TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

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Bertrand's ballot problem: for $p > q \geq 0$ the orderings in which the first candidate is strictly ahead throughout satisfy $(p + q) N = (p - q) (\binom{p + q}{p})$

Statement

Let $p, q \in N$ with $p > q$ . A count in which the first candidate receives $p$ votes and the second $q$ votes, the votes being read in order, is recorded by a diagonal lattice path of length $p + q$ from $(0, 0)$ whose step word has exactly $p$ letters $U$ , one for each vote for the first candidate; such a path ends at height $p - q$ (Diagonal lattice paths with steps $U = (1, 1)$ and $D = (1, - 1)$ , and the height function). The first candidate is strictly ahead throughout when the height after each of the $p + q$ votes is at least $1$ . Write

$B (p, q) : = {v \in W ((0, 0), (p + q, p - q)) : h (i) \geq 1 for 1 \leq i \leq p + q}, N : = ∣ B (p, q) ∣ .$

Then $B (p, q)$ is finite and, in $N$ ,

$(p + q) N = (p - q) (\binom{p + q}{p}) .$

Facts & Assumptions

Given: natural numbers $p > q$ , so $p \geq 1$ and $p + q \geq 1$ ; and the set $B (p, q)$ above.

[F1]

A diagonal path of length $n$ from $(0, α)$ is the same datum as a function $h : {0, \dots, n} \to Z$ with $h (0) = α$ and $h (i) - h (i - 1) \in {1, - 1}$ for $1 \leq i \leq n$ ; with $μ (n)$ the number of up-steps its endpoint height is $h (n) = α + 2 μ (n) - n$ ; and it stays strictly above the level $c$ when $h (i) > c$ for every $i \leq n$ (Diagonal lattice paths with steps $U = (1, 1)$ and $D = (1, - 1)$ , and the height function).

[L1]

For $c \in Z$ , $n \in N$ and $a > c$ , $b > c$ : if $2$ divides $n + b - a$ and $b - a \geq - n$ , and $u$ is the natural number with $2 u = n + b - a$ , then the set of paths in $W ((0, a), (n, b))$ staying strictly above level $c$ is finite and its cardinality $∣ A ∣$ satisfies $∣ A ∣ + (\binom{n}{u + a - c}) = (\binom{n}{u})$ (The reflection principle: paths from $(0, a)$ to $(n, b)$ staying strictly above level $c$ are counted by a difference of two binomial coefficients, clause 2).

[L2]

For $n, k \in N$ with $k \leq n$ : $(\binom{n}{k}) \cdot k! \cdot (n - k)! = n!$ in $N$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L3]

$n! \neq 0$ for every $n \in N$ , and $σ (n)! = n! \cdot σ (n)$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L4]

For all $m, n, k \in N$ with $k \neq 0$ : if $m \cdot k = n \cdot k$ then $m = n$ (Cancellation for multiplication by a nonzero factor).

[L5]

$(\binom{n}{k}) = 0$ for $k > n$ , and $(\binom{n}{n}) = 1$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L6]

For $f : A \to B$ : $f$ is a bijection if and only if there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ and $f \circ g = Δ_{B}$ ( $f : A \to B$ is a bijection if and only if there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ and $f \circ g = Δ_{B}$ ; such a $g$ is unique, equals the inverse relation $f^{- 1}$ , and is itself a bijection).

[L7]

If $A$ is finite and $f : A \to B$ is a bijection then $B$ is finite and $∣ B ∣ = ∣ A ∣$ (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1F1

For $v \in B (p, q)$ the first step is forced upward: $h (0) = 0$ and $h (1) - h (0) \in {1, - 1}$ , while $h (1) \geq 1$ , so $h (1) = 1$ .

1.2L2L3L4L5algebra

Two identities in $N$ , with $m : = p + q$ . First, $m (\binom{m - 1}{p - 1}) = p (\binom{m}{p})$ : since $p \geq 1$ and $p - 1 \leq m - 1$ , and since $m - 1 - (p - 1) = q$ and $m - p = q$ , [L2] gives $(\binom{m - 1}{p - 1}) (p - 1)! q! = (m - 1)!$ and $(\binom{m}{p}) p (p - 1)! q! = m!$ ; multiplying the first by $m$ and using $m \cdot (m - 1)! = m!$ from [L3] makes both left sides equal, and cancelling the nonzero factor $(p - 1)! q!$ by [L3] and [L4] gives the identity. Second, $m (\binom{m - 1}{p}) = q (\binom{m}{p})$ : when $q = 0$ both sides are $0$ , since then $p = m$ and $(\binom{m - 1}{p}) = 0$ by [L5]; and when $q \geq 1$ then $p \leq m - 1$ with $m - 1 - p = q - 1$ , so [L2] gives $(\binom{m - 1}{p}) p! (q - 1)! = (m - 1)!$ and $(\binom{m}{p}) p! q (q - 1)! = m!$ , and the same multiplication by $m$ and cancellation of $p! (q - 1)!$ gives it.

2.1F1L6L7step 1.1construct

Shifting the index by one is a bijection from $B (p, q)$ onto the set $A$ of paths in $W ((0, 1), (p + q - 1, p - q))$ that stay strictly above the level $0$ : given $h$ , put $h^{-} (i) : = h (i + 1)$ for $0 \leq i \leq p + q - 1$ , so $h^{-} (0) = 1$ by step 1.1, $h^{-} (p + q - 1) = p - q$ , consecutive values differ by $1$ in absolute value, and $h^{-} (i) \geq 1 > 0$ ; conversely, given such an $h^{-}$ , put $h (0) : = 0$ and $h (i) : = h^{-} (i - 1)$ for $1 \leq i \leq p + q$ , which has $h (1) - h (0) = 1$ and the remaining differences those of $h^{-}$ , ends at $p - q$ , and has $h (i) \geq 1$ for $i \geq 1$ . The two constructions undo one another, so [L6] and [L7] apply and $N = ∣ A ∣$ .

3.1L1step 2.1algebra

Apply [L1] with $n = p + q - 1$ , $a = 1$ , $b = p - q$ and $c = 0$ : the hypotheses $a > c$ and $b > c$ hold because $p > q$ , and $n + b - a = 2 p - 2$ is even with $2 (p - 1) = n + b - a$ , so $u = p - 1$ and $u + a - c = p$ ; also $b - a \geq - n$ since $n + b - a \geq 0$ . Hence $N + (\binom{p + q - 1}{p}) = (\binom{p + q - 1}{p - 1})$ .

4.1L4L5step 1.2step 3.1algebra∎

Multiplying step 3.1 by $m = p + q$ and substituting the two identities of step 1.2 gives $m N + q (\binom{m}{p}) = p (\binom{m}{p})$ , and since $q \leq p$ this is exactly $(p + q) N = (p - q) (\binom{p + q}{p})$ . At $q = 0$ it reads $p N = p (\binom{p}{p}) = p$ , so $N = 1$ by [L4], matching the single all-up path; at $p = 2$ , $q = 1$ it reads $3 N = (\binom{3}{2}) = 3$ , so $N = 1$ , the one path with step word $U U D$ .

Remarks

The quotient form. The identity of the statement is an identity of natural numbers. Reading each natural number as its canonical natural in $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field) and dividing by the nonzero real $p + q$ turns it into the familiar $\frac{p - q}{p + q} (\binom{p + q}{p})$ ; the multiplicative form is the one proved, and the division is legitimate only because $p + q \neq 0$ , which needs $p > q$ or at least $p + q \geq 1$ .
Why $p > q$ and not $p \geq q$ . With $p = q$ the height ends at $0$ , so the last vote brings the count level and the first candidate is not strictly ahead throughout; the count is then $0$ , while the right-hand side is $0$ as well, so the identity survives but says nothing. The interesting weak form, in which the first candidate is merely never behind, is a separate statement.

Depends on

Used by

Dependency tree · two levels

52 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

C. Krattenthaler, "Lattice Path Enumeration", ch. 10 of the Handbook of Enumerative Combinatorics, §10.3 (standard reference, not scraped)
A. Postnikov (notes by A. Lin), MIT 18.212 Algebraic Combinatorics, Spring 2019 (standard reference, not scraped)

Bertrand's ballot problem: for p>q≥0p>q≥0 the orderings in which the first candidate is strictly ahead throughout satisfy (p+q) N=(p−q)(p+qp)(p+q)N=(p−q)(pp+q​)