Problem of the Week 1234

Chess in Black and White

Solution

As a solution for all odd matches, this lovely general proof by Piotr Zielinski settles the cases of matches of length 3 and 5.

Let A(k) be the probability that Alice wins game k. Then the probability that Alice loses game k is 1 − A(k).

Let P(k) be the probability that Alice wins two games in a row. To win a 2k + 1-game match, Alice has to either: (i) win two consecutive games in the first 2k games, or (ii) win the two last games. To make these two events disjoint, we can additionally demand that in (ii), she loses the third last game and doesn't score two consecutive wins in the first 2k − 2 games.

In other words:

P(2k + 1) = P(2k) + A(2k) A(2k + 1) (1 − P(2k − 2)) (1 − A(2k − 1))

But P(2k), P(2k − 2), and A(2k) * A(2k + 1) do not depend on white/black choice because of symmetry. They all involve an even number of consecutive games; swapping white/black is equivalent to playing these games in the opposite order. So P(2k + 1) is maximized when 1 − A(2k − 1) is, meaning that Alice should play the third-last game with black, in turn meaning that she should start with black.

[Back to Problem 1234]

© Copyright 2017 Stan Wagon. Reproduced with permission.

2 March 2017