Grinstead-Snell-Probability · Problem 4.1.44

Conditional Probability

Mathematics · problem-of-points-service-conventions

Problem

C. L. Anderson has used Fermat's argument for the problem of points to prove the following result due to J. G. Kingston. You are playing the game of points (see Exercise 40) but, at each point, when you serve you win with probability pp, and when your opponent serves you win with probability pˉ\bar{p}. You will serve first, but you can choose one of the following two conventions for serving: for the first convention you alternate service (tennis), and for the second the person serving continues to serve until he loses a point and then the other player serves (racquetball). The first player to win NN points wins the game. The problem is to show that the probability of winning the game is the same under either convention. (a) Show that, under either convention, you will serve at most NN points and your opponent at most N−1N - 1 points. (b) Extend the number of points to 2N−12N - 1 so that you serve NN points and your opponent serves N−1N - 1. For example, you serve any additional points necessary to make NN serves and then your opponent serves any additional points necessary to make him serve N−1N - 1 points. The winner is now the person, in the extended game, who wins the most points. Show that playing these additional points has not changed the winner. (c) Show that (a) and (b) prove that you have the same probability of winning the game under either convention.
Intuitive explanation

The key insight is that both serving conventions can be "equalized" by extending the game to a fixed structure where you always serve exactly NN points and your opponent always serves exactly N−1N-1 points. Since the winner of the original game must also win the extended game (you can't lose points you've already won), the probability of victory remains unchanged—even though the serving order differs between conventions. This transforms a problem about different serving patterns into one about a single, symmetric outcome.

Intuitive explanation

The key insight is that both serving conventions can be "equalized" by extending the game to a fixed structure where you always serve exactly NN points and your opponent always serves exactly N−1N-1 points. Since the winner of the original game must also win the extended game (you can't lose points you've already won), the probability of victory remains unchanged—even though the serving order differs between conventions. This transforms a problem about different serving patterns into one about a single, symmetric outcome.

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