# Exercise C.2.10

$\star$ A prison warden has randomly picked one prisoner among three to go free. The other two will be executed. The guard knows which one will go free but is forbidden to give any prisoner information regarding his status. Let us call the prisoners X, Y and Z. Prisoner X asks the guard privately which of Y or Z will be executed, arguing that since he already knows that at least one of them must die, the guard won't be revealing any information about his own status. The guard tells X that Y is to be executed. Prisoner X feels happier now, since he figures that either he or prisoner Z will go free, which means that his probability of going free is now $1/2$. Is he right, or are his chances still $1/3$? Explain.

Obviously the event took place before prisoner X obtained that knowledge and this should not impact his chances in anyway.

But words are cheap, let's use math:

• y is the event of the guard saying prisoner Y will be executed
• X, Y and Z are the events of each prisoner going free
• If X is going free, the guard has a 50% of saying Y is executed

$$\Pr\{y|X\} = \frac 1 2 \\ \Pr\{y\} = \Pr\{y \cap X\} + \Pr\{y \cap Y\} + \Pr\{y \cap Z\} = \frac 1 3 \cdot \frac 1 2 + \frac 1 3 \cdot 0 + \frac 1 3 \cdot 1 = \frac 1 6 + \frac 1 3 = \frac 1 2 \\ \Pr\{X|y\} = \frac{\Pr\{y|X\} \cdot \Pr\{X\}}{\Pr\{y\}} = \frac{\frac 1 2 \cdot \frac 1 3}{\frac 1 2} = \frac 1 3$$

This one is also very popular