Alternative Monty Hall Discussion

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Here is a simpler analysis of the Monty Hall problem.

First, it's useful to know a basic fact about conditional probability. For two events X and Y, since

Pr(Y|X) = \frac{ Pr (X \cap Y) }{Pr(X)},

we can write

Pr( X \cap Y) = Pr(X) Pr(Y|X)

OK, now consider the Monty Hall problem. There are three doors, call them A,B, and C. Let's consider the case where the contestant initially chooses door A. We assume that:

  • The prize is placed randomly behind a door, with equal probability.
  • If the prize is behind door A (the door the contestant chose initially), Monty Hall opens door B or C with equal probability.
  • If the prize is behind door B, Monty Hall always opens door C.
  • If the prize is behind door C, Monty Hall always opens door B.

We want to turn this into probability statements. Let P denote the location of the prize, and let H denote the door opened by Monty Hall. Then:

Pr(P=A) = Pr(P=B) = Pr(P=C) = \frac{1}{3}

In addition,

Pr(H=B|P=A) = Pr(H=C|P=A) = \frac{1}{2}

Pr(H=B|P=B) = 0, \quad Pr(H=C|P=B) = 1,

Pr(H=B|P=C) = 1, \quad Pr(H=C|P=C) = 0.

A little thinking should convince you that this completely specifies probabilities of all possible outcomes.

Now, consider the two possible strategies for the contestant. The first strategy is to stick with her original choice (A). Then the probability of winning the prize is:

Pr(P=A) = \frac{1}{3}.

Now consider the strategy of changing your choice after Monty Hall has opened one of the other doors. Let's say Monty opens door B. (A similar argument will apply if he opens door C.) If the contestant changes to door C after Monty opens door B, what is the probability of getting the prize?

Pr(P = C | H = B) = ???

Using our initial result, we can write

Pr(P=C|H=B) = \frac{ Pr( \{P=C\} \cap \{H=B\} ) }{Pr(H=B)} = \frac{ Pr(H=B|P=C) Pr(P=C) }{ Pr(H=B) }

The numerator is

Pr(H=B|P=C) Pr(P=C) = 1 \cdot \frac{1}{3} = \frac{1}{3}.

For the denominator, note that we can write

Pr(H=B) = Pr( \{ H=B\} \cap \{ P=A \} ) + Pr( \{ H=B \} \cap \{ P=B \} ) + Pr ( \{ H=B\} \cap \{P=C\}).

This in turn equals

\begin{align} & Pr(H=B|P=A) Pr(P=A) + Pr(H=B|P=B) Pr(P=B) + Pr( H=B|P=C) Pr(P=C) \\ & = \frac{1}{2} \cdot \frac{1}{3} + 0 + 1 \cdot \frac{1}{3} = \frac{1}{2}. \end{align}

So putting this all together we get

Pr(P=C|H=B) = \frac{1/3}{1/2} = \frac{2}{3}.

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