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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.

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    What is asked to prove/calculate? – 2017-01-28
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    The conditional probability that $P_2$ loses in the second round given that $P_1$ wins in the third round. – 2017-01-28

3 Answers 3

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$2n$ 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) $= \dfrac{2n}{8n-1}$

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    Interesting: two upvoted questions with different answers ... – 2017-01-28
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    @EthanBolker The probability that any player loses on the 2nd round is 1/4 (1/2 to win on 1s $\times$ 1/2 to lose on second). For large $n$, $P_1$'s influence on $P_2$'s performance is almost none, the conditional probability should converge to 1/4. This solution's probability does. Plus the reasoning is simple and sound. I'd say this one is correct. – 2017-01-29
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    Could you elaborate on How did you jump to $\frac{2n}{8n-1}$? – 2017-01-29
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    @Vedvart1: Half the players lose in each round; e.g. $4n$ players lose in the first round, $2n$ in the second round. Since $P_1$ doesn't lose in the second round, the loser has to be among $8n-1$ players other than $P_1$. Each of them have the same *Pr* of losing, hence $P_2$ has a *Pr* of $\frac{2n}{8n-1}$ of losing the second round. – 2017-01-29
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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.

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    Interesting: two upvoted questions with different answers ... – 2017-01-28
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    The probability that any player loses on the 2nd round is 1/4 (1/2 to win on 1s $\times$ 1/2 to lose on second). For large $n$, $P_1$'s influence on $P_2$'s performance is almost none, the conditional probability should converge to 1/4. This solution's probability converges to 1, both not to 1/4 and failing a sanity check. So *this solution can not be correct* – 2017-01-29
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    Hmm, i definitely see your reasoning, but i can't see specifically where my answer fails except the answer. – 2017-01-29
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    @Vedvart1 If you do find your error, or someone else does, please leave your answer up, with an appropriate comment, since often there is a lot to be learned from honorable failed attempts. I'll upvote yours for the attempt alone. – 2017-01-29
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    My best guess is that there's some factor I'm not taking into consideration, but I can't see it for the life of me, if anybody does please respond, now I'm dying to figure this out lol – 2017-01-29
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    Why is $P(X_{2,2}) = \frac12$ for round $2$. He would lose if he played $P_1$ – 2017-01-29
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    When there are $8n$ players, you have only $8n-1$ opponents ! I have posted a longer non-technical answer, may be it'll help you spot your mistakes(s). – 2017-01-31
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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 $Pr = \dfrac{4n-1}{8n-1}\cdot\dfrac{2n}{4n-1} = \dfrac{2n}{8n-1}$