Knock out tournament question (asked before on MSE)... not getting the correct answer Knock out tournament 1
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 $P_1$, $P_2$, $P_3$, .....$P{_8}{_n}$ 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.

The answer given there is ${{2n}\over{8n-1}}$. However, I got something different, so I'm asking for your help where I went wrong. The probability of $P_1$ winning in the first two rounds is $\left({1\over2}\right)^2 = {1\over4}$. The probability of both $P_1$ and $P_2$ winning in the first round is$$\left({1\over2}\right)\left({{8n-2}\over{8n-1}}\right)\left({1\over2}\right)\left({{8n-3}\over{8n-2}}\right) = \left({1\over4}\right)\left({{8n-3}\over{8n-1}}\right).$$The probability of $P_1$ winning and $P_2$ losing in the second round is$$\left({1\over2}\right)\left({1\over{4n-1}}\right) + \left({1\over2}\right)\left({{4n-2}\over{4n-1}}\right)\left({1\over2}\right)\left({{4n-3}\over{4n-2}}\right) = {1\over4}.$$Therefore our desired probability should be$${{{\left({1\over4}\right)\left({{8n-3}\over{8n-1}}\right)}\left({1\over4}\right) }\over{{1\over4}}} = {{8n-3}\over{4(8n-1)}},$$which clearly does not equal ${{2n}\over{8n-1}}$. Where did I go wrong?
 A: If we randomly pair $8n$ players and then randomly select the winner in each pair, it's the same as we randomly choose $4n$ of the players as winners.
For the first player to win the first round, they need to get to $4n$ winner places out of $8n$ players, so P(P1 wins round 1) = 1/2. If we know that player 1 won round 1, then player 2 must get to $4n-1$ places out of $8n-1$ players left. Thus, the probability they both win is:
$$
P(\text{both win round 1}) = \frac12\frac{4n-1}{8n-1}.
$$
If we know that both players made it to the second round, then the same logic applies, but now player 2 must get to $2n$ losers rather than $2n-1$ winners:
$$
P(\text{player 1 wins, player 2 loses round 2}|\text{both win round 1}) = \frac12\frac{2n}{4n-1}.
$$
Since ‘player 1 wins, player 2 loses round 2‘ implies ‘both win round 1’, we conclude that
$$
P(\text{player 1 wins, player 2 loses round 2}) = \frac12\frac{4n-1}{8n-1} \times \frac12\frac{2n}{4n-1} = \frac14 \frac{2n}{8n-1}.
$$
Adjust for P(player 1 wins two rounds), and you get your answer.
Finally, you don't need to do all of the above and come to the conclusion directly like in the answer you cited. Player 2 must go to $2n$ losers of round 2 from $8n-1$ players of round 1.
