Question 1.16.2 (Markov chains, Part IB, 1991, 307D) Three girls A, B and C are playing table tennis. In each game, two of the girls play against each other and the third girl does not play. The winner of any given game n plays again in game n + 1. The probability that girl x will beat girl y in any game that they play against each other is sx/(Sx+Sy) for x, y = {A,B,C}, x‡y, where SA, SB, SC represent the playing strengths of the three girls. In what proportion of games does each girl play, in the long run?

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Question 1.16.2 (Markov chains, Part IB, 1991, 307D)
Three girls A, B and C are playing table tennis. In each game, two of the girls play
against each other and the third girl does not play. The winner of any given game
n plays again in game n + 1. The probability that girl x will beat girl y in any game
that they play against each other is sx/(Sx+Sy) for x, y = {A,B,C}, x‡y, where sa,
SB, SC represent the playing strengths of the three girls.
In what proportion of games does each girl play, in the long run?
Transcribed Image Text:Question 1.16.2 (Markov chains, Part IB, 1991, 307D) Three girls A, B and C are playing table tennis. In each game, two of the girls play against each other and the third girl does not play. The winner of any given game n plays again in game n + 1. The probability that girl x will beat girl y in any game that they play against each other is sx/(Sx+Sy) for x, y = {A,B,C}, x‡y, where sa, SB, SC represent the playing strengths of the three girls. In what proportion of games does each girl play, in the long run?
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