Discrete Stochastic Processes

Lesson 7 of 10

The Reflection Principle and the Ballot Problem

A candidate who wins 52 to 48 has only a 4% chance of leading all the way through the count, and reflecting paths at their first touch of zero shows why.

The question

An election ends 52 to 48. The ballots are counted one at a time in a random order. What is the chance that the winner led at every moment of the count, from the first ballot to the last?

Most people guess something large, since 52 against 48 is a clear win, but the answer is 4%.

Record the count as a walk. Each vote for A is a step up and each vote for B a step down. After all 100 votes the walk sits at 52−48=452 - 48 = 4. "A always strictly ahead" means the walk stays above 00 at every step after the start. The ballot theorem says that if A gets aa votes and B gets b<ab < a, then

P(A strictly ahead throughout the count)=a−ba+b.P(\text{A strictly ahead throughout the count}) = \frac{a-b}{a+b}.

That is the final margin divided by the total number of votes. For 52 and 48 it is 4/1004/100. A landslide of 70 to 30 gives 0.40.4, and a narrow win of 501 to 499 gives 0.0020.002. In a close race the loser almost always draws level at some point.

A small case you can check by hand: A gets 3 votes and B gets 2. There are (53)=10\binom{5}{3} = 10 equally likely orders, and only AAABB and AABAB keep A strictly in front, so the probability is 2/10=1/5=(3−2)/52/10 = 1/5 = (3-2)/5.