Lets say a certain type of lottery ticket advertises that one out of every five tickets is a "winner", then what are the odds that if you buy 5 of those tickets, at least one is a winner.

I thought about this fairly hard and figured that your chance of having at least one winning ticket increase by 20% of the odds you have left to have a non winning ticket. For example...


1 ticket: 100/5 = 20% chances a winner
2 tickets: 100/5 + ((100-20)/5) = 36% at least one is a winner
3 tickets: 36 + ((100-36)/5) = 48.8% at least on is a winner
4 tickets: 48.8 + ((100-48.8)/5) = 59.04% " "
5 tickets: 67.232%

I don't expect this to be right but i like to think on the right track. If i could have this explained in a simple algorithm that would be appreciated. Thank you!

We don’t have details on just how winners are determined so we will assume that there are so many tickets sold that someone having won does not change the chances that you will win. Under this reasonable assumption, if 20% is the constant probability of winning, then 80% is the chance a ticket will lose. Losing all 5 tickets you buy has probability 0.80^5 = 0.328. Therefore, the probability of winning at least one ticket is 1- 0.328 = 0.672, which is what you got through an unusual way of using inclusion-exclusion.

In general, the binomial distribution will give you the .probability of exactly r winners out of 5 tickets:

P(r winners) = [5!/r!*(5-r)!] 0.2^r * 0.8 ^(5-r)