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 |

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:

- If Monty is a helpful person and wants you to win, then you should always switch, because the fact that Monty offered you a second chance means that your first choice was a goat.
- If Monty is mean or received pressure from his tv station, you should always stick. The fact that Monty offered you a second chance only means that he wants to lure you away from your correct first choice. On the other hand, if Monty is truly devious, he could on very rare occasions open a door even if the contestant's initial choice was incorrect, just to fool future contestants.

- Instead of one door being opened and thus eliminated from the game, it may equivalently be regarded as combining two doors into one, as a door containing a goat is essentially the same as a door with nothing behind it. In essence, this means the player has the choice of either sticking with their original choice of door, or choosing the sum of the contents of the two other doors. Clearly, the chances of the prize being in the other two doors is twice as high. Notice how the above assumptions play a role here: The reason switching is equivalent to taking the combined contents is that Monty Hall is
*required*to open a door with a goat. - It may also be easier for the reader to appreciate the result by considering a hundred doors instead of just three, with one prize behind only one of the doors. After the player picks a door, Monty opens 98 doors with goats behind them. Clearly, there's now a very high chance (precisely 99/100) that the prize is in the other door Monty did not open.
- It may also help to think in terms of why Monty chooses a particular door. Before Monty opens any doors, the contestant knows that each door is equally likely to have the prize behind it, therefore in about one third of the games the contestant will be lucky and choose the door with the prize. However Monty's choice of door adds information which the contestant can make use of. Monty will not choose the contestant's door, but which of the other two he chooses depends upon what is behind the contestant's door.

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.

- John Tierney "Behind Monty Hall's Doors: Puzzle, Debate and Answer?",
*The New York Times*July 21, 1991, Sunday, Section 1; Part 1; Page 1; Column 5 - Bapeswara Rao, V. V. and Rao, M. Bhaskara: "A three-door game show and some of its variants", The Mathematical Scientist 17 (1992), no. 2, pages 89-94