Probability of Winning in SquashDate: 06/23/99 at 04:45:00 From: Ashley Subject: Probability (sports application) I have tried to start this question and I am finding it rather difficult. I heard about you from a friend; so I hope you can help me please... here goes. In the game squash, the player who has the serve wins a point if he wins the rally. If he loses the rally, he loses the serve (but the score is not changed) and the other player serves to try to win the point. The aim is to be the first to reach 9 points. If the score reaches 8 all, the player receiving the serve can choose to play to either 9 or 10 points. Consider two players A and B. Let p(A) be the probability that A wins the point when serving so that q(A) = 1-p(A) is the probability that B wins the serve from A. Similarly define p(B) and q(B), and suppose that the score has reached 8 all with player A to serve. A question of interest is whether it is more advantageous for player B to call for play to go to 9 or 10 points. a) Suppose p(A) = p(B) = 0.5 Determine whether player B has a better chance of winning if she calls for 9 or 10. b) Repeat part (a) for various values of p(A) and p(B) and draw up a table showing under what conditions it is better for player B to go to 9 rather than 10 points. If you didn't understand that, then p(A) = A serves and wins the point q(A) = A serves and loses, therefore B wins serve p(B) = B serves and wins the point q(B) = B serves and loses, therefore A wins serve Date: 06/23/99 at 14:16:42 From: Doctor Anthony Subject: Re: Probability (sports application) >a) Suppose p(A) = p(B) = 0.5 Determine whether player B has a better chance of winning if she calls for 9 or 10. List the probability that B wins in the two possible situations. B calls for 9 points: --------------------- For B to win overall he must win,win win,lose, win,win, win,lose, win,lose, win,win, and so on. As we can see, B's probability of winning is the infinite series with terms 1/4 + (1/4)^2 + (1/4)^3 + ... to infinity = (1/4)/[1 - 1/4] = (1/4)/(3/4) = 1/3 It follows that A has 2/3 probability of winning if B has called for 9 points. We see therefore that if scores are equal, the server has 2/3 probability of winning if one more point is required. An alternative and neater way to do this is to consider what happens in the next one or two points. Let p = probability that B wins from 8,8 with A serving. B could have win,win prob = 1/4 win,lose (prob = 1/4 and we are back to 8,8 with A serving) prob B wins is p So p = 1/4 + (1/4)p (3/4)p = 1/4 p = 1/3 as we found before. B calls for 10 points: --------------------- We consider various probabilities B gets to 9 before A. Probability = 1/3 A gets to 9 before B. Probability = 2/3 If now the score is 9,8 in B's favor (with B serving), then for the next one or two points we can have: win prob = 1/2 and B wins the series. lose,win, prob = 1/4 and we are back to 9,8 with B serving lose,lose, (prob = 1/4 score now 9,9 with A serving) prob B wins is 1/3 Let p = probability that B wins from 9,8 with B serving. Then we see that p = 1/2 + (1/4)p + (1/4)(1/3) (3/4)p = 1/2 + 1/12 = 7/12 and p = 28/36 = 7/9 So the probability that B will reach 9,8 first and go on to win = 1/3 * 7/9 = 7/27. We now consider what happens if A gets to 9,8 first with A serving: In the next two points we could have for B: win,win (prob = 1/4 score now 9,9 with B serving) prob B wins is 2/3 win,lose prob = 1/4 and we are back to 9,8 with A serving Let p'= probability that B wins from 9,8 in A's favor with A serving Then p' = (1/4)(2/3) + (1/4)p' (3/4)p' = 1/6 p' = 2/9 (This checks with the complementary value 7/9 that we had earlier.) So the probability that B will win with this sequence (A gets to 9,8 first) is (2/3)(2/9) = 4/27 Total probability that B will win if he chooses 10 points = 7/27 + 4/27 = 11/27. Comparing this with 1/3 (= 9/27) if he chooses 9 points, we see that he should ask for 10 points. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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