If each of your selections is a 17% chance of winning, then the chance of backing at LEAST one winner out of 5 bets is about 60%.

This is calculated by working out the probability of backing NO winners in 5 bets and is equal to .83^5 = .3939 = 39.39%. Therefore if there is a 39.39% chance of backing NO winners in 5 bets then there must be a 60.61% chance of backing one or more winners. This is based on the assumption that each bet is independant of any other.

See JFC's post for breakdown of chances as they are correct!

[ This Message was edited by: AssumeTheCrown on 2004-08-31 14:00 ]