Eight players `P_(1), P_(2), …, P_(8)` play a knock-out tournament. It is known that whenever the players `P_(i)` and `P_(j)` play, the player `P_(i)` will win if `i lt j`. Assuming that the players are paired at random in each round, what is the probability that the player `P_(4)` reaches the final?

Explanation

Let A be the event of `P_(1)` winning in third round and B be the event of `P_(2)` winning in first round but ossing in second round. We have
`P(A)(""^(8n-1)C_(n-1))/(""^(8n)C_(n))=1/8`
`P(BnnA)`
= Probability of both `P_(1) and P_(2)` winning in first round `xx` Probability of `P_(1)` winning and `P_(2)` losing in second round `xx` probability of `P_(1)` winning in third round
`(""^(8n-2)C_(4n-2))/(""^(8n)4_(n))xx(""^(4n-2)C_(2n-1))/(""^(4n)C_(2n))xx(""^(2n-1)C_(n-1))/(""^(2n)C_(n))=(n)/(4(8n-1))`
Hence, `P((B)/(A))=(P(BnnA))/(P(A))=(2n)/(8n-1)`
Alternate solution:
Probability than `P_(2)` wins in first round given `P_(1)` wins is
`P((B)/(A))=(P(BnnA))/(P(A))=(2n)/(8n-1)`
In second round, probability that `P_(2)` loses in second round given `P_(1)` wins in
`1-(2n-1)/(4n-1)=(2n)/(4n-1)`
Hence, probability than `P_(2)` loses in second round, given `P_(1)` wins in third round is 2n/(8n-1).