I encountered the following problem a few weeks ago: suppose that Alice and Bob want to play a match consisting of a series of games, where the first player to win $n$ games wins the match.  This is often referred to as a “best-of-$(2n-1)$ series,” with many examples in a variety of sports, such as the World Series in baseball ($n=4$), sets and matches in tennis, volleyball, etc.

There is a problem, though: each game must be played at either Alice’s or Bob’s home field, conferring a slight advantage to the corresponding player.  Let’s assume that the probability that Alice wins a game at home is $p$, and the probability that Alice wins a game away (i.e., at Bob’s home field) is $q.

(Note that this asymmetry may arise due to something other than where each game is played.  In tennis, for example, the serving player has a significant advantage; even against an otherwise evenly-matched opponent (i.e., $p+q=1$), values of $p$ may be as large as 0.9 at the highest levels of play; see Reference (1) below.)

What is the probability that Alice wins the overall match?  Of course, this probability surely depends on how Alice and Bob agree on who has “home-field advantage” for each game.  Let’s assume without loss of generality that Alice plays at home in the first game, and consider a few different possibilities for how the rest of the series plays out:

1. Alternating: Alice plays at home in odd-numbered games, and away in even-numbered games.  (This is similar to a set in tennis.)
2. Winner plays at home: The winner of the previous game has home-field advantage in the subsequent game.  (This is similar to a set in volleyball.)
3. Loser plays at home: The loser of the previous game has home-field advantage in the subsequent game.
4. Coin toss: After Alice’s first game at home, a fair coin toss determines home-field advantage for each subsequent game.

I’m sure there are other reasonable approaches I’m not thinking of as well.  It is an interesting exercise to compute the probability of Alice winning the match using each of these approaches.  Certainly they yield very different distributions of outcomes of games: for example, the following figure shows the distribution of number of games played in a World Series, between evenly-matched opponents with a “typical” home-field advantage of $p=0.55$.

Distribution of number of games played in a best-of-7 series with p=0.55, q=0.45.

The motivation for this post is the observation that, despite these differences, approaches (1), (2), and (3) all yield exactly the same overall probability of Alice winning the match!  This was certainly not intuitive to me.  And the rule determining home-field advantage does matter in general; for example, the coin toss approach in (4) yields a different overall probability of winning, and so lacks something that (1), (2), and (3) have in common.  Can you see what it is?

References:

1. Newton, P. and Keller, J., Probability of Winning at Tennis I. Theory and Data, Studies in Applied Mathematics, 114(3) April 2005, p. 241-269 [PDF]
2. Kingston, J. G., Comparison of Scoring Systems in Two-Sided Competitions, Journal of Combinatorial Theory (Series A), 20(3) May 1976, p. 357-362
3. Anderson, C. L., Note on the Advantage of First Serve, Journal of Combinatorial Theory (Series A), 23(3) November 1977, p. 363
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