Knock out tournament In a knockout  tournament  $2n$  equally skilled  players;  $S_1, S_2,\ldots,S_{2n}$ are  participating.  In  each round  players are  divided  in pairs at  random &  winner  from each pair  moves in the next  round.  If   $S_2$ reaches the semifinal then find the probability that  $S_1$ wins the tournament.
I understood the problem but not able to approach.
 A: Recognize that every group of four players is equally likely to make it to the semi-finals.  Recognize further that every choice of three additional people apart from $S_2$ are equally likely to be with $S_2$ in the semi finals.

 Informally, this is because of symmetry.  There is nothing giving $S_1,S_3,S_4$ any more or less chance than $S_i,S_j,S_k$, so by a change of labels we see the probabilities must be the same.  Alternatively, we could instead temporarily assume each match happens in a specific sequence one at a time.  We can then look at the sequence of losers.

$Pr(S_1~\text{wins}|S_2~\text{in semis})=\frac{Pr(S_1~\text{wins}\cap S_2~\text{in semis})}{Pr(S_2~\text{in semis})}$
From the earlier observations, we can see that $Pr(S_2~\text{in semis})=\frac{\binom{2n-1}{3}}{\binom{2n}{4}}=\dfrac{\frac{(2n-1)!}{3!(2n-4)!}}{\frac{(2n)!}{4!(2n-4)!}}=\frac{4}{2n}$
Looking at the numerator more closely:
$Pr(S_1~\text{wins}\cap S_2~\text{in semis})=Pr(S_1\text{wins}\cap S_1~\text{in semis}\cap S_2~\text{in semis})$
$=Pr(S_1~\text{wins}\mid S_1~\text{in semis}\cap S_2~\text{in semis})Pr(S_1~\text{in semis}\cap S_2~\text{in semis})$
The chance that $S_1$ wins given that $S_1$ was in the semis regardless of who else is in the semis is $\frac{1}{4}$.  The chance that $S_1$ and $S_2$ are in the semis together is similar to previously $\frac{\binom{2n-2}{2}}{\binom{2n}{4}}=\frac{4\cdot 3}{2n(2n-1)}$
Putting all of this together then we have:
$$Pr(S_1~\text{wins}\mid S_2~\text{in semis})=\frac{1}{4}\cdot\frac{4\cdot 3}{2n(2n-1)}/\left(\frac{4}{2n}\right)=\frac{3}{4(2n-1)}$$

Note: this answer works regardless of the nature of $2n$, but the tournament itself might not exist in the first place if $2n$ is not a power of $2$.  This suggests either games do not happen concurrently, meaning that some players may have to win many times more than other players, or you have a typo and you meant for it to be $2^n$ instead.  If it is intended to be $2^n$, all of the arguments still work, but it will simplify instead to $\frac{3}{4(2^n-1)}$.  This matches the answer given in the comments on the other post.
A: If player $S_2$ has taken one of the $4$ spots in the semi-finals then the probability that $S_1$ takes one of the other $3$ spots is
$$P(\text{$S_1$ in semis }|\text{ $S_2$ in semis})=\frac{3}{2n-1}$$
since each of the remaining $2n-1$ players are all equally skilled.
The probability that $S_1$ then goes on to win both the semi-final (probability $1/2$) and the final (probability $1/2$) is $(1/2)^2=1/4$, hence
$$P(\text{$S_1$ wins final }|\text{ $S_2$ in semis})=P(\text{$S_1$ in semis }|\text{ $S_2$ in semis})\cdot\frac{1}{4}$$
$$\implies P(\text{$S_1$ wins final }|\text{ $S_2$ in semis})=\frac{3}{4(2n-1)}\tag{Answer}$$
