Markov

Appeared in Journal of Chinese Statistical Association, 2004

Optimal Decision for the Squash Player
Jan Veˇeˇ∗ c r Columbia University, Department of Statistics, New York, NY 10027, USA First version: October 15, 2001 This version: February 8, 2005 Abstract. The player of squash has a choice between two options when his/her opponent changes the score from 8:7 to 8:8. The player to receive hasto decide whether to play until 9th or until 10th point. We observe in this paper that the score of the squash game can be modelled as a Markov chain. Thus using standard techniques of Markov processes, we can derive the optimal strategy for the squash player. Key words: Markov Chains, Optimal Control, Squash Game. Mathematics Subject Classification: 60J20, 91A35.

1

Introduction

WorldSquash Singles Rules 2001 define squash game in the following way: “The game of Singles Squash is played between two players, each using a racquet, with a ball and in a court. Only the server scores points. The server, on winning a rally, scores a point; the receiver, on winning a rally, becomes the server. The player who scores nine points wins the game, except that on the score reaching eight-allfor the first time, the receiver shall choose, before the next service, to continue that game either to nine points (known as “Set one”) or to ten points (known as “Set two”).” When the score reaches 8:8 for the first time, the receiver has to choose between “Set one” and “Set two”. The objective of this article is to determine the optimal strategy for this decision. This situation can be modelled asa Markov Chain. For the choice of “Set one”, we may consider the state space to be: {Win, Loss, 8S:8, 8:8S}. We have to keep information about who serves, so 8S:8 means that the first player serves, and 8:8S means that the second player (the opponent) serves. Win means that the first player wins and Loss means that he/she loses the game. For the choice of “Set two”, we have the following statespace: {Win, Loss, 9S:9, 9S:8, 8S:8, 8S:9, 8:9S, 8:8S, 9:8S, 9:9S}. In both cases, states {Win, Loss} are recurrent, all other states are transient. The decision is made at the point when the score is 8:8S, i.e., the second player has the serve, but the first player has an option to choose between “Set one” and “Set two” when this score is reached for the first time. From each possible score there is acertain probability that the first player would either win or lose the rally, thus changing the state of the Markov chain. For illustration, from the 8:8S state, the game can move to the 8S:8 state, which means that the first player won the rally and thus the serve without changing the score. If the first player loses the rally, the second player would make a point and keep the serve, changing thescore to 8:9S. In the “Set one” option, the game would be over since it is played only until one of the players reaches the 9th point, resulting in the loss of the whole game for the first player. In the “Set two” option, the game goes on even from 8:9S state until one of the players reaches the 10th point.
∗ I would like to thank Mark Broadie for pointing out his previous independent work on asimilar topic. See Broadie, M., Joneja, D. (1993) “An Application of Markov Chain Analysis to the Game of Squash”, Decision Sciences, Vol. 24, No. 5, 1023 – 1035.

1

Eventually, the Markov chain would reach one of the two recurrent states, which represent the win or the loss of the first player. Thus our problem can be viewed as computing the probability of being absorbed in the win situationgiven that we start at the state 8:8S. When the probability of winning is bigger in “Set one”, the player to receive should choose to play to the 9th point, otherwise he/she should choose to play to the 10th point.

2

General Results from Markov Chain Theory

General theory of Markov chains explains how to compute the probability that the chain ends up in a particular recurrent state under...