The Prisoner’s Dilemma

Problem

Three prisoners, \(A\), \(B\) and \(C\), with apparently equally good records have applied for parole. The parole board has decided to release two of the three, and the prisoners know this but not which two. A warder friend of prisoner \(A\) knows who are to be released. Prisoner A realizes that it would be unethical to ask the warder if he, \(A\), is to be released, but thinks of asking for the name of one prisoner other than himself who is to be released. He thinks that before he asks, his chances of release are \(\frac{2}{3}\). He thinks that if the warder says "\(B\) will be released", his own chances have now gone down to \(1/2\), because either \(A\) and \(B\) or \(B\) and \(C\) are to be released. And so \(A\) decides not to reduce his chances by asking. However, \(A\) is mistaken in his calculations. Explain.

Solution

Prisoner \(A\)’s mistake is not considering the entire sample space of pairs of prisoners that could be released, which is \(\Omega = \{AB, AC, BC\}\). In the case of \(AB\), the warden will say that \(B\) will be released, and \(A\) has a \(100\%\) chance of release. Similarly, in the case of \(AC\), the warden will say that \(C\) will be released, and \(A\) has a \(100\%\) chance of release. In the case of \(BC\), the warden will say that either \(B\) or \(C\) will be released, and \(A\) has a \(0\%\) chance of release. Each outcome has a \(\frac{1}{3}\) chance of occurring. When combined, this gives the overall chance of release with the warden’s statement as:

\[P(A\;Released) = \left(\frac{1}{3}\right)1 + \left(\frac{1}{3}\right)1 + \left(\frac{1}{3}\right)0 = \frac{2}{3}\]

Thus, asking the warden has no effect on prisoner \(A\)’s chances of being released.