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In a series of seven games, the first team to win four games wins the series. Both the team have equal probability of winning one game. What is the probability that the team that wins the first game will win the series?

My answer: in case of zero losses: 1/8

       in case of one loss: 3/16

       in case of two losses: 6/32

       in case of three losses: 10/64

And then you just add them up. The only thing is that the book's answer for two losses is 5/32. Since there are four positions for losses and you are choosing two, it should be 6 according to me.

Where am I going wrong?

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  • $\begingroup$ I think you are correct. Possibly a typo in the book. $\endgroup$
    – saulspatz
    Sep 25, 2019 at 17:50

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Since there are four positions for losses and you are choosing two, it should be 6 according to me.

I don't see your logic. If there are two losses, then the series lasts $6$ games, and the first game is given as being a win for the team in question. So there are $5$ possible locations for the two losses.

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  • $\begingroup$ They win the first game and the last game. So there are only four positions left, isn't that correct? $\endgroup$
    – user708015
    Sep 25, 2019 at 19:46
  • $\begingroup$ Maybe I should clarify that the series will end once one team is able to score 4 wins. The first line of the question states that. $\endgroup$
    – user708015
    Sep 25, 2019 at 19:47

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