Bertrand's ballot theorem


In combinatorics, Bertrand's ballot problem is the question: "In an election where candidate A receives p votes and candidate B receives q votes with p > q, what is the probability that A will be strictly ahead of B throughout the count?" The answer is
The result was first published by W. A. Whitworth in 1878, but is named after Joseph Louis François Bertrand who rediscovered it in 1887.
In Bertrand's original paper, he sketches a proof based on a general formula for the number of favourable sequences using a recursion relation. He remarks that it seems probable that such a simple result could be proved by a more direct method. Such a proof was given by Désiré André, based on the observation that the unfavourable sequences can be divided into two equally probable cases, one of which is easily computed; he proves the equality by an explicit bijection. A variation of his method is popularly known as André's reflection method, although André did not use any reflections.

Example

Suppose there are 5 voters, of whom 3 vote for candidate A and 2 vote for candidate B. There are ten possibilities for the order of the votes cast:
For the order AABAB, the tally of the votes as the election progresses is:
CandidateAABAB
A12233
B00112

For each column the tally for A is always larger than the tally for B so the A is always strictly ahead of B. For the order AABBA the tally of the votes as the election progresses is:
CandidateAABBA
A12223
B00122

For this order, B is tied with A after the fourth vote, so A is not always strictly ahead of B.
Of the 10 possible orders, A is always ahead of B only for AAABB and AABAB. So the probability that A will always be strictly ahead is
and this is indeed equal to as the theorem predicts.

Equivalent problems

Rather than computing the probability that a random vote counting order has the desired property, one can instead compute the number of favourable counting orders, then divide by the total number of ways in which the votes could have been counted. The total number of ways is the binomial coefficient ; Bertrand's proof shows that the number of favourable orders in which to count the votes is . And indeed after division this gives.
Another equivalent problem is to calculate the number of random walks on the integers that consist of n steps of unit length, beginning at the origin and ending at the point m, that never become negative. Assuming n and m have the same parity and n ≥ m ≥ 0, this number is
When m = 0 and n is even, this gives the Catalan number.

Proof by reflection

For A to be strictly ahead of B throughout the counting of the votes, there can be no ties. Separate the counting sequences according to the first vote. Any sequence that begins with a vote for B must reach a tie at some point, because A eventually wins. For any sequence that begins with A and reaches a tie, reflect the votes up to the point of the first tie to obtain a sequence that begins with B. Hence every sequence that begins with A and reaches a tie is in one-to-one correspondence with a sequence that begins with B, and the probability that a sequence begins with B is, so the probability that A always leads the vote is

Proof by induction

Another method of proof is by mathematical induction:
A simple proof is based on the beautiful Cycle Lemma of Dvoretzky and Motzkin.
Call a ballot sequence dominating if A is strictly ahead of B throughout the counting of the votes. The Cycle Lemma asserts that any sequence of A's and B's, where, has precisely dominating cyclic permutations. To see this, just arrange the given sequence of A's and B's in a circle and repeatedly remove adjacent pairs AB until only A's remain. Each of these A's was the start of a dominating cyclic permutation before anything was removed. So out of the cyclic permutations of any arrangement of A votes and B votes are dominating.

Bertrand's and André's proofs

Bertrand expressed the solution as
where is the total number of voters and is the number of voters for the first candidate. He states that the result follows from the formula
where is the number of favourable sequences, but "it seems probable that such a simple result could be shown in a more direct way". Indeed, a more direct proof was soon produced by Désiré André. His approach is often mistakenly labelled "the reflection principle" by modern authors but in fact uses a permutation. He shows that the "unfavourable" sequences consist of an equal number of sequences that begin with A as those that begin with B. Every sequence that begins with B is unfavourable, and there are such sequences with a B followed by an arbitrary sequence of B's and p A's. Each unfavourable sequence that begins with A can be transformed to an arbitrary sequence of B's and p A's by finding the first B that violates the rule and deleting it, and interchanging the order of the remaining parts. To reverse the process, take any sequence of B's and p A's and search from the end to find where the number of A's first exceeds the number of B's, and then interchange the order of the parts and place a B in between. For example, the unfavourable sequence AABBABAA corresponds uniquely to the arbitrary sequence ABAAAAB. From this, it follows that the number of favourable sequences of p A's and q B's is
and thus the required probability is
as expected.

Variant: ties allowed

The original problem is to find the probability that the first candidate is always strictly ahead in the vote count. One may instead consider the problem of finding the probability that the second candidate is never ahead. In this case, the answer is
The variant problem can be solved by the reflection method in a similar way to the original problem. The number of possible vote sequences is. Call a sequence "bad" if the second candidate is ever ahead, and if the number of bad sequences can be enumerated then the number of "good" sequences can be found by subtraction and the probability can be computed.
Represent a voting sequence as a lattice path on the Cartesian plane as follows:
Each such path corresponds to a unique sequence of votes and will end at. A sequence is 'good' exactly when the corresponding path never goes above the diagonal line y = x; equivalently, a sequence is 'bad' exactly when the corresponding path touches the line y = x + 1.
For each 'bad' path P, define a new path P′ by reflecting the part of P up to the first point it touches the line across it. P′ is a path from to. The same operation applied again restores the original P. This produces a one-to-one correspondence between the 'bad' paths and the paths from to. The number of these paths is and so that is the number of 'bad' sequences. This leaves the number of 'good' sequences as
Since there are altogether, the probability of a sequence being good is.
In fact, the solutions to the original problem and the variant problem are easily related. For candidate A to be strictly ahead throughout the vote count, they must receive the first vote and for the remaining votes they must be either strictly ahead or tied throughout the count. Hence the solution to the original problem is
as required.
Conversely, the tie case can be derived from the non-tie case. Note that the number of non-tie sequences with p+1 votes for A is equal to the number of tie sequences with p votes for A. The number of non-tie votes with p + 1 votes for A votes is, which by algebraic manipulation is , so the fraction of sequences with p votes for A votes is.