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Monty Hall problem

The Monty Hall problem is a riddle in elementary probability that arose from the American game show Let's Make a Deal with host Monty Hall. In spite of being an elementary problem, it is notorious for being the subject of controversy about both the statement of the problem and the correct answer.

The problem is as follows: At the end of the show, a player is shown three doors. Behind one of them, there's a prize for him to keep, while the other two contain goats (signifying no prize to be won). Although the show host knows what is behind each door, of course the player does not. After the player makes a first choice, Monty opens one of the two other doors, revealing a goat. He then offers the player the option to either stick with the initial choice or switch to the other closed door. Should the player switch?

The classical answer to this problem is yes, because the chances of winning the prize are twice as high when the player switches to another door than they are when the player sticks with their original choice. This is because upon the original choice, the player has only a 1/3 chance of choosing the door with the prize; this probability does not change when Monty opens a door with a goat. Hence the chances of winning the prize are 1/3 if the player sticks to their original choice, and thus 2/3 if the player switches.

Table of contents
1 Assumptions
2 Aids to understanding
3 Variants
4 Origins
5 Anecdotes
6 References
7 External Links


The classical answer presented above makes two assumptions that are rarely made explicit:

If one of these assumptions is violated, the answer is different. In the first case, it could be that Monty does not always open a losing door. Maybe in some shows he does, and on other occasions he does not, he simply gives the contestant whatever is behind his first choice door. If that is the case, it all depends on Monty's character:

In the second case, Monty may open an unpicked door at random, rather than always opening a door with a goat behind it. In this case, if Monty happens to open a door with a goat behind it, then both switcher and sticker have a 50% chance of winning, so it doesn't matter what you do. This is because a correct initial guess means that Monty is certain to open a door with a goat behind it, whereas if the initial guess was incorrect there is only a 50% chance that Monty will open a door with a goat behind it.

Aids to understanding

In about one third of the games, the contestant's chosen door will be the prize door and Monty will choose one of the other two doors at random. A sticking contestant will win and a switching contestant will lose when this happens, no matter which door Monty picks, since the unchosen door must be a losing door.

In about two thirds of the games, the contestant's door will be a losing door and Monty will have to choose the other losing door. Therefore the unchosen door is the prize door in these cases. A sticking contestant will lose and a switching contestant will win when this happens.

A contestant may now choose to stick or to switch, knowing that the unchosen door is a loser in one third of the games and a winner in two thirds of the games. Knowing this, a rational contestant will always switch.

Empirical proof of the Monty Hall problem for a Perl program which demonstrates the result.


In the original game show, there were in fact two contestants. Both of them chose a door; they were not allowed to choose the same one. Monty then eliminated a player with a goat behind their door (if both players had a goat, one was eliminated randomly, without letting the players know about it), opened the door and then offered the remaining player a chance to switch. Should the remaining player switch?

The answer is no. The reason: a switcher in this game will lose if and only if either of two initial choices of the two contestants was correct. How likely is that? Two-thirds. A sticker will win in those 2/3 of the cases. So stickers will win twice as often as switchers.

There is a generalization of the original problem to n doors: in the first step, you choose a door. Monty then opens some other door that's a loser. If you want, you may then switch your allegiance to another door. Monty will then open an as yet unopened losing door, different from your current preference. Then you may switch again, and so on. This continues until there are only two unopened doors left: your current choice and another one. How many times should you switch, and when, if at all?

The answer is: stick all the way through with your first choice but then switch at the very end. This was proved by Bapeswara Rao and Rao.

An unrelated paradox involving infinite sequences of actions is sometimes called the Monty Hell problem.


The game used in the Monty Hall problem is similar to three card monte, a gambling game in which the player has to find a single winning card among three face-down cards. As in the Monty Hall problem, the dealer knows where the winning card is and tries to trick the player into picking the wrong card. As the card is often a Queen court card, it is also known as Find the Lady

An older paradox in probability theory involves three prisoners, one of whom (already chosen at random but unknown to the prisoners) is to be executed in the morning. The first prisoner begs the guard to tell him which of the other two will go free, arguing that this reveals no information about whether he will be the victim; the guard responds by claiming that if the prisoner knows that a specific one of the other two prisoners will go free it will raise the first prisoner's subjective chance of being executed from 1/3 to 1/2. The question is whether the analysis of the prisoner or the guard is correct.


After this problem's solution was discussed in Marilyn vos Savant's "Ask Marilyn" question-and-answer column of Parade magazine in 1990, many readers including several math professors wrote in to declare that her solution was wrong.


External Links