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Thayer Watkins
Silicon Valley
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 The Monty Hall Game

Monty Hall was the host of a television game show in which contestants were allowed to choose one of three doors. Behind each door was a prize but one prize was very good and the other two were not so good.

Let us say that behind one door was a car and behind the other two were goats. After the contestant chose a door Monty Hall would reveal a goat behind one of the doors not chosen by the contestant. He would then give the contestant the opportunity to switch his or her choice. Intuition says that it shouldn't matter whether the contestant switches or not. Intuition is however wrong in this case. To understand why we need to consider the concept of conditional probability.

The conditional probability of A given B, P[A|B], is the probability that A has or will occur given that B has occurred. In some cases it is possible to compute conditional probabilities directly. In other cases, it is easier to determine a conditional probability indirectly by means of the formula

## Analysis of the Monty Hall Problem Using Conditional Probability

Take a typical situation in the game. Suppose the contestant has chosen Door 3 and Monty Hall reveals that there is a goat behind Door 2. Let us now compute the conditional probability that the car is behind Door 1. Let A stand for the condition that there is a car behind Door 1 and the contestant has chosen Door 3. Let B stand for the condition that Monty Hall has revealed that there is a goat behind Door 2 given that the contestant has chosen Door 3. The probability of A and B ( P[A and B] ) is just (1/3)(1/3)=1/9 because if the car is behind Door 1 and the contestant has chosen Door 3 it is 100 percent certain that Monty Hall will show what is behind Door 2. The problem is the computation of the probability of being shown a goat behind Door 2 given that the choice was Door 3. This situation can arise in two ways. One, when the car is behind Door 1 and another when the car is behind Door 3. The first way has a probability of 1/9, as has already been computed above. In the second way, the host could reveal either what is behind Door 1 or Door 2. If he is equally likely to choose either of these doors then the prob ability of the second way is (1/2)(1/9). Therefore the probability of there being revealed a goat behind Door 2 when the con testant has chosen Door 3 is

#### (1/9) + (1/2)(1/9) = (3/2)(1/9).

This is the probability of B; i.e., P[B].
The conditional probability we want, P[A|B], is given by

#### P[A|B] = P[A and B]/P[B] = (l/9)/(3/2)(1/9) = 2/3.

This means that the conditional probability that the car is behind Door 3 given that the contestant has chosen Door 3 and has been shown that there is a goat behind Door 2 is only l/3. Therefore it is worthwhile to switch.

This other conditional probability can be calculated either as

#### 1 - P[A|B],

or from the probability the car is behind Door 3 given that the contestant has chosen Door 3 and has been shown a goat behind Door 2. This latter probability is (1/3)(1/3)(1/2) so when it divided by P[B] one gets l/3.