Computational divertissements

Statistical independence and the Monty Hall Problem

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I want to use the notions of statistical independence and conditional probabilty to present a straightforward solution to the Monty Hall problem.

Notation: probablists often write AB or even A,B for the intersection of sets A and B. Both are shorthand for $A \cap B$. The comma version looks better when the events have names longer than single letters.

Definition: Events A and B are said to be independent if P(AB) = P(A) P(B). In words, the probability that both A and B occur is the product of their respective probabilities.

The solution to the MH problem that I will propose hinges upon the fact that if an event C has probability 1, then C is independent of any event E.

&&Claim: P(C) = 1 \Rightarrow P(EC) = P(E)P(C) = P(E)&&

This follows from facts that usually are demonstrated or assumed as axioms in an intro to probability:

Let's establish the claim. Let $P(C)=1$ and hence $P(\bar C)=0$. For any event E, &&E = E \cap (C \cup \bar C) = EC \cup E\bar C&&

The rightmost union is a disjoint union, so &&P(E) = P(EC) + P(E\bar C) &&

But $P(E\bar C) = 0$ because $E\bar C \subseteq \bar C$ so $P(E\bar C) \le P(\bar C) = 0$. It follows that $P(EC) = P(E),$ as claimed. If an event C is certain, i.e. P(C) = 1 then C is independent of any event E.

Now let's talk about the Monty Hall problem.

Monty shows you N doors. Behind one of the doors is a prize, behind the rest there is nothing. You win if you can guess which door conceals the prize. The probability that you initial guess is correct is 1/N.

But there is a twist to the game. After you make your guess Monty opens all the doors but two, the one you chose and one other of his choice. After this big reveal Monty offers you the opportunity to switch doors. This is the problem, should you switch? A lot of people argue that there are two doors left, the prize is behind one, so the odds are 50-50. They conclude that it does not matter whether you switch.

Let's think about the game. From the outset Monty knows where the prize is. He can always choose doors so that the prize is not behind any of the N-2 doors opened during the reveal. In fact, he always does this, that is how the game is played.

One of two things happens during the reveal. If you did happen to pick the right door then Monty opens all but one of the remaining doors, it does not matter which. In the more likely case that you did not guess correctly Monty's hand is forced. He cannot open either your door or the prize door. In that (more likely) case the unopened door other than yours hides the prize.

Let's do the numbers. Before the reveal we had P(Win)=1/N, where W is the event that your initial guess was correct. Does the reveal change that? We need to compute the winning probability conditional on the reveal, P(Win|Reveal).

The general formula for conditional probabilities is P(A|B) = P(A,B)/P(B). So P(Win|Reveal) = P(Win,Reveal)/P(Reveal). Recall that Monty always does the reveal, that is how the game is played. So P(Reveal)=1. Consequently P(Win,Reveal) = P(Win) P(Reveal)/P(Reveal) = P(Win). This means that P(Win|Reveal) = P(Win) = 1/N. The probability that the original guess was correct is still 1/N.

Some people see two doors after the reveal, and reason that the odds must be 50-50. They believe (at least implicitly) that the reveal changed the win probability of the original guess from 1/N to 1/2. Not so; that probability is not changed by the reveal. But the reveal does change the number of doors remaining in the game, from N to 2. The original choice of door continues to win with probability 1/N, and the set of doors other than the original continues to win with probability 1-1/N. But now there is only one other door in that set. That remaining door wins with probability 1-1/N. It is better to switch.