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Chess Players Truel
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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 3rd player. The player who first gets 2 wins is the winner of the series. The player B determines who will play the 1st game. Find the best choice for B. In general: if probability of A to win B is 
Copyright © 1996-2008 Alexander Bogomolny
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 3rd player. The player who first gets 2 wins is the winner of the series. The player B determines who will play the 1st game. Find the best choice for B. In general: if probability of A to win B is 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 PAB and PBC. We assume the outcomes of all games are independent.
PAB = p·(1 - r)·q·(1 - p) + (1 - p)·q + (1 - p)·(1 - q)·r·(1 - p).
PBC = q·(1 - p) + q·p·(1 - r)·q + (1 - q)·r·(1 - p)·q.
The task is to compare the two expressions PAB and PBC assuming
So here are the two expressions to compare:
In the difference PBC - PAB, the common term (1 - p)·q cancels out leading to
which, since p + q > .5 + .5 = 1, is positive. Note that this is true even without the condition
Copyright © 1996-2008 Alexander Bogomolny |
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