Probability of meeting your friend on a tournament. There are $2^n$ players in a knock-out chess tournament. They all have identical skill. This means for any two pairings the probability of either winning is $\frac{1}{2}$.
The table is assigned uniformly randomly. You and your friend are both competing. What is the probability you guys meet?
Method one
Draw out a knock-out table in a rooted binary tree sense. Without loss of generality put yourself at the bottom left leaf. We can condition the probability you and your friend meet on the friends initial location.
Notice we have $2^n-1 $ positions they could be in. One of the positions is adjacent to you in which case the probability you meet is $1$. Two of these positions are only one game away from in which case you must both win your games to meet, with probability $\frac{1}{4}$. Four of these positions are two games away from you in which case you meet if you both win two games. $\frac{1}{2^4}$. And so on.
Formally: Probability of meeting = $\sum\limits_{i = 0}^n \frac{2^i}{2^n-1} \cdot (\frac{1}{2^i})^2$ = $\frac{1}{2^n-1}\sum\limits_{i = 0}^n \frac{1}{2^i} = \frac{1}{{2^n}-1} \cdot \frac{\frac{1}{2^{n+1}} -1}{1-\frac{1}{2}} = \frac{1}{2^{n-1}}$
This method is the nuts and bolts. A little bit of fiddle with geometric series but fairly intuitive.
Method two
This method seems much nicer. I thought it was legitimate and it gets the same answer but I found a big hole in it explained at the end.
Let the positions of the contestants be $C_1 , C_2, C_3 , ... , C_{2^n} $. Let $M_{i,j}$ denote the event that contestant $i$ meets contestant $j$ By full symmetry the probability of $M_{i,j}$ is the same for all $i \not= j \in [1,2^n] $.
We have a total of $2^n - 1$ games in the tournament as each game removes a single player.
We have a total of $2^n$ choose $2$ =  $\frac{2^n \cdot (2^n - 1)}{2} = 2^{2n-1} -2^{n-1}$ couplings of $M_{i,j}$.
($\star$) Matches played are like picking $2^n -1$ balls from a bag with $2^{2n-1} -2^{n-1}$  balls in it. So the probability you and your friend meet is $\frac{2^n -1}{2^{2n-1} -2^{n-1}} = \frac{1}{2^{n-1}}$.
Nifty huh, we get the same answer as before but much cleaner. However this method is wrong!
Step $(\star)$ is a violation. We cannot for example pick the balls $M_{1,2} , M_{1,3} , M_{1,4} , ..., M_{1,2^n}$ as contestant $1$ cannot play everyone!
Can someone help me make the second method work? Im sure its not far off. Or is it just a coincidence that it works?
 A: In a $2^n =k$ player knockout tournament,  you will have a total of   $k-1$ matches to knock out all but the winner
Since with random pairing and even odds, all pairings are equally likely,
P(you and your friend meet) $= \dfrac{k-1}{\binom{k}2} = \dfrac{2}{k}$
A: Here is another way to get the answer.
In a knockout tournament with $2^n$ players, so that each player plays at most $n$ rounds, you can expect to play
$$2-{1\over2^{n-1}}$$
games. This is easy to see by induction. It's certainly true for $n=1$, because there's just one game. And when you go from $2^n$ to $2^{n+1}$ players, you'll start in a bracket with $2^n$ players, in which you can (inductively) expect to play $2-1/2^{n-1}$ games, with the prospect of playing one additional game (namely the championship final) only if you win that bracket, which happens with probability $1/2^n$, for an expected total of $2-{1\over2^{n-1}}+{1\over2^n}=1-{1\over2^n}$ games.
Now $2-1/2^{n-1}=(2^n-1)/2^{n-1}$, and there are $2^n-1$ other players, so the probability your friend will be among the opponents you can expect to face is $1/2^{n-1}$.
