In a two-candidate election A receives 7 votes and B receives 3. The votes are counted in a uniformly random order. What is the probability that A is strictly ahead of B at every point of the count?

In a two-candidate election A receives 7 votes and B receives 3. The votes are counted in a uniformly random order. What is the probability that A is strictly ahead of B at every point of the count?

Approach: Use the reflection principle to count the counting orders that reach a tie, or apply the cycle lemma to the ten cyclic rotations of any fixed vote sequence.

2/5. The ballot theorem gives (a - b)/(a + b) = (7 - 3)/10 = 2/5 = 0.4. There are C(10,3) = 120 equally likely counting orders and exactly 48 of them keep A strictly ahead. The reflection principle proves this by pairing each order that touches a tie with a reflected order beginning with the other candidate, which counts the bad orders exactly. The cycle lemma gives the same result more directly: among the 10 cyclic rotations of any fixed sequence, exactly a - b = 4 keep A strictly ahead throughout, and each rotation is equally likely under a uniform random ordering. The weaker question of A never trailing has probability (a + 1 - b)/(a + 1) = 5/8.

Follow-up: What is the probability that the count is tied at some point when A wins 60 votes to 40?

Key concepts: ballot theorem, reflection principle, cycle lemma, random ordering.