A Probability and Statistics Companion, John J Kinney Copyright 2009 by John Wiley & Sons, Inc in Java

Create QR Code ISO/IEC18004 in Java A Probability and Statistics Companion, John J Kinney Copyright 2009 by John Wiley & Sons, Inc
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To write out the points where B wins the series, interchange the letters A and B above Note that the number of ways in which the series can be played in n games is easily counted The last game must be won by A, say, so in the previous n 1 games, A must win exactly three games and this can be done in n 1 ways For 3 example, there are 5 = 10 ways in which A can win the series in six games So 3 there are 4 1 + 5 1 + 6 1 + 7 1 = 35 ways for A to win the series and so 3 3 3 3 70 ways in which the series can be played These points are not equally likely, however, so we assign probabilities now to the sample points, where either team can win the series: P(4 game series) = p4 + q4 P(5 game series) = 4p4 q + 4q4 p P(6 game series) = 10p4 q2 + 10q4 p2 P(7 game series) = 20p4 q3 + 20q4 p3 These probabilities add up to 1 when the substitution q = 1 p is made We also see that P(A wins the series) = p4 + 4p4 q + 10p4 q2 + 20p4 q3 and this can be simpli ed to P(A wins the series) = 35p4 84p5 + 70p6 20p7 by again making the substitution q = 1 p This formula gives some interesting results In Table 61, we show p, the probability that team A wins a single game, and P, the probability that team A wins the series
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Seven-Game Series Table 61 p 040 045 046 047 048 049 050 051 052 053 054 055 060 070 080 P(A wins the series) 02898 03917 04131 04346 04563 04781 05000 05219 05437 05654 05869 06083 07102 08740 09667
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It can be seen, if the teams are fairly evenly matched, that the probability of winning the series does not differ much from the probability of winning a single game! The series is then not very discriminatory in the sense that the winner of the series is not necessarily the stronger team The graph in Figure 61 shows the probability of winning the series and the probability of winning a single game The maximum difference occurs when the probability of winning a single game is 0739613 or 0260387 The difference is shown in Figure 61 What is the expected length of the series To nd this, we calculate A = 4 p4 + q4 + 5 4p4 q + 4q4 p + 6 10p4 q2 + 10q4 p2 +7 20p4 q3 + 20q4 p3
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Figure 61 Probability of winning the series and the probability of winning a single game
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This can be simpli ed to A = 4 1 + p + p2 + p3 13p4 + 15p5 5p6 A graph of A as a function of p is shown in Figure 62
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575 55 525 5 475 45 425 02 04 p 06 08 1
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The graph shows that in the range 045 p 055, the average length of the series is almost always about 58 games
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Since the probability of winning the series is not much different from that of winning a single game, we consider the probability that the winner of the rst game wins the series From the sample space, we see that P(winner of the rst game wins the series) = p3 + 3p3 q + 6p3 q2 + 10p3 q3 and this can be written as a function of p as P(winner of the rst game wins the series) = p3 (20 45p + 36p2 10p3 ) A graph of A as a function of p is shown in Figure 63