If it is given that $P_1$ wins in the third round.Find the probability that $P_2$ loses in the second round. $8n$ players $P_1,P_2,P_3,....,P_{8n}$ play a knock out tournament.It is known that all the players are of equal 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.Find the probability that $P_2$ loses in the second round.
I know that as all the players are of equal strength,so their probability of winning the rounds are equal,but i dont know how to solve this question.I tried many attempts but failed.Please help me.
Thanks
 A: If $P_1$ wins in the third round, that means there is one player whom $P_1$ eliminates in the first round. The probability that this is $P_2$ is $1/(8n-1)$. There is also one player whom $P_1$ eliminates in the second round, and the probability that this is $P_2$ is again $1/(8n-1)$. Finally, there is one player whom this latter player eliminated in the first round, and again the probability that $P_2$ is this player is $1/(8n-1)$. The remaining $8n-4$ players do not interact with $P_1$ up to the second round and thus have the standard $\frac14$ probability of losing in the second round. Thus the probability that $P_2$ loses in the second round is
$$
\frac1{8n-1}+\frac14\frac{8n-4}{8n-1}=\frac{2n}{8n-1}\;.
$$
The same result can be obtained more directly by noting that $2n$ players lose in the second round and $P_1$ is not one of them, so the probability for $P_2$ to be one of them is $2n/(8n-1)$.
A: Let $A$ be the event $P_2$ loses in the second round, and $B$ the event $P_1$ wins in the third round. We want $\Pr(A\mid B)$, which is $\Pr(A\cap B)/\Pr(B)$. Finding $\Pr(B)$ is easy. So we want $\Pr(A\cap B)$.
For $A\cap B$ to happen, both players must get through the first round. In particular, $P_1$ must not be matched against $P_2$. The probability of this is $\frac{8n-2}{8n-1}$. And they must both win their matches, probability $\frac{1}{4}$. So the probability they both get through the first round is $\frac{1}{4}\cdot \frac{8n-2}{8n-1}$. Call this number $a$. 
On to the second round! Now two things can happen. Maybe $P_1$ is matched against $P_2$ and wins. The probability of this is $\frac{1}{2}\frac{1}{4n-1}$. And then $P_1$ must win her third round match, probability $\frac{1}{2}$. This gives a total contribution of 
$$a\frac{1}{2}\frac{1}{4n-1}\frac{1}{2}$$
to $\Pr(A\cap B)$.
Or maybe $P_1$ is matched against someone else in the second round, and wins while $P_2$ loses, and then $P_1$ wins her third round match. This gives a contribution of 
$$a\frac{4n-2}{4n-1}\frac{1}{2}\frac{1}{2}\frac{1}{2}.$$
