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 1 3 to 1 2 because he would then be one of two prisoners. What do you think of the jailer’s reasoning?

From the question, we observe that 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.

We have to find the jailor's reasoning.

Let's consider $A,B$and $C$are the three events.

The probability of each prisoners will be die is equal and they are given below:

$P\left(A\right)=\frac{1}{3}$

$P\left(B\right)=\frac{1}{3}$

$P\left(C\right)=\frac{1}{3}$

If $A$is to die he could be told either $B$ is to be forced or $C$is to be forced, each with a probability of $\frac{1}{2}$.

Consider $D$remains the event that represents the jailor told $B$to be freed.

That is, the conditional probability of jailor told $B$to be freed given $A$dies is as follows.

$P\left(D\right|Adies)=\frac{1}{2}$

If $B$is to die, $A$would not be told $B$is to be freed.

Therefore, the conditional probability of jailor told $B$to be freed given $B$dies as follows.

$P\left(D\right|Bdies)=0$

If $C$is to be die, $A$would must be told $B$is to be freed. Therefore, the conditional probability of jailor told $B$told to be freed given $C$dies as follows:

$P\left(D\right|Cdies)=1$

By using Bayes theorem, to calculate the probability of prisoner $A$dies given that jailor told prisoner $B$to be freed.

$P(A\text{dies}\mid D)=\frac{P\left(A\text{dies}\right)P(D\mid A\text{dies})}{\left\{\begin{array}{l}P\left(A\text{dies}\right)\times P(D\mid A\text{dies})+P\left(B\text{dies}\right)\times \\ P(D\mid B\text{dies})+P\left(C\text{dies}\right)\times P(D\mid C\text{dies})\end{array}\right\}}$

$=\frac{\frac{1}{3}\times \frac{1}{2}}{\left(\frac{1}{3}\times \frac{1}{2}\right)+\left(0\times \frac{1}{3}\right)+\left(1\times \frac{1}{3}\right)}$

$=\frac{\frac{1}{6}}{\frac{1}{6}+0+\frac{1}{3}}$

Simplify,

$=\frac{\frac{1}{6}}{\frac{1}{2}}$

$=\frac{2}{6}$

Therefore,

$=\frac{1}{3}$

From the probability, we observe that the Jailor is wrong.