Bertrand's ballot theorem
theorem that, in an election where candidate A receives 𝑝 votes and candidate B receives 𝑞 votes (𝑝>𝑞), the probability that A will be strictly ahead of B throughout the count is (𝑝−𝑞)/(𝑝+𝑞)
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 under the assumption that votes are counted in a randomly picked order?" The answer is
p
−
q
p
+
q
.
{\displaystyle {\frac {p-q}{p+q}}.}
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 (the case where B receives the first vote) 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.
Bertrand's ballot theorem is related to the cycle lemma. They give similar formulas, but the cycle lemma considers circular shifts of a given ballot counting order rather than all permutations.
Example
Suppose there are 5 voters, of whom 3 vote for candidate A and 2 vote for candidate B (so p = 3 and q = 2). There are ten equally likely orders in which the votes could be counted:
AAABB
AABAB
AABBA
ABAAB
ABABA
ABBAA
BAAAB
BAABA
BABAA
BBAAA
For the order AABAB, the tally of the votes as the election progresses is:
For each column the tally for A is always larger than the tally for B, so A is always strictly ahead of B. For the order AABBA the tally of the votes as the election progresses is:
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
2
10
=
1
5
,
{\displaystyle {\frac {2}{10}}={\frac {1}{5}},}
and this is indeed equal to
3
−
2
3
+
2
{\displaystyle {\frac {3-2}{3+2}}}
as the theorem predicts.
This brief starts where responsible research should: with the source description of “Bertrand's ballot theorem” as theorem that, in an election where candidate A receives 𝑝 votes and candidate B receives 𝑞 votes (𝑝>𝑞), the probability that A will be strictly ahead of B throughout the count is (𝑝−𝑞)/(𝑝+𝑞). Everything that follows is an evidence route, not borrowed authority.
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The subject matters to the history & society register because the source frames it as theorem that, in an election where candidate A receives 𝑝 votes and candidate B receives 𝑞 votes (𝑝>𝑞), the probability that A will be strictly ahead of B throughout the count is (𝑝−𝑞)/(𝑝+𝑞). Its deeper value depends on whether names, dates, institutions and citations support that framing.
Chronology, provenance and viewpoint should be read together before a broad social or political interpretation is accepted. The source revision retrieved here is dated Sep 21, 2026. The linked authority identifier is Q2253746. None of the 0 selected statements returned an explicit reference. The first chronological checks are 1878 and 1887.
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