In 1878 W. A. Whitworth imagined an election between two candidates. A receives *m* votes, B receives *n* votes, and A wins (*m*>*n*). If the ballots are cast one at a time, what is the probability that A will lead throughout the voting?

The answer, it turns out, is given by the pleasingly simple formula

Howard Grossman offered the proof above in 1946. We start at *O*, where no votes have been cast. Each vote for A moves us one point east and each vote for B moves us one point north until we arrive at E, the final count, (*m*, *n*). If A is to lead throughout the contest, then our path must steer consistently east of the diagonal line *OD*, which represents a tie score. Any path that starts by going north, through (0,1), must cut *OD* on its way to *E*.

If any path does touch *OD*, let it be at *C*. The group of such paths can be paired off as *p* and *q*, reflections of each other in the line *OD* that meet at *C* and continue on a common track to *E*.

This means that the total number of paths that touch *OD* is twice the number of paths *p* that start their journey to *E* by going north. Now, the first segment of any path might be up to *m* units east or up to *n* units north, so the proportion of paths that start by going north is *n*/(*m* + *n*), and twice this number is 2*n*/(*m* + *n*). The complementary probability — the probability of a path not touching *OD* — is (*m* – *n*)/(*m* + *n*).

(It’s interesting to consider what this means. If *m* = 2*n* then p = 1/3 — even if A receives twice as many votes as B, it’s still twice as likely that B ties him at some point as that A leads throughout.)