Six to one, chance the leader never lets up

Question

Candidate P receives 6 votes and candidate Q receives 1 vote. The 7 ballots are counted one at a time in a completely random order.

What is the probability that P is strictly ahead of Q at every point during the count?

Accepted Answer

With each P ballot as +1 and the single Q ballot as -1, we want the running total strictly positive throughout. The ballot theorem gives P(	ext{always strictly ahead}) = \frac{p - q}{p + q} = \frac{6 - 1}{6 + 1} = \frac{5}{7} \approx 0.714. With only one Q ballot, you can check the formula directly. P is strictly ahead the whole way exactly when the Q ballot is counted neither first (which would put Q ahead) nor second (which would tie the score at 1 to 1). So the single Q must fall in one of the last 5 of the 7 positions, and since it is equally likely to sit in any position, that probabili…

Six to one, chance the leader never lets up

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