Monty Hall problem

The Monty Hall problem is a puzzle in game theory involving probability that is loosely based on the American game show Let's Make a Deal. The name comes from the show's host, Monty Hall. In this puzzle a player is shown three closed doors; behind one is a car, and behind each of the other two is a goat. The player is allowed to open one door, and will win whatever is behind the door. However, after the player selects a door but before opening it, the game host (who knows what's behind the doors) must open another door, revealing a goat. The host then must offer the player an option to switch to the other closed door. Does switching improve the player's chance of winning the car? The answer is yes — switching results in the chances of winning the car improving from 1/3 to 2/3.

Aids to understanding

The most common objection to the solution is the idea that, for various reasons, the past can be ignored when assessing the probability. Thus, the first door choice and the host's reasoning about which door he opens are ignored. Because there are two doors to choose from, there is then a fifty-fifty chance of choosing the right one.

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Although ignoring the past works fine for some games, like coin flipping, it doesn't work for all games. The most notable counterexample is card counting in some card games, which allows players to use information on past events to their advantage. Past information helps also in the Monty Hall problem, as shown below.

Related Topics:
Coin flipping - Card counting

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Venn diagrams

The probability that the car is behind the remaining door can be calculated with the Venn diagrams below. After choosing door 3, for example, the player has a 1/3 chance of having selected the door with the car, leaving a 2/3 chance between the other two doors. Note that there is a 100% chance of finding a goat behind at least one of the two unchosen doors because there is only one car.

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The host now opens door 1. Since the host must always open a door revealing a goat, and does not open a door at random, opening this door does not affect the chance that the car is behind the originally chosen door which remains 1/3. The car is not behind door 1, so the entire 2/3 probability of the two unchosen doors is now carried only by door 2, as shown below. Another way to state this is that if the car is behind either door 1 or 2, by opening door 1 the host has revealed it must be behind door 2.

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A more formal probability diagram is shown below.

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Increasing the number of doors

It may be easier to appreciate the result by considering a hundred doors instead of just three. In this case there are 99 doors with goats behind them and 1 door with a prize. The player picks a door; 99% of the time, the player will pick a door with a goat. Thus, the chances of picking the winning door at first are very small: only 1%. The game host then opens 98 of the other doors revealing 98 goats and offers the player the chance to switch to the other unopened door. On 99 out of 100 occasions the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. At this point a rational player should always switch.

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Combining doors

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. Notice how the above assumptions play a role here — the reason switching is equivalent to taking the combined contents is that the game host is required to open a door with a goat.

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Bayes's theorem

Analysis of the problem using Bayes's theorem has the least reliance on verbiage and the most on formal mathematics. It also makes explicit the effect of the assumptions given earlier. Consider the position when door 3 has been chosen and no door has been opened. The probability that the car is behind door 2, p(C2), is plainly 1/3, as it may equally well be in any of the three places. The probability that the game host will open door 1, p(O1), is 1/2 since there is equal probability the car is behind door 1 (forcing the host to open door 2) or door 2 (forcing the host to open door 1) and if the car is behind neither door by the given assumptions the host opens one of them randomly. But when the car is behind door 2, the game host will certainly open door 1, by the assumptions; that is, p(O1|C2) = 1. Hence the probability that the car is behind door 2 given that the game host opens door 1 is

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:

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P(C2|O1) = rac{P(O1|C2) P(C2)}{P(O1)}

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