The Flippant Juror

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 $\frac{1}{2}p + \frac{1}{2}p = p$ .

This is problem 3 of Frederick Mosteller’s “Fifty Challenging Problems in Probability”. ↩︎