Suppose that two evenly matched teams are playing in the WorldSeries. On the ave

Question

Suppose that two evenly matched teams are playing in the WorldSeries. On the average, how many games will be played?(The winneris the first team to get four victories.) Assume that each game isan independent event. (The answer is 5.8125,how can I get this answer?) Suppose that two evenly matched teams are playing in the WorldSeries. On the average, how many games will be played?(The winneris the first team to get four victories.) Assume that each game isan independent event. (The answer is 5.8125,how can I get this answer?)

Explanation / Answer

We know it has to end in somewhere from 4-7 games. Socompute the probabilities of each of these events happening. Use 1 to denote the first team winning and 2 to denote the secondteam winning. Note that the odds that any given n-term stringwill occur is 1/2n. 4 games It will either be 1111 or 2222, both of which have a 1/16chance of occuring, so there is a 1/8 chance it will end in 4games. 5 games It will be one of the following: 21111, 12111, 11211, 11121,12222, 21222, 22122, or 22212, each of which have a 1/32 chance ofoccuring, so there is a 1/4 chance that it will end in 5games. 6 games Now, instead of listing all of the possibilities (which waseasy for the last two) we will count the number of possibiliesusing combinatorial methods. We know the last game has to bewon by the winning team. In the first five games the winningteam has to win three times and lose twice. Thus the numberof ways is 5choose3 multiplied by two (because either team can bethe winner) so there are 5!/(3!2!)*2=20 ways for this tooccur. Each of them happens with a 1/64 chance, giving aprobability of 5/16 of this occuring. 7 games In the same way as the last one, the number of possibilitiesis going to be 6choose3*2=6!/(3!3!)*2=40 ways. Each of thesehappen with a 1/128 chance, so there is a 5/16 chance of thishappening. To find the expected value of the number of games it takes toend, sum the products of the number of games and their respectiveprobabilities. 4(1/8)+5(1/4)+6(5/16)+7(5/16)=5.8125 In the same way as the last one, the number of possibilitiesis going to be 6choose3*2=6!/(3!3!)*2=40 ways. Each of thesehappen with a 1/128 chance, so there is a 5/16 chance of thishappening. To find the expected value of the number of games it takes toend, sum the products of the number of games and their respectiveprobabilities. 4(1/8)+5(1/4)+6(5/16)+7(5/16)=5.8125

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