Election count, staying ahead the whole way

Question

In a small election, candidate P receives 7 votes and candidate Q receives 5 votes. The 12 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

Track the running lead: +1 for each P ballot, -1 for each Q ballot. The count is a random walk of 7 up-steps and 5 down-steps, and we want it to stay strictly above zero from the very first ballot to the last. The ballot theorem answers exactly this: P(	ext{always strictly ahead}) = \frac{p - q}{p + q}, where p and q are the two vote totals. With p = 7 and q = 5, \frac{7 - 5}{7 + 5} = \frac{2}{12} = \frac16 \approx 0.167. The very first ballot must be a P (chance 7/12) just to get above zero, and staying above zero the rest of the way trims the probability down to 1/6. Takeaway. By the ballo…

Election count, staying ahead the whole way

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