Okay so I have this problem. 5 card poker Assume that ace of hearts, king of hearts and queen of spades are on the table. Jessica has ace of diamonds and queen of clubs in her hand. How many different pair of cards could Adam have that gives a better pokerhand then Jessica's?
A "pair" is considered any 2 cards.
How am I supposed to think here? I know absolutely nothing about poker.
This is a pretty poor problem imo, since it says 5 card poker, but clearly they are talking about texas holdem, which while popular, is still going to be confusing for most people to answer.
Anyway, Jessica has 2 pair of aces and queens, the better hands that are possible that beat her are
2 pair, Aces and kings: $\binom{2}{1} \cdot \binom{3}{1}$ since there are 2 aces to choose from, and 3 kings
3 of a kind: $\binom{2}{2} + \binom{3}{2} + \binom{2}{2}$ There are 2 aces left, you need both. There are 3 kings left, you need 2 of them, or there are 2 queens left and you need both.
Straight: $\binom{4}{1} \cdot \binom{4}{1}$ To make a straight you need a run of 5 cards. The only way to do that here is if you have a Jack and a Ten. There are 4 jacks and you need 1, and 4 tens and you need one.
It's not possible in this scenario to have any better poker hand than a straight, so you just have to sum all those up now. 6+(1+3+1)+16 = 27 different pair hands that beat Jessica