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Problem Statement:

5 players of equal strength play one game with each other. $P(A)$=probability that at least one player wins all matches he play.

I have to find $P(A)$.

My Approach:

I took cases in which a particular player wins $1,2,3,4$ and $5$ games, respectively,

As there can be $5$ games for a particular player,

$P(Winning)=P(Losing)=\frac{1}{2}$

Hence, $P(A)=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{4}+ \binom{5}{1}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{4}+ \binom{5}{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{3} +\binom{5}{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{2}+ \binom{5}{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{}=\frac{31}{32}$

But the answer is given $P(A)$=$\binom{5}{1}\frac{2^6}{2^{10}}=\frac{5}{16}$

What am I doing wrong?

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There are only four games for a particular player, not five. Let players be 1,2,3,4,5. If player 1 wins all the games, none of the others can. For example, player 2 won't be able to win all the games, as he has already lost to player 1. Similarly, if player 2,3,4 or 5 win all the games, none of the others can. Probability that any player wins all the games is (0.5)^4. Multiply with 5 for 5 cases (1,2,3,4 and 5 win all the games) to get 5/16.

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