestants will win about 67 times out of 100 by switching versus only 33 times if they stick with their original door. So, it’s the same answer and the same benefits for switching.

Here’s an analogy. Imagine we have a weighted coin where heads appears 67% of the time and tails 33% of the time. The coin is flipped and the result is hidden from the contestant, who has to pick heads or tails. Meanwhile Monty observed the outcome and knows which one the contestant should choose after each flip. However, when pondering future flips, Monty knows that contestants will win twice as often if they pick heads.

Same answer. Just different amounts of information along the way.

PLAY THE GAME! You seem to be resisting that but I think it would be informative. Play it both as Monty and the contestant and you’ll see that the percentages of wins for switching is the same. The only difference along the way is that as Monty, you’ll know the outcome before the contestant.

Kicab Castaneda-Mendez says

Happy New Year, Jim!

When I read “from the contestants stand point the probability of the car being behind any given door is 1/3” I wondered why would you include “from the contestants stand point”?

That’s when I had my “Aha!” moments for your correct solution.

Aha 1: The probability the car is behind each door is 1/3 for the contestant who has no knowledge and the probabilities are different for Monty who has perfect knowledge.

Aha 2: For the contestant, the probability the car is behind the Group 1 door is 1/3 from the moment the door was selected to the end of the game. While for Monty, who has perfect knowledge, from the beginning the probability is 1 if the car is behind the selected door and 0 if it is not.

Aha 3: While the probability that the car is behind the door in Group 2 that Monty opens stays 0 from the beginning for Monty, for the contestant the probability changes from 1/3 before the door is opened to 0 after it is opened.

Aha 4: For the contestant, the probability the car is behind the Group 2 door that is not opened is 2/3. While for Monty, who has perfect knowledge, the probability that the car is behind that door is 0 if the car is not behind that door or 1 if it is.

But then I remembered that when I asked if the probability for the unopened door from Group 2 doors was the same for everyone, you said it was. Was that a typo or,

Q1: Are the probabilities for each door different for the contestant and Monty (depending on knowledge or lack thereof)?

Q2: If the probabilities are not different than what is the one probability that the car is behind the Group 1 door that applies to both people who don’t know whether the car is behind that door and people who do know?

Jim Frost says

Hi Kicab,

Happy New Year! And, this is a great way to celebrate with you understanding the Monty Hall Problem! I’m glad we could work through this together.

You’re right, understanding it from Monty’s Perspective is crucial. That’s why I always pointed out how he was using a non-random process that affects the probabilities that the contestant faces. He has perfect knowledge and probabilities (other than 0 and 1) don’t apply to him.

I must’ve misunderstood your question about probabilities for everyone. I thought you were asking about all contestants who play. All contestants have the same probabilities. But, no, as I state throughout my comments to you and in the blog post, Monty has perfect knowledge. The probabilities do not apply to him. His intentional actions based on perfect knowledge alter the final probabilities for the contestant.

It’s as if you flipped a coin but you knew the outcome. For someone without perfect knowledge, the odds are 50/50. But for someone who knew the outcome, it’s 100% for one side of the coin and 0% for the other!

Ilpoj says

I don’t get it why some people make counter arguments for this article, even after it is proved in the article itself. And it is correct of course. I too tested this in real life for 52 times (Which is enough FOR ME, in addition to seeing that the 2/3 win change when choosing to change is true) and the result was 32 wins and 20 losses ~2/3 = 66,666%.

Jim Frost says

Hi, I think some people just can’t get past the fact that there are two doors and, in their minds, it has to be 50/50. However, after understanding the process, it becomes clear that’s not the case!

I think you can probably get the gist of the solution empirically with fewer than the 100 trials I recommend. That’s my recommendation because I’m thinking in terms of a hypothesis test where I want a 90% chance (actually 93.7% statistical power) of being able to reject the notion that the probability of winning by switching is 50%. To have a ~90% chance of being able to accomplish that with a formal hypothesis test, you need about 100 trials.

girdhari lal saini says

Each door has a probability of 0.33 for having a prize. If we stick not to switch then probability of winning is 0.33. The combined probability of two door having the prize is 0.66 and as per the problem Monty always opens empty door hence its switching the door make 66.666…percent chances

Jim Frost says

Yep! Exactly right! 🙂

Kicab Castaneda-Mendez says

Jim, thanks for your patience. I’m sorry it appears I am not focusing on the process or don’t understand it, because I think I do. But you can confirm or tell me where I am wrong. Because I think I understand the process is why I was asking for confirmation at different points of the process.

First, here is how I saw and still do see the Monty Hall process:

Step 1: A car (the prize) is placed behind one of three doors and a goat is placed behind each of the other two doors. After the placements, Monty knows which door has the car behind it and which doors have a goat behind them [“because he has perfect knowledge”].

Step 2: A contestant who does not know which door has the car behind it selects a door, hoping to select the door with the car behind it and win the car.

Step 3: Monty, knowing which two doors have a goat behind them and which has the car behind it, deliberately and knowingly opens a door that does not have the car behind it. If the contestant selected the door with the prize behind it, Monty has a choice of two doors to open; otherwise, only one choice when the contestant selects a door with a goat behind it.

Step 4: Monty offers the contestant the option of switching to the other unopened door not selected.

Problem question: What should the contestant do?

Second, I have always agreed (I realize I didn’t state this—sorry) that the simulation correctly shows that if the game is played many, many times and the car is randomly placed behind a door with each door equally likely to be selected for the car, then switching at step 4 wins 2/3 of the time.

Third, your table of the nine sequences of the process does show that 2/3 of the time switching wins.

Fourth, it is the two sentences I asked about in my last response that I still do not understand what you mean. It may be because I may not have been clear in what I was asking. So, let me try again. At Step 2 the contestant will select a door. Let’s assume that the pick is random with equal probability of picking each door.

Question 3: Do you see a difference between

(a) the probability a door has a car behind it is 1/3

(b) the probability of picking the door that contains the car is 1/3?

If yes, why? If no, why not? Thanks.

Jim Frost says

Hi Kicab,

Ok, that’s great, thanks for writing that out. I just wanted to be sure that we are working from the same page.

There’s just one thing I’d add to your process, which dovetails nicely with your question.

Part of the process is that there are two groups of doors in this game that are created when the contestant picks the initial door. Group one contains the contestant’s single door, which has a 1/3 chance of having the car. And Monty’s group of two doors, which has a 2/3 chance of having the car. We know that there is a zero probability that Monty will open a door with the car. Consequently, if his initial set of two doors contains the car (2/3), the door with car will 100% be Monty’s unopened door (2/3). Given this inevitability, the 2/3 probability for the set of two doors still applies to Monty’s single unopened door after he opens the other one. I think that’s just the final step of understanding you need to understand the solution.

So for your Question 3, yes, from the contestants stand point the probability of the car being behind any given door is 1/3. And their chances of picking the the correct door is also 1/3. And, as mentioned, that means the other two doors combined have a 2/3 probability. Given Monty’s non-random actions, that 2/3 probability is fully transferred to his final unopened door. Again, if either of his two doors have the car, that door WILL definitely, positively be the door that he let’s you switch to at the end. I think it’s just that last portion you’re having difficulty with.

Keep in mind that it’s Monty’s perfect knowledge combined with the rule of never opening a door with a car that confers the 2/3 probability to the his single, unopened door at the end.

If you want to understand this more intuitively, I highly recommend that you play this game with a friend using cards as I write about in my other post about the Monty Hall Problem. In that game, be Monty. I think that you’ll find that buy playing the game and seeing it from Monty’s view that if his doors contain the car that his knowledge and rules constrain him to opening a particular door and forcing the car to be behind his final door. So, if you don’t fully understand it from what I’ve written, that’s what I want you to do next. Set aside an hour or so, get a friend, play it out 100 times, and record the results for always switching. I did this with my daughter and she caught on!

I think by seeing it from Monty’s point of view, it’ll become clear!

Happy New Year!

Kicab Castaneda-Mendez says

Jim, thanks again for the details. I understood your explanation but am confused by these two statements:

1. “Any door that you pick has a 1/3 chance of containing the prize.”

2. “the door he opens by definition has zero probability from the beginning [of having the car behind it].”

I will state my confusion as a deductive argument so you can tell me which premises are false and which inferences are invalid, and why.

Deduction

Step 1 (Premise 1): “the door he [Monty] opens by definition has zero probability from the beginning” of having the car behind it.

Step 2 (Inference 1): Therefore, from the beginning the probability of winning by selecting that door was 0.

Step 3 (Inference 2): Therefore, from the beginning not “Any door that you pick has a 1/3 chance of containing the prize [car behind the door].” In particular, the door Monty opens did not have a 1/3 chance of containing the prize from the beginning.

Where does the logic fail and why?

Jim Frost says

Hi Kicab,

Your sequence starts at the wrong point. The game begins with the random placement of the car behind a door. There are three doors and the contestant can pick one at random. Consequently, the contestant has a 1/3 chance of picking the correct door at the outset because he has absolutely no information about where the car is located.

The next step is when Monty opens a door. I’ll answer you question with a question. Assume you’re Monty. You have perfect knowledge about where the car is located. You also strictly adhere to your rule of not opening a door that has the car. Given those two assumptions, what are the chances that you’ll open the door with the car?

So, yes, both are true. The contestant has a 1/3 chance of picking the door that contains the car at the start of the game. And then Monty has a zero probability of opening the door with the car. Monty is really operating outside the realm of probability because there’s no chance involved because he has perfect knowledge.

Now, it seems like you’ve been spinning your wheels thinking about in your own way. And, I also want to be sure that you are taking in my comments. So, before I answer any more questions from you, I’ve got an assignment for you! I want you to write in your own words a summary of Monty’s process. You can refer back to my previous replies to you. But, I want you to focus on Monty’s process, how it ensures that if one of his two doors has the car, that door becomes the last unopened door. And, how switching flips the outcome compared to staying with your original door. Focus on the process and then work in the probabilities as needed. This is all stuff I’ve covered in my previous replies to you.

So write that up so I can see that you’re absorbing and at least understanding my comments to you. You haven’t touched on my comments at all to show that you’re taking it onboard and I want to be sure before spending more time answering endless questions from you. And, again, you’re spinning your wheels trying to come at it from that single angle you’re fixed on. I think working on it from a process standpoint first (maybe putting yourself in Monty’s shoes would be helpful) and then adding in the probabilities will help you understand it from another standpoint. Your most recent questions show me that you don’t understand the process, which is making it hard for you to understand the probabilities. So, focus on the process first and then I think the probabilities will come to you more easily.

After you do that, I’d be willing to answer more questions if you have any.

Joe W. says

Hi Jim,

I just posted a comment and then read through the other comments and I feel like you might have been really close to explaining that plants prefer water over Brawndo (because you can talk to them). I hope you get that reference from Idiocracy, if not go watch it, I have a feeling you will truly appreciate that movie!

Joe W says

I enjoyed reading this article, I was looking for a good explanation to share with my family and friends who want to be “Monty Hall Problem Deniers”. I understand why someone without a math or science background might be confused, but I really don’t understand how educated people have a hard time understanding this. It’s not that hard. It didn’t make immediate sense to me, but I understood that the difference between a 50%/50% chance ( which is what most people believe the answer to be at first glance), and a 33%/66% chance are large enough to prove in a relatively small simulation. I ran 100 tests in Excel and sure enough, 33%/66% +/- 1%. Then I did 1,000 tests in Excel and the results were nearing perfect in accordance with the law of large numbers. My point is, if such a simpleton like me can figure it out, why would it stump a statistician? Surely they would see that they can just run a simulation to prove the answer. Whether you call it the law of large numbers or the scientific method, we are taught methods to test our theories. I was appalled when I found an article from a Math Institute (written in 2015) that was arguing for the wrong side of the problem.

https://ima.org.uk/4552/dont-switch-mathematicians-answer-monty-hall-problem-wrong/

My aha moment came when constructing the simple test in excel. I knew what the results were going to be before I ran the tests. I was thinking of how many columns I will need when it dawned on me, only two. The first door you pick and the winning door. The door Monty opens is irrelevant when making the simulation. Either the first door you picked is the winner, or you would win by switching doors. When you understand that statement and what it means, it is easy to see that your odds are 33%/66%.

Jim Frost says

Hi Joe,

Thanks so much for you nice comment, and for recommending Idiocracy. I haven’t seen but I’ll check it out!

Well that’s one thing I don’t understand from the deniers. It’s not hard to simulate the process and, run it many times and observe the results, never mind the probability calculations! In my other post about the Monty Hall Problem, I run both a computer simulation and a physical simulation using cards and both confirm the results. Although, I was convinced by the math. I also agree that understanding that switching doors causes you to flip the outcome is a key point. And, you have a 2/3 chance of picking the wrong door (losing) at the outset, which means you have a 2/3 chance of winning by switching!

I haven’t read the article you link too. I’ll have to check it out but I’m shocked that a Math Institute would write something like that!

Thanks again and Happy New Year!

GL says

I wrote to you last night about being able to explain the solution to another using your section above. This morning I did so with my wife. She had heard of the problem before but had forgotten the solution. She said it was the best explanation she had ever heard. I laughed and said that that’s I wrote the guy last night.

Jim Frost says

Hi, that so makes my day! Thanks for letting me know! Now you just need Monty Hall to offer you a door to switch to! 🙂

GL says

Your “Here’s how it works” section is, by far, the best simple explanation of this problem I’ve seen. I think I have a grasp enough of it to explain to another person. Thanks!

Kicab Castaneda-Mendez says

Thanks for the detailed explanation, Jim.

I understand what you wrote so I must be misunderstanding something else. There are two items where that might have occurred.

Item 1. I have understood that after a door is opened revealing no car behind it, “the contestant has a 2/3 chance of winning by switching” as meaning that the probability the car is behind the door the contestant can switch to, is 2/3.

Question 1a. If this is not correct, what is the probability the car is behind the door the contestant can switch to?

Question 1b. Is this probability from 1a the same for everyone, including for example those who think the probability is 1/2?

Item 2. The probability of winning by selecting the door that was opened after it was opened is 0, of course. You state that before the door is opened the probability of winning by selecting that door is 1/3.

Question 2. Why does the probability change from 1/3 to 0 by opening the door? I see a couple of possible reasons but want to make sure I understand. For example, it could be because of additional information or it could be because in fact the car was not behind the door or some other reason.

Jim Frost says

Hi Kicab,

1a. You are correct. The probability that the car is behind the door that the contestant can switch to is 2/3.

1b. This probability is the same for everyone. Believing that the probability is 1/2 does not change the underlying reality. This illustrates how knowledge is power! As you gain more information and understand how a process/system works, you can better predict the outcomes using accurate probabilities.

2. When the game starts, we have no information other than that the car was placed randomly behind one of the doors. Any door that you pick has a 1/3 chance of containing the prize. This is the only truly random part of the process.

There are two sets of doors in this game. The door you choose, which has a 1/3 chance. And Monty’s two doors, which collectively has a 2/3 chance of having the car. At this point, Monty makes a non-random choice to open a door that does not have the car. There’s zero probability that the door he opens has the prize because he has perfect knowledge and yet his set of two doors still has a 2/3 chance because that was the random part of the process at the beginning. Consequently, we know that there is a 2/3 chance the car is behind Monty’s unopened door.

So, the probability really isn’t changing from 1/3 to 0. Monty’s knowledge is perfect so the door he opens by definition has zero probability from the beginning.

Kicab Castaneda-Mendez says

First, I apologize for my tone and attitude. That was not my intention. But there is no reason to be that way. I’m sorry.

Second, I don’t believe my first comment had that attitude or tone but it’s not always easy to see things from others’ perspectives.

Third, my comments are not opinions but facts.

Let’s start the discussion and I will ask questions so I can better understand where my errors are. I think these are three facts that determine the solution.

Fact 1: There are nine possible combinations of door for the car and door for the contestant before the game starts.

Fact 2: There are 3 combinations are possible after the car is placed behind a door. If that is not a fact, why not?

Fact 3: There is one combination possible after the car has been placed behind a door and the contestant selects a door. If that is not a fact, why not?

Jim Frost says

Hi Kicab,

I’m glad we can move forward more constructively. What you need to realize is that the Monty Hall Problem is solved mathematically and there is one correct solution. The mathematically correct solution is that you win 2/3 of the time by switching. Additionally, computer simulations have played this game hundreds of thousands of times and have confirmed that mathematical solution. You say that your comments are “facts.” However, you need to realize that you’re essentially arguing that 2 + 2 = 5 because you’re disagreeing with the mathematically correct solution. Instead of digging in on the wrong answer, I’d recommend opening your mind and trying to understand how the correct solutions works.

I’ll get to your questions in just a minute. But, I’d highly recommend trying to understand the two explanations I provided in my previous reply to you. You haven’t discussed them at all and I’m pretty sure that you have not tried working through them. Take a moment to consider them. It shows you the HOW and WHY you win more often by switching. Right now you’re stuck spinning your wheels by coming at it from one direction. Try understanding the solution using the other explanations I mention.

You seem to be misunderstanding the table. For your first “fact,” you’re a bit off. The table does NOT represent 9 possible combinations of “the car and door before the game starts.” Instead, the table lists out all possible sequences and possible outcomes. All of these outcomes can happen. There are 9 possible sequences the game can take. These incorporate the initial placement of the prize, your initial choice, the door that Monty opens, and you’re decision to stay or switch. All of those factors boil down to the 9 possible sequences and outcomes in that table.

Given the placement of the car and your initial door choice, the last two columns indicate the outcomes when you stay or switch. As I mentioned, notice how switching ALWAYS flips the outcome. If you win by staying, then you lose by switching. If you lose by staying, then you win by switching. The table makes it clear that you win twice as often by switching doors (6 times vs 3).

For your fact 2 and 3, the car can be placed behind one of the three doors. And, yes, after the contestant picks a door, the contestant either has picked the correct door or not–although the contestant does not know which condition is correct at this point. However, using simple probability, the chances of picking the correct door initially is only 1/3. Right there you know that the probability of winning by sticking with that initial door can only be 1/3. However, it doesn’t end there because Monty offers you the option of switching doors.

Now, you need to take the next step and understand the how and why it works. Upon the initial selection, the contestant only has a 1/3 chance of winning if they stay with their original door. However, switching will flip the outcome, which means the contestant has a 2/3 chance of winning by switching. That’s what I mean by trying to understand what I wrote about how switching flips the outcome. And you need to understand Monty’s process whereby if the prize is behind one of his two doors (which happens 2/3 of the time), then that door ultimately becomes Monty’s unopened door to which he lets you switch.

The real trick to this problem is understanding that Monty has full knowledge and bases his actions on that knowledge. When you understand that his behavior is not random but based on his insider information and the rule that he will never reveal the prize, you can use that understanding to improve your chances of winning. You seem to be missing that entire angle.

Finally, if you aren’t connecting with the math and logic of how it works, I recommend you try it empirically. Read my other post on Revisiting the Monty Hall Problem. I show a computer simulation of the outcome. Additionally, I perform an empirical experiment that confirms the results. If nothing else convinces you, grab a friend and play the game 100 times as I describe in that post. One of you is Monty and the other is the contestant who always switches. Record the win/loss outcomes. You’ll see how the winning percentage converges on 66.6% when you always switch.

Kicab Castaneda-Mendez says

I have tried posting four responses but they fail to show. Here’s number five. I believe it’s because they show why the solution of 2/3 probability of winning by switching is wrong. The explanations have several fatal mistakes. Here is one.

Mistake 1. Calculating the probabilities using all possible combinations. There are nine possible combinations of door for the car and door for the contestant. However, there are only three possible combinations after the car door has been selected and there is only one possible combination after the contestant selects door. Hence, calculation of the probability of winning by switching using all nine combinations when only one combination exists is a mistake. This is easily seen by looking at the column of Win/Lose for each combination. There are no 1/3s or 2/3s or any number. Win or Lose means with probability 1.

Jim Frost says

Hello Kicab,

I’ve seen your comments, and no, I haven’t approved them because of your attitude and tone. I’m okay with differences of opinion. However, this is a mathematical problem with a correct solution. There’s no room for your opinion on it! You come in here claiming there many “fatal mistakes.” Next time, try coming here with some humility and a mindset for trying to understand. Yes, this problem is counter-intuitive. That’s why I call it a statistical illusion.

Did you really think that you found something that one else did over the past several decades? Statisticians have looked at this, and we’re unanimous in that switching doubles your odds. But, apparently, you found something we all missed?! And think about it a bit. Computers can simulate this game and play it fairly. And guess what. Simulations that play the game hundreds of thousands of times always show that you double your chances of winning by switching. Explain that!

So, even if you don’t believe the math (which is conclusive), the computer simulations also come to the same conclusion. In another of my posts about the Monty Hall Problem, I show one such computer simulation. I also conduct an empirical experiment and play the game 100 times. You can try them yourself.

About your comment relating to the 9 rows in the table. It’s straightforward. There are 9 possible combinations of choices and outcomes. As you can see, of these 9 outcomes, you win 3 times when you don’t switch and 6 times when you do switch. Using elementary math (6 / 3 = 2), you find that you double your chances of winning by switching.

I’m going to provide the two simplest explanations I can think of to explain it to you.

1) As you can see, when you switch, it causes the opposite outcome. For example, if you pick door 1 and the prize is behind door 1, you win if you stay but lose if you switch. At the beginning of the game, you have a 2/3 chance of picking the incorrect door. That initial door you choose probably does not have the prize. Consequently, switching doors reverses that outcome. In this case, having an initial 2/3 chance of losing means that switching gives you a 2/3 chance of winning. That’s about as simple as I can put it.

2) For a slightly more complex reason, consider that Monty uses a non-random process. That process isn’t obvious with only 3 doors but becomes apparent with more doors. Imagine that there are 101 doors and 1 prize. You pick one door, and that leaves 100 doors for Monty. It’s almost guaranteed that the prize is behind one of Monty’s doors (100/101). Next, Monty systematically opens 99 of his 100 doors. During this process, he carefully does not reveal the prize. He gets to his final door, which he does not open, and offers you the chance to switch. Do you? If you do switch, you’ll increase your chances of winning by 100 times!

The trick is understanding that Monty intentionally avoids opening the door with the prize. With his 100 doors, the prize is almost certainly behind one of them. Using his rules, he cannot open that door. Hence, it becomes the last, unopened door at the end of his process.

The same thing applies to 3 doors, but Monty’s intentional selection, non-random process is less obvious because he only opens one door.

If you have genuine questions and an open mindset for learning, I’ll discuss them with you. But if you come here claiming everyone else is wrong and that there are fatal mistakes, I will not discuss it with that type of tone.

Ronald says

Hi, Keegan Hall

Sorry, but I must tell you that you are wrong. If the other person (the host) randomly reveals one door from the other two and just by chance it results to have a goat, then the chance to have the car is 1/2 for each remaining door, no longer 1/3 vs 2/3.

Firstly, you can see it in an intuitive way: If the other person is doing it without knowledge, then it makes no difference if you (the player) are who also makes the revelation, because both are doing it randomly. For example, you could pick door 1 and then decide to reveal door 3. But in this way what you are doing is basically selecting which two doors will remain closed, I mean, it is the same as if you said: “I want to keep closed doors 1 and 2”, and then you revealed door 3. Now, if door number 3 results to have a goat, which one do you think is more likely to have the car, door 1 or door 2? According to your reasoning, since you declared door 1 first, it must be only 1/3 likely because it is the equivalent to the staying door. But what about if you had said it in the opposite order: “I want to keep closed doors 2 and 1”? Does it seem logical that door 1 has 2/3 probabilities now? So does the order in which you declare them determine which is more likely to have the car? The doors don’t know that order.

You can also visualize it in the frequentist way. If you played the game 900 times, in the first selection you would get the door that hides each content in about 1/3 of them, so about 300 times goat1, about 300 goat2, and about 300 the car. Then, since the host will randomly open a door from the rest, he has 1/2 chance to reveal each of the other two remaining contents. The possible cases are:

1) In 300 games you pick the door that hides the car.

1.1) In 150 of them the host reveals goat1.

1.2) In 150 of them the host reveals goat2.

2) In 300 games you pick the door that hides goat1.

2.1) In 150 of them the host reveals goat2.

2.2) In 150 of them the host reveals the car.

3) In 300 games you pick the door that hides goat2.

3.1) In 150 of them the host reveals goat1.

3.2) In 150 of them the host reveals the car.

So, if a goat results to be revealed, you know you are not in case 2.2) and neither in 3.2). That restricts you to a subset of 600 games. You win by staying in 300 (cases 1.1) and 1.2) and by switching in 300 (cases 2.1) and 3.1), so each strategy wins 1/2 of the time of this subset of 600 games.

This is different as if the host had known the positions and followed the rule of always revealing a goat. In all the 300 games of case 2) he would have revealed the other goat, and the same in case 3), so you would win by switching in 600 games (case 2) and 3) that are 2/3 of the total 900, and by staying in 300 games (cases 1.1) and 1.2), that are 1/3 of the total 900.

Peter Huang says

Thx for your reply. Yeah, the thing about being suspicious makes a good point. The MHP is interesting because it is an exercise on how the ‘correct’ answer disgrees with our intuition.

However, our intuition is based on real-life experiences of being offered alternatives. Why are am being offered a change?

I guess what I am saying is that the MHP is disappointing to the extent that part of the trick of the problem is that the assumption for the problem (host behaviour) is different than real world behaviour that forms our intuition.

Of course the MHP is interesting b/c of the math.

I just find the problem flawed.

Jim Frost says

Hi Peter,

I guess I’d have a different take on it. I absolutely love the Monty Hall Problem for several reasons.

One reason is that it symbolizes life for me. This problem contains a mixture of luck, process knowledge, and information. By understanding the process and deducing what Monty knows, you can use that information to make an informed choice that increases your chances of winning. Yet, there is still luck involved. So you can make the correct choices that maximize your chances of winning but you can still lose. It’s life in a nutshell. Those who understand the system and use the available information can improve their expected outcomes. I also appreciate the psychological factors that make this solution feel intuitively wrong.

For me, I don’t see the assumptions as flawed. Probability questions always have assumptions that are necessary to calculate an answer. These are just the assumptions for this given problem. Of course, if you change the assumptions, you’ll change the correct answers. Given the assumptions as stated, the best approach is to switch doors.

Peter says

What has always bugged me about the Monty Hall problem is that it makes certain assumptions that make the math work.

Some of the assumptions we can agree on. For example Monty will never reveal the prize. (which incidentally implies that Monty knows what is behind the doors)

However, it is also assumed that Monty _must_ reveal a door. Why is this important?

Well, Monty can decide which he reveals a door, he may do so conditionally based on his knowledge of what is behind the door.

So what?

Well, Monty’s behaviour can range from ‘evil monty’ (only offers a switch if you originally chose correctly), ‘nice monty’ (only offers a switch if you originally chose wrong), game theory equilabrium monty (offers choice randomly 2/3 of the time you chose correctly, 1/3 of the time you chose wrongly) etc.

Based on Monty’s behaviour, the correct action is to not switch, switch, switch/not (equally) respectively.

Monty Hall himself has said that a choice was not always offered.

To make my point in another way.

We bet $1000 dollars on the colour of a randomly chosen card. You say red. I look at the card.

_Then_ I say. I will let you switch if you want.

What do you do?

Jim Frost says

Hi Peter,

I agree that Monty can behave in different ways. But the assumptions for the classic Monty Hall problem is that he’ll always open a door, he’ll never open a door showing the prize, and he’ll always offer the chance to switch. Those are the assumptions for this puzzle. Of course, if you change the assumptions, the answers changes.

I remember as a kid that I’d occasionally watch the game show. It was usually when I was home sick from school. Whenever Monty offered the chance to switch, I did think he was trying to trick them into unwittingly giving up the prize! I was suspicious! However, at least for this puzzle, we’re assuming he always offers the chance to switch.

Keegan Hall says

He does not “remove” a door from the equation, figuratively he does by eliminating something you would logically choose but he doesn’t take a door out of the problem. I think that may be where you’re getting confused

Keegan Hall says

Intent has absolutely NOTHING to do with the Monty hall problem if he still opens a door you have not selected that has a goat behind it. Okay if you pick a door yeah it has a 1/3 chance of being right and the other two doors have a collective 2/3 chance of being right. Now, he opens one of the 2/3 doors and there is a goat behind it; now right here I want you to imagine that you can still choose the open door okay… so you know that the two doors that you have not chosen have a collective 2/3 chance of having a car, but you don’t want to choose the open door because you know it’s a loser. BOTH doors still have a 2/3 chance of being right (if you ignore the fact that you can clearly see a goat) you just don’t want to choose the open one.

The problem works in real life, as long as the 2/3 goat door is opened it doesn’t matter if it’s a video game; a host opening it accidentally; or your friend running up on stage, the problem always works!

Keegan Hall says

Because you as the person who picked the door would NEVER choose the one that you originally selected. Your friend on the other hand would IF he doesn’t know who one you picked; if he saw you pick a door I would still be 2/3 vs 1/3.

Joe Johnson says

The fact that Monty knows the location of the car is precisely what leads to the results not being 50/50.

When he chooses door one, the contestant knows the odds are 1/3 that the car is behind door one, and 2/3 that the car is behind either door two or door three. The contestant already knows that at least one of the doors two or three has a goat behind it, so Monty revealing that to be the case doesn’t tell the contestant anything that changes the probabilities.

A good way to more intuitively see this is to imagine the problem with, say, a million doors. If the contestant picks door one, he has only a one in a million chance of being right. Out of the remaining doors at least 999,998 have goats behind them — and Monty knows which those may be, so can always open those to show the goats. But if you’re the contestant, and you know Monty knows where things are, this doesn’t really tell you anything you didn’t already know. Door one still has only a one in a million chance of having the car, and the odds are 999,999 in a million it’s elsewhere. That Monty knows which doors don’t have the car behind them doesn’t change that.

Another way to think about this is not to confuse it with the case of the contestant accidentally discovering that a goat is behind door number three. Suppose the contestant accidentally heard the goat make a noise behind the door — a random event rather than one Monty planned. Then you’d be right to think the car has an equal chance of being behind door one or two. But Monty doesn’t open door three as a random event that might reveal a goat or reveal a car — if the car had been behind door three, he would have opened door two instead. All he’s doing is affirming what was already known to the contestant when he picked door number one and initially computed the odds: that at least one of the remaining doors has a goat behind it.

Graeme says

Here is the problem I have with the Monty Hall problem. It is flawed. Yes the probability rules show an increase in chance of winning if you change doors, however if this is a real world situation Monty Hall knows where the car is.

Basically at the start you have a 33.3% chance of selecting the correct door. Monty Hall opens a wrong door. Meaning 1 door will have the car, and the other door will have nothing. So in the real world situation it is 50/50 if the car is behind your door or not. The door Monty Hall opens should not be factored into the probability as he knows it contains nothing.

This question is an example of the model not being a true reflection of observable results.

Zachary Dorman-Jones says

I think that’s a very smart question. The answer is that the knowledge of which door you picked is essential to getting the best odds. Like the Monty Hall problem itself, it becomes more intuitive when you try it with more doors. With a little analysis, it is obvious that with the rules you outlined, Ted has a 50/50 chance of choosing the correct door, no matter how many original doors there were. Whether we started with 3 doors or 1000, for Ted it comes down in the end to a random choice between two doors.

It’s not a paradox, and it’s a little easier to demonstrate why if we take Ted out of the equation, and use 1000 doors. So you pick a door, and then Monty reveals 998 doors without a car. Regardless of which one you picked, one of the two doors that Monty did not reveal has the car. So now we have a new rule that you must determine which of the remaining two closed doors to open by coin flip (this is exactly equivalent to the role of Ted). Monty gave you critical knowledge by opening those 998 doors: either you picked the correct door initially–very unlikely–or the car is behind the other door. The coin flip is forcing you to disregard this knowledge. It’s not a paradox that you have a different probability of success now, because we have significantly altered the game, so we can expect the probability calculations to be different as well.

I like this question because it really gets to the heart of why the Monty Hall problem is so counterintuitive: most people who have trouble with the solution do so because it’s hard to grasp how the scenario is different from a simple 50/50 choice. Once you overcome that hurdle, you really understand the problem, and have gained an insight into probability theory.

livedarklions says

There’s no paradox. You have better information than Ted. Thus, you are more likely to select the correct box. The probability that any given box contains the car doesn’t vary, it’s either 1 or 0. What does vary is the probability that you or Ted will select the correct box.

J. Tripper says

I think the issue here is that your adding a piece to the Monty problem that isn’t stated. The Monty problem only factors from Ted’s friend, the original “contestant”, perspective of 1 in 3 doors. The variables can and will likely change when adding or subtracting to the original stated position.

Kicab Castaneda-Mendez says

I have three questions:

1. What is the difference between a non-random process and a random process?

2. How do non-random processes get probabilities assigned to the outcomes?

3. All processes have steps. In the Monty Hall game, what are the specific steps of the non-random process (to which you reference)?

Randy Weatherford says

I helped people understand this problem by changing the problem to 10 doors. You pick 1 door, Monty then shows you 8 empty doors, should you switch? Pretty obvious now. 🙂

Jim Frost says

Hi Randy,

I agree with that. Somewhere way back in these comments another reader and I discuss this approach. It seems to work better with more doors because it emphasis that there is a non-random process at work, which affects the probabilities. That process is still present with 3 doors but its implications are not as obvious.

Joe Johnson says

I’m not sure if it’s clear, but my comments posted on 10/3 were a response to the “monkey wrench” brought up by Mike in his post of 8/29.

Joe Johnson says

The probabilities are still one third and two thirds, but Ted doesn’t have the same amount of information that you do and therefore can’t correctly compute the probabilities. In particular, Ted doesn’t know what door you picked. Let’s suppose you picked door one, after which Monty opened door three to show it had no prize. If Ted knew you picked door one, and knows Monty’s choices in opening a non-prize door had to be either door two or door three, his analysis would be the same as yours. But for all Ted knows you picked door two. In that case, door one would have the two thirds chance of having the prize. Without knowing which door you picked, he can only guess which door, one or two, has the advantage. Seeing the problem through Ted’s eyes, it’s a toss-up. But we know more than Ted since we know you picked door one, and that the real probabilities are one third for door one, and two thirds for door two.

Matt Brown says

oops, should be “and that 1 minute elapses in 1/2 mile”…

Matt Brown says

It’s 50-50 once we know that Monty is going to open a door with a goat (that trashes randomness). As someone replied (paraphrasing) since we know from the get-go that Monty will always open a dud door, then we should know that the 2 doors we didn’t choose are ‘linked’ doors, and in effect act as a single door…

This reminds me another classic teaser where: if a car goes 30mph for 1/2 mile, how fast does the car have to go to average 60mph for 1 mile? The convoluted answer is that 30mph=2 miles/minute and that 1 minute elapses in 1 mile, while 60mph=1 mile/minute, whereby no more time is available to ‘average’ 60mph across entire 1 mile, thus goofy answer is infinite. Hmmm, every cop I’ve run this past said tell that to the judge…

Jim Frost says

Hi Matt, I’m pretty sure you meant it’s 33/67 because of the linked doors. It’s because it’s not random that you’re twice as likely to win by switching.

Henryk Pliszczak says

I don’t see here problem at all. It is only to options here. 1 option the door you chose- 33.3333% chance. 2 option the two other doors 66.6666% chance, the two doors works like one because one is open. Is nothing to even thing about. Just switch and take the 66.666% chance

Jim Frost says

Yes, exactly!

Clint says

Here is my issue with the question that tripped me up, while initially got that Monty would use his knowledge to not pick the door with the car behind, i actually never thought it would get to that point and here is why. No where in the problem does it state that Monty will always offer to switch doors, my brain actually said that he will only offer to switch when i have choosen the right door and he knows it.

So then he would wait for me to choose, if i choose correctly he will then make the offer for me to switch. If i do i lose, if i don’t i win.

If on my first choice i don’t pick the car he stays silent. And so in this situation i lose because he never gives me the option to switch.

I can see the logic, thought not stated that if he choosing it is logical he would not choose the door with the car behind it. What i cannot see(agree with) is him always offering the option to switch because it is not stated he always offers it, and thus my logical assumption was he would only offer it when i have choosen correctly and he wants another chance at making me choose incorrect.

Is there a fault in my logic i am missing somewhere?

Quin Bagwell says

Whether Monte knows or not doesn’t change the odds, as long as he opens a door that doesn’t contain the prize. The odds are still 2/3 that it is behind the one he didn’t open.

Ronald says

Hi, Mike.

You must distinguish two different things: one is the frequency with which each option will be the correct one, and another is the frequency with which a player selects the correct one. The switching door is the correct option 2/3 of the time, and the staying one is only 1/3. So, if you are the original player who knows which option is which and you always decide to stick with your first option, you will win 1/3 of the time.

Instead, a person that does not know which door is which cannot force his selection to always be the switching door. That person has 1/2 probability to select each. If played several times, about half of them would select the switching door, and the staying door in the other half, so that person ends winning 50% of the time but not because each door wins 50% of the time, but because the advantage that gets with one is compensated with the disadvantage that gets with the other. They average:

1/2 * 2/3 + 1/2 * 1/3

= 1/2 * (2/3 + 1/3)

= 1/2.

To put another more clear example: imagine a psychopath raped lots of persons and forced them to play this macabre game to survive. Each person will have two bottles, one green and the other red, where the green one always contains a deadly poison and the red one always contains a natural juice. Each person is forced to drink one of of those bottles, but has no clue about which has the poison, so the person has 50% chance to select either and so 50% chance to survive. No one can see what other person selected, so each selection will be completely random and only about half of the total persons will survive. But that does not mean that the green bottle is killing 50% of the persons that drink it. It kills 100% of the persons that drink it, only that about half of the total population selects it:

1/2 * 100% + 1/2 * 0%

= 1/2

Mike says

Okay, thank you for the great explanation. I just have one more monkey wrench to throw in. Let’s say you go through the first round with Monty. You’ve picked your door and he’s taken one away. You decide 5o stick with the door you got. Great. Now, they bring your friend Ted in from offstage, and also ask him to pick the door with the car. Ted has no idea what door you picked or what door has been left over by Monty. There are two doors, one with the car and one without. Does Ted have a 50/50 chance of picking the right door? Won’t he pick your door half the time and be right half the time? Can one door still represent a 33% chance of being right and the other a 66% of being right, bur because Ted has no knowledge of what transpired, he we still win the car half the time? Hopefully you can see the paradox I am trying to draw out. How can Ted be a winner half the time but you only be a winner a third of the time if he’s just as likely to pick your door as the one with the car behind it?

IH8ALOTOFPEEPOLE says

i agree

Mike says

The colour on square A and B is the same, but by having them adjacent to in case A a lighter colour it looks darker and in case B it’s adjacent to a darker colour and that makes it look lighter.

Or in short, don’t believe you’re lying eyes, trust what you can measure or calculate.

Dylan says

I see it this way…

Let’s say:

Door 1 = Donkey

Door 2 = Donkey

Door 3 = Car

If you choose door 1 he will be FORCED to open door 2. He can’t open 1 since you picked it and can’t open 3 since it has the prize.

You switch = you win

You don’t = you loose

If you choose door 2 he will be FORCED to open door 1. He can’t open 2 since you picked it and can’t open 3 since it has the prize.

You switch = you win

You don’t = you loose

If you choose door 3 he won’t be forced, he will have a choice between door 1 and 2…

You switch = you loose

You don’t = you win.

If you picked a donkey with your first pick and then switched, you 100% won the prize. Since there is 2 donkeys and 1 prize, the chance of you picking a donkey is 2/3. Meaning that you obviously have a higher chance of winning the price when switching.

This was super hard for my little brain to understand. Lol

Jim Frost says

Hi Dylan, that’s a great explanation of how it works!

Stephen Longbotham says

The problem never states Monty Hall’s reasoning as to which door he opens. Since the problem does not state that Monty Hall opens a door that he knows does not have the prize, a person could assume Monty Hall was just opening any random door. In that case, there is a one out of three chance the door he opens has the prize behind it. If Monty Hall opens a random door and it happens to be one of the doors with no prize behind it, the contestant can not increase his chances to win the prize by switching choices. For the problem to work, Monty Hall has to know which door has the prize behind it and the contestant has to know Monty would not open the door with the prize behind it.

Jim Frost says

Hi Stephen,

It’s a reasonable assumption that Monty knows the location of the prize and won’t randomly open a door that could contain the prize. Yes, that’s part of the trickiness of the solution!

You’re correct, if Monty did open the doors randomly, the contestant doesn’t gain by switching. However, 1/3 of the time Monty would open the door with the prize. That just didn’t happen on the show and it would nullify the whole contest.

For your last part, for the problem to work, only Monty needs to know which door contains the prize. The contestant doesn’t have to know that Monty knows. It’s Monty’s knowledge and non-random choice that drives the probabilities in this scenario. The contestant’s knowledge only determine whether the contestant can take advantage of the process to increase their chances of winning. If the contestant doesn’t know, there’s still an advantage to switching, but the contestant won’t be able to take advantage of it knowingly. It would be like flipping a coin that has a 2/3 chance of heads and 1/3 chance of tails. If you knew, you’d pick heads. However, if you didn’t know, it’s still advantageous to pick heads, you just wouldn’t know that was the better choice!

Jos Cambell says

The difficulty in understanding of this problem is that you end up focusing on the final choice, not the choice that brought you there (some life lesson etc… etc…)

i.e. Whatever I choose initially, I have a 50:50 shot of having it right by the time I get to choose to either hold or switch.

The crux is that you are more likely to choose a goat with your first choice (2/3) (the other goat always removed after) and therefore more likely to need to switch to win, due to that first choice disadvantage.

This used to bake my noodle, then I got real drunk and it made perfect sense, then I signed up to some weird website to post a comment. You people are weird, I like it.

JUST * (@JUSTTWITT) says

This is correct. I have the same conclusion too.

Eric says

I’m confused. Brain hurts, but I love it.

Scratch the 100 door scenario. It’ll help me visualize better if it’s three doors. Everything else is the same.

If it’s a three card game and the second contestant is brought out after a goat is revealed and then the contestant is told to pick a door are his chances at that point 50/50? At that point I see the contestant basically playing a new game independent of the “switch or stay game” which he knows nothing about. Now it’s just pick one of two cards.

Jim Frost says

Hi Eric,

I’m glad you love it even though your brain hurts! I think those are some of the best puzzles!

In this case, you have to remember that what the contestant thinks is irrelevant to the outcome. The important part is understanding Monty’s process–and that hasn’t changed. The contestant might have a misunderstanding of that process, which doesn’t affect the true probabilities, but it will affect the contestants decision making. So, yes, he is left with two cards and he might think the odds are 50/50, but in actuality they are still 33% and 67%. He might not think it’s to his advantage to switch, but it still is. You have to understand the true underlying process to have the best chances.

That really ties into the larger picture of using accurate data to make decisions–which is why I love statistics. You use the best methodologies to collect the best data available, apply the correct analyses, and that gives you the best chances of making the correct decision. If you have incorrect data, that can lead you to making bad decisions. And that’s the situation the new contestant is in if he thinks it’s 50/50. He might make a poor decision because of bad data.

The universe doesn’t behave as you think it ought to behave. It behaves as it actually behaves and it’s up to you understand it correctly and make decisions accordingly.

Bob Z says

I really like your explanation – and there are plenty out there. Plus there are few enough outcomes that you can easily build a table, which you did, illustrating all possible choice and results. You can’t refute that. Similar to your last poster, I read you could perform the same exercise with a deck of cards. You could take the deck of 52 and ask somebody to pick a card. What are the odds they picked the Ace of Diamonds? 1 in 52. What are the odds you hold the Ace in the remaining stack? 51 in 52. I could examine my stack of cards and start flipping over all cards that are not the Ace of Diamonds. No matter how many cards I flip over and show you, your 1 in 52 odds of picking it do not change….even if I flipped over all but the last one in the deck.

Jim Frost says

Hi Bob,

I think using a deck of cards is a great way to illustrate that principle! And, I do think the illusion falls apart when you add more doors, or more cards.

Eric says

Hi Jim

Could you comment on the following:

Imagine a variation on the Monty Hall Game where everything is the same except that the game is played with 100 doors and Monty opens 98 doors revealing all goats (the statistical illusion collapses with larger numbers) and right after the contestant is asked to “stay or switch” doors he leaves the studio and a new contestant walks out from an isolation booth. The new Second contestant is completely ignorant of everything that has happened between Monty and the first contestant. The 2nd contestant sees two closed doors and 98 open doors with goats. Monty tells him one thing: he has a 50/50 chance of opening one of the two closed doors and winning the prize.

Is Monty correct about the SECOND contestant’s 50/50 odds?

If contestant one stays with his door are his odds 1 in a 100?

If contestant 1 switches are his odds 99 in 100?

Jim Frost says

Hi Eric,

I agree, the illusion falls apart when you consider more doors. It becomes more obvious that Monty is selectively opening doors that don’t have the prize.

For your scenario, it’s important to remember that it’s Monty’s process that counts rather than what the contestant thinks happens. Knowledge about the process helps you make the correct decision. Inaccurate knowledge won’t change the actual probabilities but will change the contestants decision making process and, in this case, lead him/her astray.

So, to answer your questions, Monty is incorrect when he says it’s 50/50. And yes, if the contestant stays with his door, there is 1/100 chance of winning. And, there is a 99/100 chance of winning by switching.

Of course, the contestant won’t know what Monty’s process was and would likely take Monty at his word. That could lead the contestant astray. But, the underlying probabilities are still 1/100 and 99/100. Knowledge is key!

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.