In a prison there were three prisoners, $A_1$, $A_2$ and $A_3$. A draw had been made to give two of them a pardon. $A_1$ asked a guard (who knew the draw) the name of another prisoner who had been pardoned. The guard said that he couldn't do that, otherwise the probability that $A_1$ would be pardoned would change from $\frac{2}{3}$ to $\frac{1}{2}$. Is the guard correct? (sorry for my english)
At first the guard seems to be wrong to me.
But I've given a proof that he's correct:
Let $A_i$ be the event "$A_i$ has been pardoned".
Suppose, without loss of generality, that the guard told $A_1$ that $A_3$ has been forgiven.
$$P(A_1|A_3)=\frac{P(A_1\cap A_3)}{P(A_3)}$$
Note that $P(A_1\cap A_3)$ is the probability of $A_2^c$, which is $\frac{1}{3}$. And $P(A_3)=P(A_1^c)+P(A_2^c)=\frac{2}{3}$.Therefore,
$$P(A_1|A_3)=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$$
What's wrong?
The answer is no. The guard is not correct because the draw had already been made. If two names were being pulled from a hat and the first one was $A_3$, then the probability that $A_1$ would come up on the next draw would be $\frac{1}{2}$.
Note, however, that $A_1$'s probability of being pardoned is still $\frac{2}{3}$. This is because there is a probability of $\frac{1}{3}$ of him being pardoned on the first draw followed by a probability of $\frac{2}{3}\cdot\frac{1}{2} = \frac{1}{3}$ of being selected on the second draw for a total probability of $\frac{2}{3}$.