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.

First off, an even number of players is a necessary but not sufficient condition for a full balanced tournament. Fir this you need $P=2^R$ where $R$ is the number of rounds (including finals, semis etc). I will assume this is what you are after.
Without loss of generality, let Player 1 occupy the first slot in the draw.
For the players to meet in the first round Player 2 must be in Slot 2, for round 2 - slot 3 or 4, round 3 - slot 5 to 8, etc. In general, for the players to meet in round $r\in 1,\dots,R$, there are $2^{r-1}$ slots available. Given that Player 2 is randomly allocated to one of $2^R-1$ slots, the probability of a potential meeting in round $r$ is 
$$S(r)=\frac{2^{r-1}}{2^R-1}$$
The probability that each player makes round $r$ is $\left(\frac{1}{2}\right)^{r-1}$.
The probability that both do is therefore $\left(\frac{1}{2}\right)^{2(r-1)}$
So, the probability that they meet in round $r$ is 
$$\begin{align}
M(r)&=\left(\frac{1}{2}\right)^{2(r-1)}\frac{2^{r-1}}{(2^R-1)}\\
&=\left(\frac{1}{2}\right)^{(r-1)}\frac{1}{(2^R-1)}
\end{align}$$
And that they meet in any round is
$$\begin{align}
M&=\sum_{r=1}^R \left(\frac{1}{2}\right)^{r-1} \frac{1}{(2^R-1)}\\
&=\sum_{r=0}^{R-1} \left(\frac{1}{2}\right)^{r} \frac{1}{(2^R-1)}\\
&=\frac{1-\left(\frac{1}{2}\right)^R}{1-\frac{1}{2}}\frac{1}{(2^R-1)}\\
&=2\frac{2^R-1}{2^R}\frac{1}{(2^R-1)}\\
&=2^{1-R}\\
\end{align}$$
Sanity check!
For 2 players $R=1, p=1$; 4 players $R=2, p=\frac{1}{2}=\frac{1}{3}\text{Rd1}+\frac{1}{6}\text{Rd2}$; 8 players $R=2, p=\frac{1}{4}=\frac{1}{7}\text{Rd1}+\frac{1}{14}\text{Rd2}+\frac{1}{28}\text{Rd3}$ - Sanity check passed.
