Explaining Monty Hall problem by Bayes' theorem

Question

The Monty Hall problem is the famous puzzle that, in a game of choosing the right box containing the treasure, one should switch to another box when told that one of the unchosen ones is empty. My friend tries to use Bayes' theorem to explain this puzzle, though reaching a contradictory statement. For case of 3 boxes, his argument is like this: let $C$ denote the event that the unchosen box is empty, and $A$ the event that the chosen one has the treasure. Then by bayes' theorem,

$$\mathbf{P}(A|C) = \frac{\mathbf{P}(A \cap C) }{\mathbf{P}(C) } = \frac{\mathbf{P}(A) }{\mathbf{P}(C) } = \frac{1/3}{2/3} = \frac{1}{2}$$

which is wrong. So he asked me which part is mistaken, and I think it should be the probability of $\mathbf{P}(C)$, which surprisingly should be $1$. The reason is that, no matter we have made the correct choice or not, there always exists an unchosen box not containing the treasure. So the event $C$ actually doesn't tell us any information and thus $\mathbf{P}(A|C) = \mathbf{P}(A) = \frac{1}{3}$ which is correct. Is my rationale sound, or is there any better explanation?

Answer 1

Here are several correction:

• $P(A|C)=1$ because if the one we did not choose is empty then the one we chose must contain the treasure
• $P(C)=\frac{1}{3}$ because the probability of treasure among the boxes we did not choose is $\frac{2}{3}$ so the probability of treasure not in these boxes is the complement

In the end you did find the probability of getting treasure if you do not switch i.e. the probability that the box we chose contain treasure and the one we did not is empty $P(A\cap C)=\frac{1}{3}$