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In a knockout tennis tournament of 2n contestants, the players are paired and play a match. The losers depart, the remaining 2n−1 players are paired, and they play a match. This continues for n rounds, after which a single player remains unbeaten and is declared the winner. Suppose that the contestants are numbered 1 through 2n, and that whenever two players contest a match, the lower numbered one wins with probability p. Also suppose that the pairings of the remaining players are always done at random so that all possible pairings for that round are equally likely.

What is the probability that player 1 wins the tournament?

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Final answer:

To find the probability that player 1 wins the tournament, we can consider the possible outcomes for each round and calculate the overall probability using p to the power of n, where n is the number of rounds.

Step-by-step explanation:

To find the probability that player 1 wins the tournament, we can consider the possible outcomes for each round. In the first round, player 1 must win their match. The probability of this happening is p. In the second round, player 1 must win the match to move on, and so on for each subsequent round. Therefore, the overall probability that player 1 wins the tournament can be calculated as p to the power of n, where n is the number of rounds.

For example, if there are 4 rounds, the probability would be p^4. If there are 5 rounds, the probability would be p^5, and so on.

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