Chess Players Truel

Players A, B, and C play a series of chess games. Assume that A is the strongest player and C is the weakest one. Assume that there is no tie for each game. The winner of each game will play with the 3^rd player. The player who first gets 2 wins is the winner of the series. The player B determines who will play the 1^st game. Find the best choice for B. In general: if probability of A to win B is p > .5, probability B to win C is q > .5, probability A to win C is r, and r > p, evaluate chances of B.

Solution

The simplest way to approach the problem is by constructing an event tree. We are going to have three of them. The trees grow downwards starting with one of the pairs AB, BC or AC. Each node is a game, except for the terminal ones shown in squares. These are the winners of the tournament who collected two wins first. For the first two trees we'll find the probability of B being the winner. These will be denoted P_AB and P_BC. We assume the outcomes of all games are independent.

P_AB = p·(1 - r)·q·(1 - p) + (1 - p)·q + (1 - p)·(1 - q)·r·(1 - p).

P_BC = q·(1 - p) + q·p·(1 - r)·q + (1 - q)·r·(1 - p)·q.

The task is to compare the two expressions P_AB and P_BC assuming p, q, r > .5 and r > p. You can play with the applet below to verify that always P_AB < P_BC which we shall prove below. The result is imminently plausible as it seems quite reasonable for B to start with playing the weakest player (C) first.

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So here are the two expressions to compare:

P_AB = p·(1 - r)·q·(1 - p) + (1 - p)·q + (1 - p)·(1 - q)·r·(1 - p).
P_BC = q·(1 - p) + q·p·(1 - r)·q + (1 - q)·r·(1 - p)·q.

In the difference P_BC - P_AB, the common term (1 - p)·q cancels out leading to

P_BC - P_AB

= [p·q·(1 - r) + (1 - p)·(1 - q)·r]·(p + q - 1)

which, since p + q > .5 + .5 = 1, is positive. Note that this is true even without the condition r > p.

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