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,


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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.