Question

In a knockout tournaments $2^n$ 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 $1/84$. The value of $n$ equals__________.

Answer 1

There are $2^n$ teams total

4 player reaches semi-final

so probability of winning of S1=1/4

Probability that someone else wins=3/4

No one is special so they will have same chance of winning.

Since there are $2^n-1$ of them, the probability that one of them S1 wins=$\frac{3}{4(2^n-1)}=\frac{1}{84}{2}^n-1=632^n=64n=6$