# The Ballot Box

## Problem

In an election, \(2\) candidates, Albert and Benjamin, have in a ballot box \(a\) and \(b\) votes respectively, with \(a > b\). For example, \(a=3\) and \(b=2\). If ballots are randomly drawn and tallied, what is the chance that at least once after the first tally the candidates have the same number of tallies?

## Solution

Every drawing of the \(a+b\) votes will fall into one of three categories:

The first tallied vote is for Benjamin; in this case, there is a tie at some point, since Albert ultimately wins

The first tallied vote is for Albert, and he stays ahead during the entire drawing

The first tallied vote is for Albert, but he does not stay strictly ahead during the entire drawing; there is a tie at some point

We will denote votes for Albert and Benjamin as *A* and *B*, respectively. Then, an example of a drawing in the \(1\)^{st} category with \(a=3\) and \(b=2\) is the following:

\[BBAAA\]

In the above example, the \(4\)^{th} vote tallied is the first position of a tie. As noted earlier, every drawing in the \(1\)^{st} category will contain a tie at some point. If we "invert" every vote at or before the position of the first tie (*A* becomes *B*, and vice versa), this drawing becomes the following:

\[AABB|A\]

We observe now that this "inverted" drawing falls into the \(3\)^{nd} category. This is not an anomaly: there is a bijection between the \(1\)^{st} category and \(3\)^{rd} category created by following this inversion rule. It follows that the number of drawings in each of these \(2\) categories is the same, and that their probabilities are also the same. This probability is the probability of the first vote tallied being for Benjamin, which is \(\frac{b}{a+b}\).

Using this observation, we can calculate the probability of a random drawing having a tie at some point. Every drawing in the \(1\)^{st} category leads to a tie, since Albert must ultimately win (\(a > b\)). By definition, the \(3\)^{rd} category also consists entirely of drawings with ties. Thus, the overall probability of a tie is:

\[P(Tie) = \frac{2b}{a+b}\]

This approach of applying inversions to solve the problem is called a *Proof by Reflection*.