Count paths that maintain a lead, distinguish strict from weak boundaries, and use reflection to explain why a fixed 5–3 finish gives probabilities 1/4 and 1/2.
Ballot theorem and lattice-path counting
The ballot theorem counts orderings in which one running tally stays ahead of another, given their final totals. Lattice-path counting represents those orderings as unit steps and counts the paths satisfying a boundary rule. For a quant, this separates “finished with a gain” from “stayed above a threshold throughout”: an endpoint alone cannot answer a path-risk question.
Suppose candidate A receives a votes and candidate B receives b votes, with a>b≥0. If every ordering of these fixed totals is equally likely, the strict ballot theorem gives
P(A strictly leads after every vote)=a+ba−b.
The count starts after the first vote; before counting, both tallies are zero. “Strictly leads” excludes even a temporary tie. This is the convention in Curtis McMullen's Harvard probability notes, §III.
Five votes against three: the boundary changes the answer
Fix five A votes and three B votes. Choosing the three positions occupied by B gives the binomial coefficient
(38)=56
possible letter sequences. The voters need not be individually labelled: each A/B sequence corresponds to the same number of permutations of individually labelled ballots.
Encode A as +1 and B as −1. After k votes, let Sk be A's tally minus B's. Every sequence starts at S0=0 and ends at S8=2, but the histories differ:
Event at every count k=1,…,8
Valid orders
Probability
Strict lead: Sk>0
14
14/56=1/4
Weak lead: Sk≥0
28
28/56=1/2
The weak event contains the strict event. Its additional 14 orders touch zero after counting begins without ever going negative. The remaining 28 orders fall behind at least once.
Three eight-vote paths all finish with A ahead by two: one stays strictly positive, one touches zero, and one visits minus one. Of 56 orders, 14 have a strict lead, 14 have only a weak lead, and 28 fall behind.
Original exact enumeration of fixed 5–3 totals. The three displayed paths illustrate the disjoint groups; the bar counts all 56 orders.
Here a binomial coefficient with lower index −1 means zero. The weak formula is the classical boundary count in Goulden and Serrano, §1. Substituting the strict formula into a question that permits ties loses valid paths.
Count the paths that fail
A weakly admissible path never visits −1. Consider instead a bad 5–3 path and locate its first visit to −1. Swap A and B in the prefix ending at that visit. That prefix had one more B than A, so swapping it adds one A and removes one B from the full sequence. The transformed order has six A votes and two B votes.
This map is reversible. Any 6–2 path ends at +4, so it must first visit +1 somewhere. Swap the signs through that first visit to recover the original bad 5–3 path. Thus the bad paths correspond one-to-one to all 6–2 paths:
Nweak=(38)−(28)=56−28=28.
A bad 5 A, 3 B sequence first reaches minus one at vote three. Swapping its first three signs produces a 6 A, 2 B sequence first reaching plus one at vote three and ending at four. The reversible map counts 28 bad orders.
Original illustration of the first-passage sign swap. Dashed vertical lines mark the prefix cut; the transformed endpoint differs from the original endpoint.
This is a reflection argument: a boundary-constrained count becomes an unrestricted count with altered endpoints or step totals. For a strict lead, the first vote must be A. Of the (ba+b−1) orders beginning with A, reflection removes (b−1a+b−1) that later touch zero. Hence
Nstrict=(ba+b−1)−(b−1a+b−1)=a+ba−b(ba+b).
With five A and three B votes, this is 35−21=14.
An alternative drawing puts B votes on the horizontal axis and A votes on the vertical axis, using right and up steps. The same weak condition becomes “never below the diagonal.” When a=b=m, its count is the Catalan number(m2m)/(m+1). Strict leadership through the final vote is then impossible, because the endpoint is a tie.
A conditional path calculation, not a trading forecast
For an illustrative gross P&L with unit gains and unit losses, fixing five gains and three losses fixes the terminal result at +2 units. Under uniform ordering, the chance of remaining strictly above the initial P&L after each observation is 1/4; allowing a return to the initial P&L raises it to 1/2.
The same ordering law arises from a simple random walk with independent steps and a common up-probability p∈(0,1), conditional on these totals. Each eligible path has the same weight p5(1−p)3, which cancels when computing the conditional probability. The conditioned steps themselves are dependent: observing one gain leaves fewer gains available in the remaining positions.
Without conditioning on the endpoint, that cancellation does not supply the survival probability. Unequal step sizes, changing probabilities or dependence in the original process also require a different model. A barrier option additionally needs its price dynamics, monitoring schedule and payoff; the ballot ratio alone is not an option valuation or a continuous-time crossing formula.
For the detailed discrete reflection proof and its connection to first returns, see SelfMadeQuant's The Reflection Principle and the Ballot Problem. The linked page is a public preview; the full lesson requires membership.
Practice: a final tie
Four A votes and four B votes are counted in uniformly random order. What is the probability that A never falls behind, including the final count? Would requiring a strict lead give the same answer?
Answer. There are (48)=70 orders. The weak count is 70−(38)=14, so the probability is 14/70=1/5. A strict lead after every vote has probability zero: the eighth vote ends in a tie. A final tie does not force an earlier deficit, and a weak lead permits returns to equality.
