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.
 A: $2n$ players have to lose in the second round, and $P_1$ is not among them while $P_2$ is,
thus $\Bbb P(P_2$ loses in the second round$ | P_3$ wins third round) $= \dfrac{2n}{8n-1}$  
A: Not sure about my casework and such, but my thinking is that if $X_{n,k}$ is the probability of $P_n$ winning round $k$, then we know this: if $X_{1,3}$ is true then $X_{1,2}$ must be true, i.e. $P_1$ won in round 2. Then $P(X_{2,2}) = \frac{1}{2}$ if $P_2$ didn't play $P_1$ in round 1, but $P(X_{2,2})=0$ if they did play each other. We also know that there are $\frac{8n}{2}$ players in the $2^{nd}$ round, leaving $\frac{8n}{2}-1$ players for $P_2$ to play against. Assuming he won round 1 and weighting each probability by its frequency,
$$P(X_{2,2} | X_{2,1})=\frac{1}{2}*\frac{\frac{8n}{2}-1}{\frac{8n}{2}}+0*\frac{1}{\frac{8n}{2}} = \frac{8n-2}{16n} = \frac{4n-1}{8n}$$
But we also know the $P(X_{2,1})$ is $\frac{1}{2}$. Therefore
$$P(X_{2,2} | X_{2,1})=\frac{P(X_{2,1} \text{ and } X_{2,2})}{P(X_{2,1})} = \frac{P(X_{2,1})*\frac{4n-1}{8n}}{\frac{1}{2}} = P(X_{2,1})\frac{4n-1}{4n}$$
$$P(X_{2,1})=\frac{1}{2}*\frac{8n-1}{8n} \implies P(X_{2,2} | X_{2,1})=\frac{(8n-1)(4n-1)}{64n^2}$$
EDIT: Fixing arithmetic errors gave me $\frac{4n-1}{8n}$ which is closer, but still only converges to 1/2, and in no way conforms to true blue anil's answer (which is the correct one).
EDIT 2: I think I've taken into account $P_2$ playing $P_1$ in round 1, but $\frac{(8n-1)(4n-1)}{64n^2}$ seems further off and still converges to 1/2.
A: Another (longer) way to solve, round by round to help Vedavart1 find the flaw in his casework.
It is sufficient to track just $P_2$, it is given that $P_1$ won the third round.
$\underline{Round\;1}$
To win, $P_2$ can't play against $P_1$, thus $\Bbb P(P_2$ wins) $=\dfrac{8n-2}{8n-1}\cdot \dfrac12 = \dfrac{4n-1}{8n-1}$
$\underline{Round\;2}$ 
Having got through to round $2$, $P_2$ needs to lose among $4n$ players
This requires $\Bbb P(P_2$ plays $P_1$ and loses) or $\Bbb P(P_2$ plays someone else and loses)
$= \dfrac1{4n-1}\cdot 1 + \dfrac{4n-2}{4n-1}\cdot \dfrac12 = \dfrac{2n}{4n-1}$
$\underline{Final:result}$
By the multiplication law, required $Pr = \dfrac{4n-1}{8n-1}\cdot\dfrac{2n}{4n-1} = \dfrac{2n}{8n-1}$   
A: 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) = ${\frac{2n}{8n -1 }}$.
