# Confusing probability.

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?