A well-recognized fact of tennis is that the best players are more likely to lose in smaller tournaments than larger ones. For example, from 2004-2022, the “Big 3” — Federer, Nadal, and Djokovic — won 84% of the Grand Slams (61/72) but just 61% of the ATP 1000 Masters tournaments (100/163).
A common and intuitive explanation for this discrepancy is that better players “pull through” when it counts for more. The idea here is that top players may put forth less effort in smaller tournaments, either because the stakes are lower or because they are saving themselves for larger tournaments. This theory certainly appears to have purchase in such leagues as the NBA, where players like Lebron James and Kobe Bryant are known to improve their quality of play in the playoffs.
While perhaps true, these theories miss a much simpler explanation for the discrepancy in winning percentages. Unlike Grand Slams which are played as best of five sets, smaller ATP events are played as best of three sets — crowning the first player to win 2 sets. This small difference may seem inconsequential, but in stochastic environments, reducing the sample size can significantly increase the variance of match outcomes. Just as a flipping 60% heads on a fair coin is more common on 10 flips than 100 flips, winning a match is a surer bet for a better player over a longer frame of reference.
In the remainder of this post, I derive formally how match length affects winning probability under some simplifying assumptions. Turning to the data, I explore how far this model goes in explaining the discrepancy in tournament performance. The basic result: not very far! Using fitted data, match length explains just 10% of the decreased performance exhibited by top players in Masters tournaments relative to Grand Slams. The remaining 90% is explained by other factors, such as diminished playing ability or compositional changes in opponent quality.
Suppose a tennis player (call him Roger) wins a set against another player (call him Rafael) with probability $p$. Further assume — contra reality — that sets are independent of one another: that is, the outcome of one set has no bearing on the next. We are interested in how the probability of winning any given set translates into the probability of winning the match as a whole.
Clearly, this probability depends on the length of the match. On a best of three (BO3) set match, there are 3 possible outcomes for Roger in terms of sets won ($W$) and lost ($L$): $\{ WW,WLW,LWW\}$. Given that $W$ occurs with probability $p$, the probability of $\{ WW \}$ is $p^2$ and the probability of $\{WLW\}$ and $\{LWW\}$ are each $p^2(1-p)$. The probability Roger wins is therefore:
$$ Pr(\text{Win Best of 3} \mid p) = p^2 + 2p^2(1-p) = p^2(3-2p) $$
Consider now a best of five (BO5) set match. Using the same logic as before, the probability that Roger wins is the sum of probabilities that Roger wins in three, four, and five sets respectively:
$$ Pr(\text{Win Best of 5} \mid p) =p^3 + 3p^3(1-p)+6p^3(1-p)^2 $$
There is a nice general form for this distribution. The probability a player wins a best of $n$ set match (where $m \equiv (n+1)/2$ is the number of sets that a player must reach to win) is given by:
$$
\begin{align} Pr(\text{Win Best of }n| p)
&= \sum_{j=0}^{n-m} { m+j-1 \choose j } p^{m}(1-p)^j
\end{align}
$$
where ${\cdot \choose \cdot}$ is the binomial coefficient function. (If you want a fun statistical exercise, try deriving this result yourself. Here’s a hint.)
There are a couple of things to note about this distribution. First, as $n\to\infty$, better players — those with $p>0.5$ — win the match almost surely. Worse players — those with $p<0.5$ — lose the match almost surely: