Probability Problem Number 3 of Frederick Mostellers Book

In 1916 Frederick Mosteller published a collection of problems with solutions in his famous book „Fifty Challenging Problems in Probability„.

Problem Number 3 is called „The Flippand Juror“ and is described with the following lines:

A three-man jury has two members each of whom independently has a 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 a probability p of making the correct decision. Which jury has the better probability of making the right decision?

Three-man jury:

The table lists the possibilities and probabilites to receive a correct decision.

Juror 1 Juror 2 Juror 3 (flips) Majority Decision Probability
C C C C \(p \cdot p \cdot 0.5 \)
C C W C \(p \cdot p \cdot 0.5 \)
C W C C \(p \cdot (1-p) \cdot 0.5 \)
W C C C \((1-p) \cdot p \cdot 0.5 \)

The single probabilites sum up to:

\(  P_{three-man}(C) = p \cdot p \cdot 0.5 + p \cdot p \cdot 0.5 + p \cdot (1-p) \cdot 0.5  + (1-p) \cdot p \cdot 0.5 = p \)

The three-man jury with the flippand juror has the same probability for correct decisions p as a one man jury with a „real“ juror.

Python code to simulate the problem:

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import numpy as np
 
 
p = 0.9 # Probability for the right decision of a real juror
pf = 0.5 # Probability for the right decicion of the flippand juror
 
#judges_correct = [0,0,0]
n_sim = int(1e4) # number of simulated jury decicions
 
jury_correct_count = 0 
for i in range(0, n_sim):
    judges_correct = np.random.uniform(low=0, high=1, size=3) <= [p,p, pf] # boolean array of right/wrong decisions
 
    if sum(judges_correct) >= 2: # majority rules, jury did a right decision
        jury_correct_count = jury_correct_count + 1 # increment the counter of right jury decicions
 
 
p_jury_correct = jury_correct_count / n_sim
print(p_jury_correct)