A three-man jury has two members each of whom independently has probability \(p\) of making the correct decision and a third member who flips a coin for each decision (majority rules) A one-man jury has probability \(p\) of making the correct decision. Which jury has the better probability of making the correct decision? 1
Solution
The winning outcomes for the first scenario (three juris) are:
| outcome | probability |
|---|---|
| \(A_0B_1C_1\) | \(\frac{1}{2}(1-p)~p\) |
| \(A_1B_0C_0\) | \(\frac{1}{2}(1-p)~p\) |
| \(A_1B_1C_0\) | \(\frac{1}{2}p^2\) |
| \(A_1B_1C_1\) | \(\frac{1}{2}p^2\) |
Hence, the probability of a correct decision is given by:
\[\begin{align} 2\frac{1}{2}(1-p)~p + 2\frac{1}{2}p^2 & = (1-p)~p + p^2 \\ & = p - p^2 + p^2 \\ & = p \end{align}\]The two scenarios are strictly equivalent.
Intuition
Half the times the flippant member agrees with the first member. In these cases the conditional probability of a correct decision is \(p\). The other half it agrees with the second member, where the conditional probability is also \(p\), thus \(\frac{1}{2}p + \frac{1}{2}p = p\).
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This is problem 3 of Frederick Mosteller’s “Fifty Challenging Problems in Probability”. ↩︎