Landslide count, 8 to 3, never tied

Question

Candidate P receives 8 votes and candidate Q receives 3 votes. The 11 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

Each P ballot is a +1 step and each Q ballot a -1 step; we want the running total strictly positive throughout. The ballot theorem gives P(	ext{always strictly ahead}) = \frac{p - q}{p + q} = \frac{8 - 3}{8 + 3} = \frac{5}{11} \approx 0.455. The bigger the gap between the totals, the closer this probability climbs toward 1. Takeaway. With totals 8 and 3, the leader stays in front the entire count with probability (p - q)/(p + q) = 5/11.

Landslide count, 8 to 3, never tied

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