Question

8n players `P_(1),P_(2),P_(3),....,P_(8n)` play a knock out tournament. It is known that all the players are of equal strngth. The tournat random is held in three rounds where the players are paired at random in each rouns. If it is given that `P_(1)` wins in the third round. Find the probability than `P_(2)` loses in the second round.

Answer 1

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).