There are some mathematical theorems regarding the principles of voting systems that are well-known in popular science. Take, for example, Arrow’s impossibility theorem, which states that no perfect voting system exists when choosing among three or more options, or the closely related Gibbard-Satterthwaite theorem1. There is, however, a basic claim that is not so popular, although quite interesting in my humble opinion: May’s theorem. It states that there is no other reasonable voting system for two candidates than a simple majority rule.

Preliminaries and a formulation of May’s theorem#

A reasonable voting system for two candidates should satisfy exactly four criteria:

  • Near decisiveness — it should choose exactly one winner in every situation except when both candidates receive the same number of votes.
  • Anonymity — all voters should be treated equally.
  • Neutrality — all candidates should be treated equally.
  • Monotonicity — if candidate $A$ wins, transferring votes from candidate $B$ to candidate $A$ should not cause $A$ to lose.

Meeting three of these criteria is relatively easy; for example, a dictatorship (where the result depends on only one voter) is decisive, neutral, and monotone. We are, however, interested in systems that meet all four criteria at once.

May’s theorem says that the only voting system that is nearly decisive, anonymous, neutral, and monotone is majority rule (where the candidate with over half the votes wins).

Proof#

Our social choice function (voting system) should be anonymous, so we can disregard the choices of specific voters and care only about the total number of votes for each candidate. Let $a$ and $b$ be the numbers of votes for candidates $A$ and $B$ respectively.

Case 1: when $a + b$ is even#

In a very special case when $a = b$, every neutral voting system should conclude that there is a tie. If $a \neq b$ and $a + b$ is even, we can assume wlog that $a > b$ and prove that candidate $A$ wins. Since our voting system must be nearly decisive, it suffices to show that $B$ cannot win.

Suppose that $B$ is the winner. Since $a > b$, we can increase the number of votes for candidate $B$ so that $a = b$. We have already shown that in that case there is a tie, so this social choice function is not monotone, leading to a contradiction.

Case 2: when $a + b$ is odd#

Since $a + b$ is odd, $a = b$ is impossible, so there has to be a winner (tie is not allowed by near decisiveness). If candidate $A$ has the smallest possible majority (i.e., $a = \frac{a + b + 1}{2}$), then candidate $B$ has $a - 1$ votes.

Suppose that $B$ is the winner in this setup. By neutrality, $A$ would have also won if they had $a-1$ votes. By monotonicity, $A$ must also win with $a$ votes. This results in both candidates winning, which is a contradiction.

This implies that $A$ wins with $a = \frac{a + b + 1}{2}$ votes. Again, by the monotonicity condition, $A$ always wins if they have at least $\frac{a + b + 1}{2}$ votes, or equivalently, when $a > b$.

  1. There is a great paper by Philip J. Reny that provides formal proofs of those theorems: Arrow’s theorem and the Gibbard-Satterthwaite theorem: a unified approach. ↩︎