Probability Question about Tennis Games! $2^{n}$ players enter a single elimination tennis tournament. You can assume that the players are of equal ability.

Find the probability that two particular players meet each other in the tournament.
I could't make a serious attempt on the question, hope you can excuse me this time.

We  will make the usual unreasonable assumptions. We also assume (and this is the way tournaments are usually arranged) that the initial locations of the players on the left of your picture are chosen in advance. We also assume (not so reasonable) that the locations are all equally likely.
Work with $n=32$, like in your picture.  It is enough to show the pattern. Call the two players A and B. Without loss of generality, we may, by symmetry, assume that A is the player at the top left of your picture.
Maybe they meet in the first round. Then player B has to be the second player in your list. The probability of this is $\frac{1}{31}$.
Maybe they meet in the second round. For that, B has to be $3$ or $4$ on your list, and they must both win. The probability is $\frac{2}{31}\frac{1}{2^2}$.
Maybe they meet in the third round. For that, B must be one of players $5$ to $8$, and each player must win twice. The probability is $\frac{4}{31}\frac{1}{2^4}$.
Maybe they meet in the fourth round, probability $\frac{8}{31}\frac{1}{2^6}$.
Or maybe they meet in the final, probability $\frac{16}{31}\frac{1}{2^8}$.
Add up. For the sake of the generalization, note we have a finite geometric series.
