Knock out tournament 2 $2^n$ players of equal strength are playing a knock out tournament. If they are paired randomly in all rounds, then what is the probability that out of two particular players $S_1$ and $S_2$, exactly one will reach the semi-finals of the tournament. 
 A: Out of $2^n$ players, $4$ will reach the SF, $2^n-4$ will not.
Since each player has equal a priori chances,
P[Federer reaches, Djokovich doesn't (or vice-versa)] $=\dfrac{4}{2^n}\cdot\dfrac{2^n-4}{2^n -1}\times 2$
$\underline{Added\; explanation}$
I have simply treated it as a hypergeometric distribution (drawing w/o replacement), say drawing from $4$ blue chips and $2^n-4$ black chips.
The alternative combinatorial way for finding P($1$ blue, $1$ black) would then be  $\dfrac{\binom41\binom{2^n-4}1}{\binom{2^n}2}$
The point I am making is that though this is a knockout tournament with pairing and semi-finals and all that, it is unnecessary to look at it like that.  
A: There are $\binom{2^n}{4}$ possible choices of 4 players in the semifinals, all of which we assume are equally likely.  If exactly one of players $S_1$ and $S_2$ is in the semifinals, then there are $2 \binom{2^n-2}{3}$ possible selections. So the probability that exactly one of players $S_1$ and $S_2$ is in the semifinals is
$$\frac{2 \binom{2^n-2}{3}}{\binom{2^n}{4}} = \frac{8 (2^n-4)}{2^n (2^n-1)}$$
A: Kind of a stab at these probabilities, but here goes. Let $X_n$ be the event in which $S_n$ makes it to the finals. $$P(X_1 \text{ or } X_2) = P(X_1)+P(X_2)-P(X_1 \text{ and } X_2)$$
$$=\frac{4}{2^{n}}+\frac{4}{2^{n}}-\frac{4}{2^{n}}*\frac{3}{2^{n}-1}$$
$$=\frac{2^3}{2^{n}}-\frac{12}{2^{2n}-2^n}$$
$$=\frac{1}{2^n}\Big(2^3-\frac{12}{2^n-1}\Big)$$
$$=\frac{2^3}{2^n}\Big(\frac{(2^n-1)-12}{2^n-1}\Big)$$
$$=\frac{1}{2^{n-3}}\Big(\frac{2^n-13}{2^n-1}\Big)$$
$$=\frac{2^n-13}{2^{2n-3}-2^{n-3}}$$
A: Yeah, after @true blue anil's answer, I've realised I've calculated the probability for the finals. Let me correct myself.
There's $\frac{2^n(2^n-1)(2^n-2)(2^n-3)}{4!}$ equally likely semi-final match-ups, of which $\frac{2(2^n-2)(2^n-3)(2^n-4)}{3!}$ satisfy the condition. Hence, $$P=\frac{4!*2(2^n-2)(2^n-3)(2^n-4)}{3!*2^n(2^n-1)(2^n-2)(2^n-3)}=\frac{8(2^n-4)}{2^n(2^n-1)}=\frac{2^n-4}{2^{n-3}(2^n-1)}$$
