If $S_2$ reaches the semi-final then the probability that $S_1$ wins the tournament is $\frac{1}{20}$ In  a knockout tournament $2^n$ equally skilled players;$S_1,S_2,...,S_{2^n}$ are participating.In each round players are divided in pair at random and winner from each pair moves in the next round.If $S_2$ reaches the semi-final then the probability that $S_1$ wins the tournament is $\frac{1}{20}$.Find the value of $n.$
There will be $n$ rounds of the tournament because $2^n$ players are there.But i dont know how to solve further.Some help/hints are needed.Thanks. 
 A: Hint:  Four players reach the semi-final, each the winner of a tournament of $2^{n-2}$ players.  $S_1$ cannot be in the same bracket as $S_2$, but is a random participant from the other three brackets.  
Added: The chance that $S_1$ gets to the semis is $\frac 3{2^n-1}$ because the semi participants besides $S_2$ are chosen at random from the pool.  The chance that $S_1$ wins is then $\frac 3{4(2^n-1)}=\frac 1{20}$, so $n=4$
A: There isn't enough information to answer the question, since we don't know the probabilities of the players winning their games. I'll assume that the outcomes of all games are independent and either player has a $50\%$ chance of winning each game.
The probability that $S_2$ will win is $\frac14$. If she doesn't, every other player has the same chance.

Edit:
Here's a more explicit solution. The probability that $S_2$ will win is $\frac14$, since she reached the semifinal. Thus the probability that someone else will win is $\frac34$. No-one else is special, so they all have the same probability to win. Since there are $2^n-1$ of them, the probability that one of them, $S_1$, wins is $\frac34\left(2^n-1\right)^{-1}$. Equating this to $\frac1{20}$ then yields $n$.
