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Timothy Chow
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For a round-robin tournament, what is the favorite's least favorite size?

Suppose we have a round-robin tournament (i.e., each player plays exactly one game with each other player) with $n$ players, who are all equally skillful except for one player, the favorite, whose probability of winning a game against any other player is some fixed value $p > 1/2$. Assume that all games are independent, and also that no individual game ends in a tie. The winner of the tournament is the player who wins the most games; let us assume that if several players are tied for first place, then one of them is chosen (uniformly) at random to be the winner. Let $\pi(n,p)$ be the probability that the favorite wins the tournament.

Fact. For any fixed $p>1/2$, $\lim_{n\to\infty} \pi(p,n) = 1$.

Sketch of proof: The proportion of won games converges almost surely to $p$ for the favorite and almost surely to $1/2$ for the other players, so we can easily separate the favorite from the best of the rest; the lack of independence among the non-favorites can be finessed by using a union bound.

In light of the above Fact, it may be slightly surprising that for a fixed $p$, especially for $p$ close to $1/2$, the value of $\pi(p,n)$ actually declines for a while (as $n$ increases), and I think it may even wobble around, before eventually climbing to 1.

Intuitively, what's happening for small $n$ is that the increase in the number of competitors is increasing the chances that one of them will do well and upset the favorite, and that this is initially a more important effect than the fact that the increase in the number of games is giving the favorite an opportunity to demonstrate a skill edge.

This has led me to consider the following question.

Let $N(\epsilon)$ denote the value of $n$ that minimizes $\pi(1/2 + \epsilon, n)$. What can be said about $N(\epsilon)$ as $\epsilon\to0$?

Presumably, $\lim_{\epsilon\to0} N(\epsilon) = \infty$, but approximately how fast?

Timothy Chow
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