from those ordinals, we want to pick 2 cards that are available from a population of 4 each.we want to pick 2 ordinals from the population of 13.each one of these 3 cards can be chosen from a population of 4 suits.from the remaining 12 ordinals, we pick 3.from that ordinal, we want to pick 2 cards that are available from a population of 4.we want to pick 1 ordinal from the population of 13.We want 2 of the same ordinal (out of the 4 suits) and then the other 3 cards can't be that first ordinal, nor can we have an ordinal repeat (that would create a different hand - 2-pair or a full house). There are 13 ordinals (Ace through King) in a standard deck of cards. #C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!)# with #n='population', k='picks'#įirst let's calculate the denominator (picking 5 random cards from a pack of 52 cards): These will be combination problems (we don't care about the order of the cards dealt): When working with the probability of poker hands, we need to know the number of ways a hand can be dealt (this is the denominator) and the numerator is the number of ways to have a certain hand.

Poker ace counts as 1