# Three-prisoner problem and Bayes rule

Here is the wiki of Three-prisoner problem, in which only one prisoner is pardoned, and the Bayes solution is given in the wiki.

My problem is pretty much the same, except that only one prisoner is executed. Here is the description:

A, B, C are three prisoners, two of which will be pardoned, A asks the warden to tell him which ONE will be pardoned other than himself. A thinks that before asking, he has a chance of $\frac{2}{3}$ to be pardoned, but if he asks, the chance goes down to $\frac{1}{2}$. A's calculation is wrong, explain why.

I wanted to do it the Bayes way, let $A,B,C$ be the event that A,B,C will be pardoned repectively, and let $b$ be the event that the warden mentions B will be pardoned.

The prior here is $P(A) = \frac{2}{3}$, and after observing $b$, the posterior of $A$ is $$P(A|b) = \frac{P(b|A)P(A)}{P(b)}$$ . I think $P(b|A)P(A) = \frac{1}{2}\frac{2}{3}$, since if $A$ happens, the warden should not mention $A$, he should mention the other one who will be pardoned, which has probability $\frac{1}{2}$ for B. And $$P(b) = P(b|A)P(A) + P(b|B)P(B) + P(b|C)P(C)$$ , right? How to calculate this?

The way you've defined them, $A$, $B$, and $C$ are not disjoint events ($A$ and $C$ can happen at the same time, for example), so the equation $P(b)=\sum_{X\in{A,B,C}} P(b|X)$ doesn't work.

It would be nicer to define $A$ as "A is executed", $B$ as "B is executed", and $C$ as "C is executed". This way $A,B,C$ are disjoint and cover the whole probability space $P(A)+P(B)+P(C)=1$.

Assuming that we are looking from A's point of view, we leave $b$ meaning "A finds out that B is pardoned".

So we have $P(A)=P(B)=P(C)=\frac{1}{3}$.

$P(b|A) = \frac{1}{2}$ since the warden could mention either B or C to A.

$P(b|B) = 0$ since if B is executed, the warden won't tell A that B is pardoned.

$P(b|C) = 1$ since if C is executed, the warden has no choice but to tell A that B is pardoned.

$$P(A|b) = \frac{P(b|A)P(A)}{P(b|A)P(A) + P(b|B)P(B) + P(b|C)P(C)}$$ $$P(A|b) = \frac{P(b|A)}{P(b|A) + P(b|B) + P(b|C)}$$ $$P(A|b) = \frac{\frac{1}{2}}{\frac{1}{2} + 0 + 1}$$ $$P(A|b) = \frac{1}{3}$$

So knowing $b$ doesn't change A's chances of being executed.