Question

In a knockout tennis tournament of \(2^{n}\) contestants, the players are paired and play a match. The losers depart, the remaining \(2^{n-1}\) 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 \(2^{n}\), 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. (a) What is the probability that player 1 wins the tournament? (b) What is the probability that player 2 wins the tournament? Hint: Imagine that the random pairings are done in advance of the tournament. That is, the first-round pairings are randomly determined; the \(2^{n-1}\) first-round pairs are then themselves randomly paired, with the winners of each pair to play in round 2; these \(2^{n-2}\) groupings (of four players each) are then randomly paired, with the winners of each grouping to play in round 3, and so on. Say that players \(i\) and \(j\) are scheduled to meet in round \(k\) if, provided they both win their first \(k-1\) matches, they will meet in round \(k\). Now condition on the round in which players 1 and 2 are scheduled to meet.

Short answer

In a knockout tennis tournament with \(2^n\) contestants, the probability that player 1 wins the tournament is \(P_1 = p\), and the probability that player 2 wins the tournament is \(P_2 = 1 - p\), where p is the probability of the lower numbered player winning a match.

Step-by-step solution

Define the Problem

In this part of the problem, we want to find the probability that player 1 wins the tournament. We will use the conditional probability based on player 1's potential opponents at each round.

Calculate the Probability Tree

Since we know that player 1 wins with probability \(p\) , and there are \(n\) rounds in the tournament, we can multiply the probabilities of player 1 winning against every possible opponent. Therefore, the probability that player 1 wins the tournament is given by: \(P_1 = p(p^{n-1} +(1-p)p^{n-2} + … + (1-p)^{n-1}p^0) \). Note: Each term inside the brackets represents the probability of player 1 winning a round against a different possible opponent.

Simplify the Probability

We simplify the expression of the probability that player 1 wins the tournament: \(P_1 = p[(p^{n-1} +(1-p)p^{n-2} + … + (1-p)^{n-1}p^0)] = p[(1 - (1-p)^n )/(1 - (1-p))] = p.\) (b) Find the probability that player 2 wins the tournament. Note: The hint also suggests an alternative method, but we will proceed with our current method.

Conditional Probability

To find the probability that player 2 wins the tournament, we can base our analysis on the round in which players 1 and 2 are scheduled to meet. We will denote this probability as \(P_2\). Assuming player 2 wins against player 1 in a given round, then player 1 needs to be eliminated from the remaining rounds for player 2 to win the tournament.

Calculate the Probability Tree

Similar to player 1's case, we can calculate the probability tree for player 2: \(P_2 = (1-p)((1-p)^{n-1} + p(1-p)^{n-2} + … + (1-p)^0 p^{n-1}).\)

Simplify the Probability

We simplify the expression of the probability that player 2 wins the tournament: \) P_2 = (1-p)[(1 - (1-p)^n )/(1 - (1-p))] = 1 - p. \) So the probability that player 1 wins the tournament is \(P_1 = p\), and the probability that player 2 wins the tournament is \(P_2 = 1 - p\).