I assume he's playing each palyer one time?. If he were playing, say, player 1 we could just use a binomial to find the probability of winning n times.

The prob. of winning 1 game if each of 5 players is played once out of 5 games would be

(.25)(.55)(.45)(.25)(.05)+(.75)(.45)(.45)(.25)(.05 )+(.75)(.55)(.55)(.25)(.05)+(.75)(.55)(.45)(.75)(. 05)+(.75)(.55)(.45)(.25)(.95)=0.056

We choose one from each player as a winner and the rest as losers then add them up.

That's one way to look at it. For 2 and 3 wins, you'll have 10 arrangements to list out. That's too many to list.