Cannot Make Sense of Chess Tournament Solution I got the following question and corresponding solution, but cannot make sense of the solution, could any expert give me some advice?
A chess tournament has $2^n$ players with skills $1>2>...>2^n$. It is organized as the knockout tournament, so that after each round, only the winner proceeds to the next round. Except for the final, opponents in each round are drawn at random. Let's also assume that when two players meet in the game, the player with better skills always wins. What is the probability that players one and two will meet in the final? 
Solution:
Player one always wins. So he will be in the final. It is obvious that $2^n$ players are separated to two $2^{n-1}$ player subgroups. And each group will have one player reaching the final. For player two to reach the final, he must be in a different subgroup from player one, since any of the remaining players in $2,3,...,2^{n-1}$ are likely to be one of the $(2^{n-1}-1)$ players in the same subgroup as player one, or one of the $2^{n-1}$ players in the subgroup different from player one, the probability that player two is in a different subgroup from player one and that player one and player two will meet in the final is simply $2^{n-1}/(2^n-1)$. I'm wondering if any expert can explain why $2^{n-1}/(2^n-1)$ holds? Really appreciate it!
 A: One way to view this is as a branching tournament bracket (or a complete binary tree if you prefer) and to place the competitors randomly at each starting point, or each leaf node. Then, player 2 must be placed on the opposite side of the bracket as player 1 in order to meet in the final. We decide where player 1 gets put first. Then, that leaves $2^n -1$ places for player 2 to go. But, he must go in the other half to make it to the final, and that means $\frac{1}{2}(2^n) = 2^{n-1}$ possible places, giving us the solution $\frac{2^{n-1}}{2^n-1}$. Intuitively, for a large bracket, we would expect him to reach the final about half the time, and indeed the value is approximately $1/2$ for large $n$. This is pretty much the same as the solution above but worded a bit differently.
A: $1$ will play $n-1$ games before getting into the final. So, $1$ will meet $2$ in the final if and only if $1$ never meets $2$ in any of the earlier $n-1$ games. The probability of that is:
$$P(\text{no meet in 1st round}) \cdot P(\text{no meet in 2nd round}) \cdot ... P(\text{no meet in semi-final})$$
$$ = \frac{2^n-2}{2^n-1}\cdot \frac{2^{n-1}-2}{2^{n-1}-1}\cdot .... \cdot \frac{2^2-2}{2^2-1}$$
$$=\frac{2(2^{n-1}-1)}{2^n-1}\cdot \frac{2(2^{n-2}-1)}{2^{n-1}-1}\cdot .... \cdot \frac{2(2^2-1)}{2^3-1}  \cdot \frac{2}{2^2-1}$$
$$\require{cancel}=\frac{2\cancel{(2^{n-1}-1)}}{2^n-1}\cdot \frac{2\cancel{(2^{n-2}-1)}}{\cancel{2^{n-1}-1}}\cdot .... \cdot \frac{2\cancel{(2^2-1)}}{\cancel{2^3-1}}  \cdot \frac{2}{\cancel{2^2-1}}$$
$$=\frac{2^{n-1}}{2^n-1}$$
But, an easier way to see this is as follows. First, if we had a fixed bracket before the tournament, then it is obvious how $1$ and $2$ are going to meet in the final: $2$ should be in the 'other' half than $1$ is in. Now, there are $2^{n-1}$ players in each half, and so with $1$ in one half, there are there are $2^{n-1}$ players in the other half. And since there are $2^n-1$ players other than $1$ (i.e. there are $2^n-1$ slots to fill up), the probability is 
$$\frac{2^{n-1}}{2^n-1}$$
for $2$ to end up in one of the slots in the other half. So that is the probability for $1$ and $2$ to play in the final.
OK, but the bracket isn't fixed: the matches are only determined at the start of each round. Doesn't that make a difference? Well, no. Once all the matches have been played, we can reconstruct a bracket from that: it is as if we had a bracket the whole time, but just didn't know what it was. And so, once again, the probability of $2$ and $1$ not meeting is the probability of $1$ and $2$ being in different halves of this 'unknown' bracket, and therefore still is
$$\frac{2^{n-1}}{2^n-1}$$
A: $\displaystyle \binom{2^n-1}{2^{n-1}-1}$ equiprobable groups of $2^{n-1}$ players contain player 1, among which $\displaystyle \binom{2^n-2}{2^{n-1}-1}$ don't contain player 2. The probability is
\begin{equation}
p=\frac{\displaystyle\binom{2^n-2}{2^{n-1}-1}}{\displaystyle\binom{2^n-1}{2^{n-1}-1}}=\frac{(2^n-2)!(2^{n-1}-1)!(2^{n-1})!}{(2^n-1)!(2^{n-1}-1)!(2^{n-1}-1)!}
=\frac{2^{n-1}}{2^n-1}
\end{equation}
A: Opponents in each round are drawn at random. So in round 1, for example, the chance that player 2 does not meet player 1 is $$\frac{2^n-2}{2^n-1}=\frac{2(2^{n-1}-1)}{2^n-1}$$ Then in round 2 the chance is $$\frac{2^{n-1}-2}{2^{n-1}-1}=\frac{2(2^{n-2}-1)}{2^{n-1}-1}$$ and so on until the semifinals (round $n-1$) where the chance is $\frac{2}{2^2-1}$
When we multiply all these terms together we get $$\frac{2^{n-1}}{2^n-1}$$
