Understanding a probability paradox

Question

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer the question, pointing out that if A knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from $\frac 13$ to $\frac 12$ because he would then be one of two prisoners. What do you think of the jailer’s reasoning?

If the jailer refuses to say anything, then the probability that prisoner $A$ is excecuted is $\frac{1}{3}$. If the jailer says to prisoner $A$ that prisoner $B$ will walk free, then $2$ prisoners remain to be considered, $A$ and $C$. One dies, one does not. Heads or tails essentially. $\frac{1}{2}$ ought to be the conditional probability that $A$ dies given that $B$ walks free no? Apparently not though, allegedly the correct answer is still $\frac{1}{3}$. Even my attempt to calculate the correct answer yielded the result $\frac{1}{2}$.

Let $A_D$ and $C_D$ respectively denote the event of $A$ and $C$ dying. Let $B_F$ denote the event that $B$ walks free. Assume that the jailer tells prisoner $A$ that prisoner $B$ will walk free.

Here's my attempt.

$$P(A_D\mid B_F)=\frac{P(A_D\cap B_F)}{P(B_F)}=\frac{P(A_D\cap B_F)}{P((B_F\cap A_D)\cup (B_F\cap C_D))}=\frac{P(A_D)P(B_F\mid A_D)}{P(A_D)P(B_F\mid A_D)+P(C_D)P(B_F\mid C_D)}=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times 1}=\frac{1}{2}$$

What am I doing wrong?

Edit 1: Intuitively I am still troubled but I understand now that $B_F$ may occur even though the jailer does not necessarally say $B$.

Edit 2: I suppose that it makes some sense if the faiths of the prisoners had already been decided before prisoner $A$ asked the jailer the question. If the jailer decides to reveal one of the others who will walk free then that must've been the case.

Answer 1

Let $A,B,C$ take values $0,1$. Note that $P[B =0 \text{ or } C = 0 ] = 1$ (one of the other two will always be freed).

$P[A = 1 | B =0 \text{ or } C = 0 ] = P[A = 1]$.

Answer 2

Regardless of whether $A$ will be executed or not, the jailor can still point to $B$ or $C$ as being let free. Therefore, in principle, $A$ does not learn anything new about their chances, so they remain at $1/3$. If for example the jailor replies that $B$ will go free, the probability of $C$ being executed then rises to $2/3$; whereas if the jailor replies that $C$ will go free, the probability of $B$ being executed then rises to $2/3$, while $A$'s probability stays at $1/3$ either way.

The reasoning here is essentially that in the famous Monty Hall Problem.

Answer 3

Your mistake is that you calculate the wrong probability. The probability you need to calculate is the probability that $A$ is executed under the condition that the jailer says that B will be free. Note that those two conditions are not identical, as it is possible that $B$ gets free, but the jailer will not say that because $C$ will also be free, and the jailer decided to name $C$ instead. Note that in this case, $A$ will be executed (because the other two are not).

In short, indeed $P(A_D|B_F)=\frac12$, but $P(A_D|J_B)\ne\frac12$ where $J_B$ describes the event that the jailer says that $B$ will get free. Note that $P(J_B|C_D)=1$, but $P(J_B|A_D)=\frac12$.

Answer 4

Let $A_{E}$ indicate the event that A is executed and $C_F$ and $D_F$ indicate the events that C and D will be set free, respectively. A is concerned about the chance of him being executed after knowing whether C or D is going to be set free.

Then the probability we are looking at is $ P(A_E|C_F \cup D_F) $. Since we know that C or D is going to be set free in all scenarios where one of the three prisoners is executed, $ P ( C_F \cup D_F)= 1$

Since, $$ P(A_E|C_F \cup D_F)= \dfrac{P(C_F \cup D_F|A_E)P(A_E)}{P ( C_F \cup D_F)} $$

Whether $ A_E $ happens or not, $C_F \cup D_F$ always happens. Thus $P(C_F \cup D_F|A_E)= P ( C_F \cup D_F)= 1 $.

Hence, $P(A_E|C_F \cup D_F)= P(A_E)= \dfrac{1}{3}$