The Ballot Problem and Reflection Arguments

Bertrand's ballot formula gives the exact chance the winning candidate led throughout the count, far smaller than the final vote share suggests.

In an election where candidate A receives a votes and candidate B receives b votes, with a greater than b, and the ballots are counted in a uniformly random order, the probability that A is strictly ahead of B at every single point during the count is (a minus b) over (a plus b). For a race won 51 votes to 49, that probability is only 2 over 100, or 0.02, far smaller than the 51 percent final vote share might suggest.

The ballot problem, also called Bertrand's ballot problem, asks about the running tally as votes are revealed one at a time in a random order, rather than the final result alone. A is strictly ahead throughout means that after every single ballot is counted, from the very first one onward, A's running total is greater than B's, with no tie and no moment where B leads.

Every one of the (a plus b) choose a equally likely sequences of A votes and B votes is treated as equally probable under the random counting order assumption, since only the relative order of A votes against B votes matters to the running tally, not which specific ballot paper sits in which position.

Represent each counting order as a path: start at zero, step up one for every A vote and down one for every B vote, ending at height (a minus b) after all a plus b votes are counted. Every equally likely ordering of the ballots corresponds to exactly one such path, and the ballot problem becomes a question about what fraction of these paths stay strictly positive after the first step.

A path that stays strictly positive the whole way must start with an A vote, since starting with a B vote puts it at negative one immediately. Among paths that start with an A vote, call a path bad if it later touches zero. The reflection trick pairs every bad path with a path that starts with a B vote instead, by taking the portion of the path up to its first return to zero and flipping every step in that portion from up to down and down to up. This produces a one to one correspondence between bad A-starting paths and all B-starting paths.