**tan@iwu1d.UUCP (William Tanenbaum)** (02/29/84)

Here is a well known problem in probability which is difficult to solve by brute force but yields easily to the proper insight: ---Two candidates run for office. The winner receives p votes, the loser receives q votes. (q<p obviously). Suppose all the ballots are mixed randomly before counting, and running totals are kept as the votes are counted. What is the probability that the winner will always be ahead (never behind or tied) as the votes are counted, (the inital 0-0 tie obviously excepted)? Please prove that your answer is correct. An answer without a proof is no solution. I will post the solution in a week if no one else does first. Bill Tanenbaum A. T. & T. Bell Labs Naperville Il.

**kpv@ulysses.UUCP (Phong Vo)** (04/27/84)

William Tanenbaum has given a solution to the problem: In a two candidate election, the winner gets p votes and the loser gets q votes (p > q). The ballots are randomly shuffled and counted one at a time. What is the probability that the winner will always BE AHEAD at any point during the counting? Now the variation: What is the probability that the winner will always DO AT LEAST AS WELL AS THE LOSER at any point during the counting? That is, at any time t during the counting, if w(t) is the number of votes that the winner receives and l(t) is the number of votes that the loser receives, then w(t) >= l(t). In the original problem, it is required that w(t) > l(t).