In a knockout tournament, `2^(n)` equally skilld players, `S_(1), S_(2), …, S_(2^(n))` are participating. In each round, players are divided in pairs at random and winner from each pair moves to 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 is_______.

Answer

6

Explanation

Given `S_(2)` reaches the semifinals.
Since all other players `(2^(n)-1)` are equally likely to win the finals with proballity p. We have
`(2^(n)-1)+1/4=1`
`or (2^(n)-1)=3/4`
`or p=(3)/(4(2^(n)-1))`
If `p=1/84,then`
`1/84=(3)/(4(2^(n)-1))`
`or2^(n)-1=63`
`or 2^(n)=64`
`or n=6`