Knock out tournament 1 8n players $P_1$, $P_2$, $P_3$, .....$P{_8}{_n}$ play a knock out tournament. It is known that all players are of equal strength. The tournament is held in three rounds where the players are paired at random in each round. If it is given that $P_1$ wins in the third round then what is the conditional probability that $P_2$ loses in the second round.
I tried applying the concept of conditional probability followed by total probability theorem but somehow, there are far too many cases to consider. Any help/ suggestions/ solutions would be highly appreciated.

First Round = 8n men. Second Round or Semi-Final = 4n men. 3rd Round or Final = 2n men. The tournament Winner is $P_{_{1}}$.The Probability that he is the winner is given by $\frac{1}{Totalmen -1 } = \frac{1}{8n -1 }$ . But $P_{_{2}}$ lost in the Second Round. Now In Second Round, there are  4n men = 2n Losers + 2n Winners. Therefore, $P_{_{2}}$ is one of the losers. He can be selected in $\binom{2n}{1}$ ways.Hence, P( $P_{_{2}}$ Loses in the Second Round|  $P_{_{1}}$ wins The Tournament) = ${\frac{2n}{8n -1 }}$.
