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 for that round are equally likely. 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. 

Taking the hint I have found the probability as : 

$$\sum \_{k=1}^{n} p^{2k-2} \prod\_{i=1}^{k-1}\Big(1-\frac{1}{2^{n-i+1}-1}\Big)\frac{1}{2^{n-k+1}-1}(1-p)p^{n-k} + \bigg(1-\sum \_{k=1}^{n} p^{2k-2} \prod\_{i=1}^{k-1}\Big(1-\frac{1}{2^{n-i+1}-1}\Big)\bigg)p^n$$ 

But, I could not understand why "Imagine that the random pairings are done in advance of the tournament" is required here ?