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 (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. 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 possibilities for the order of the votes cast: 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 the "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 and this is indeed equal to formula_3 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. (This is the method used by Bertrand.) The total number of ways is the binomial coefficient formula_4; Bertrand's proof shows that the number of  favourable orders in which to count the votes is formula_5 (though he does not give this number explicitly). And indeed after division this gives formula_6. 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 formula_8. 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 (so any A becomes a B, and vice versa) 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 formula_9, so the probability that A always leads the vote is Proof by induction. Another method of proof is by mathematical induction: Proof by permutation. 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 formula_18 A's and formula_19 B's, where formula_20, has precisely formula_21 dominating cyclic permutations. To see this, just  arrange the given sequence of formula_22 A's and B's in a circle and repeatedly remove adjacent pairs AB until only formula_21 A's remain. Each of these A's was the start of a dominating cyclic permutation before anything was removed. So formula_21 out of the formula_22 cyclic permutations of any arrangement of formula_18 A votes and formula_19 B votes are dominating. Bertrand's and André's proofs. Bertrand expressed the solution as where formula_29 is the total number of voters and formula_30 is the number of voters for the first candidate. He states that the result follows  from the formula where formula_32 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 (those that reach an intermediate tie) 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 formula_33 such sequences  with a B followed by an arbitrary sequence of ("q"-1) B's and "p" A's. Each unfavourable sequence that begins with A can be transformed to an arbitrary sequence of ("q"-1) B's and "p" A's by finding the first B that violates the rule (by causing the vote counts to tie) and deleting it, and interchanging the order of the remaining parts. To reverse the process, take any sequence of ("q"-1) 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 (that is, with ties are allowed ). 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 formula_37. 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 ("p", "q ). 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 (−1, 1) to ("p", "q"). The same operation applied again restores the original "P". This produces a one-to-one correspondence between the 'bad' paths and the paths from (−1, 1) to  ("p", "q"). The number of these paths is formula_38 and so that is the number of 'bad' sequences. This leaves the number of 'good' sequences as Since there are formula_37 altogether, the probability of a sequence being good is formula_41. 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 (ignoring the first) 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 formula_43, which by algebraic manipulation is formula_44, so the "fraction" of sequences with p votes for A votes is formula_45.