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Suppose the probability for A to win a game against.B is 0.4. If A has an option of playing either a "best of 3 games" or a "best of 5 games" match against B, which option should choose so that the probability of his winning the match is higher? (No game ends in a draw).

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Case I : When A plays 3 games against B. In this case, we have n = 3, p = 0.4 and q = 0.6. 

Let x denote the number of wins. Then

Clearly, P1 > P2. Therefore, first option ie, 'the best 3 games' has higher probability of winning the match.

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