Exercise 2116Combinatorics › Counting principlesOlympiadAdded 3 Nov 2026

Bertrand's ballot problem

#ballot problem#lattice paths#Catalan

Statement

In an election, candidate $A$ receives $p$ votes and $B$ receives $q$ votes with $p>q$ . Show that the probability $A$ is always strictly ahead is $\frac{p-q}{p+q}$ .

Hint

Reflection principle: bad paths correspond to paths from $(0,-1)$ .

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Solution

Total. $\binom{p+q}{p}$ orderings.

Bad paths. By reflection, the count of "A falls behind at some point" equals paths starting at $(0,-1)$ ending at $(p-q,0)$ : $\binom{p+q}{q-1}$ .

Good. $\binom{p+q}{p}-\binom{p+q}{q-1}=\frac{p-q}{p+q}\binom{p+q}{p}$ .

Probability. $\frac{p-q}{p+q}$ .

Conclusion. The ballot theorem.

Bertrand's ballot problem

Statement

Solution

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