Vote lead, 10 to 6, ahead throughout

Question

Candidate P receives 10 votes and candidate Q receives 6 votes. The 16 ballots are counted one at a time in a completely random order.

What is the probability that P is strictly ahead of Q throughout the entire count?

Accepted Answer

Treat each P ballot as +1 and each Q ballot as -1; the count is a random walk of 10 ups and 6 downs, and we want it strictly positive the whole way. The ballot theorem gives P(	ext{always strictly ahead}) = \frac{p - q}{p + q} = \frac{10 - 6}{10 + 6} = \frac{4}{16} = \frac14. Takeaway. A wider final margin lifts the lead-throughout probability: (p - q)/(p + q) = 1/4 here.

Vote lead, 10 to 6, ahead throughout

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