Bertrand's Ballot Theorem - Proof By Induction

Proof By Induction

Another method of proof is by mathematical induction:

• Clearly the theorem is true if p > 0 and q = 0 when the probability is 1, given that the first candidate receives all the votes; it is also true when p = q > 0 since the probability is 0, given that the first candidate will not be strictly ahead after all the votes have been counted.
• Assume it is true both when p = a − 1 and q = b, and when p = a and q = b−1, with a > b > 0. Then considering the case with p = a and q = b, the last vote counted is either for the first candidate with probability a/(a + b), or for the second with probability b/(a + b). So the probability of the first being ahead throughout the count to the penultimate vote counted (and also after the final vote) is:

• And so it is true for all p and q with p > q > 0.