Sixteen people play in the quarter-finals at Wimbledon. The winner of the quarter-finals play again in the semi-final to decide who enters the finals. What is the probability that two particular people will play each other if the tournament begins with 16 players?
So I have so far (case 1 + case 2) = verse player in quarter OR verse player in semi = $\frac{1}{15} + \frac{14}{15}...$
I'm not sure what else to include in the second case
Also the third part of the question asks What is the probability when $2^n$ players begin?
The worked solutions show $\frac{1}{2^n-1}+\frac{2^n-2}{2^n-1}\left(\frac{1}{2^{n-2}}\right)^2\:=\:\frac{1}{2^{n-1}}$ which I cannot get. Even symbolab doesn't show the same simplification. As with the previous question, I do not understand the $\left(\frac{1}{2^{n-2}}\right)^2$