"I really really think I have it now.... follow w me. probability that we win the first one= 1/13. probability we will not win any of the next ones (12/13)^4 therefore probability win first one and one more (1/13)*(1-(12/13)^4)...you keep going through that equation until you run out of possibilities. so probability of winning second one and one other (1/13)*(1-(12/13)^3), win third and one other =(1/13)*(1-(12/13)^2) win last two = (1/13)*(1/13). add all of those up for you total of winning any combination of two and you get ~5.5% which is a little over 1 in 18. You should do great at 1 in 20 or higher. I know ive had a lot of answers but this one really makes the most sense I'm gonna post it on 2+2. but I can't see anyway that it's wrong. I'll let you know if they stomp me ������"