On likelihood of a Condorcet winner for uniformly random and independent voter preferences
Boris Pittel
TL;DR
The paper addresses the probability of a Condorcet winner under impartial culture as both the number of voters $m$ and candidates $n$ grow. It builds on prior work that analyzed fixed $m$ or fixed $n$ regimes and introduces a representation of random preferences via an $m\times n$ matrix of i.i.d. $[0,1]$ uniforms, enabling a combinatorial-analytic treatment of pairwise contests. Central to the method is Esseen’s inequality applied to sums of independent but non-identically distributed Bernoulli indicators, yielding precise control over the probability that a given candidate defeats all others in pairwise contests. The main results show that, uniformly in $n$, the Condorcet-winner probability $Q_{m,n}$ satisfies $Q_{m,n}=O(\exp(-n^{\varepsilon(n)})+n^2/m^{1/2})$ when $m\gg n^4$, and that for fixed $n$, $Q_n=O(n^{-\ extell})$ for any $\ell>0$. Consequently, with a very large electorate relative to the candidate pool, the probability of a Condorcet winner vanishes, strengthening prior bounds and providing a nuanced view of how $Q_{m,n}$ decays in the large-$m$, large-$n$ regime.
Abstract
We study a mathematical model of voting contest with $m$ voters and $n$ candidates, with each voter ranking the candidates in order of preference, without ties. A Condorcet winner is a candidate who gets more than $m/2$ votes in pairwise contest with every other candidate. An ``impartial culture'' setting is the case when each voter chooses his/her candidate preference list uniformly at random from all $n!$ preferences, and does it independently of all other voters. For impartial culture case, Robert May and Lisa Sauermann showed that when $m=2k-1$ is fixed ($k=2$ and $k>2$ respectively), and $n$ grows indefinitely, the probability of a Condorcet winner is small, of order $n^{-(k-1)/k}$. We show if $m, n\to\infty$ and $m\gg n^4$, then for each fixed $\ell$ the probability of a Condercet winner is at most of order $n^{-\ell} + n^2/m^{1/2}$, thus converges to zero.