A deck of cards contains $52$ cards. There are $13$ different kinds of cards, with four cards of each kind. (Among the terms commonly used instead of “kind” are “rank,” “face value,” “denomination,” and “value.”) These kinds are twos, threes, fours, fives, sixes, sevens, eights, nines, tens, jacks, queens, kings, and aces. There are also four suits: spades, clubs, hearts, and diamonds, each containing $13$ cards, with one card of each kind in a suit. A hand consists of five cards.
How many possibilities are there for the following poker hands?:
(1) Full house = $3$ of one kind and $2$ of another kind
(2) Straight = $5$ consecutive kinds (note that Ace can be high or low)
(3) $4$ of a kind
(4) Straight flush = five cards of the same suit of consecutive kinds
I calculate answers for each above, but I am not totally sure about my answer. Can someone take a look? Thanks in advance!
For (1), $C(13, 2) \cdot 2 \cdot 4^5$ since we choose $2$ kinds from $13$ kinds, and $3$ cards can have one of the kind, while the other $2$ cards can have the other kind. Since there are two choices for which kind to be for the three kinds, we multiply by $2$. Each card can be one of four suits, so we multiply $4^5$.
For (2), there are $10$ consecutive kinds. Each cards can be one of $4$ suits. Thus, $4^5 \cdot 10$.
For (3), I think there are $13$ kinds in total, so the answer is $13$.
For (4), there are $10$ consecutive kinds and four suits totally. So $10 \cdot 4 = 40$.
Best Answer
Full house: There are $\binom{13}{1}$ ways to choose the rank from which three cards are drawn. There are $\binom{4}{3}$ ways to choose three of the four suits for that rank. There are $\binom{12}{1}$ ways to choose the rank from which two cards are drawn from the remaining ranks. There are $\binom{4}{2}$ ways to choose two of the four suits of that rank. $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$$ Your answer is incorrect since you do not have four choices for the rank of each card since choosing a card of a particular rank reduces the number of choices you have for selecting another rank of that card. Suppose you select three jacks and two fives. There only four ways to select three of the four jacks since one of those four ranks must be omitted
There are only $\binom{4}{2} = 6$ ways to select two of the four fives.
Straight: In poker, a straight is usually defined as five cards of consecutive ranks, not all of the same suit, where an ace can be high or low. However, we will use the author's definition.
Choosing the rank for the low card determines which five ranks appear in the straight. The low card in a straight can be selected in ten ways (A, 2, 3, 4, 5, 6, 7, 8, 9, 10). The suit for each rank can be selected in four ways. $$\binom{10}{1}\binom{4}{1}^5$$ so your answer is correct.
Under the usual definition, we would have to exclude straight flushes, that is, a straight in which all cards are of the same suit. $$\binom{10}{1}\left[\binom{4}{1}^5 - 4\right]$$
Four of a kind: There are $\binom{13}{1}$ ways to choose the rank from which four cards are drawn. We must select all four cards of that rank. There are $\binom{12}{1}$ ways to choose the rank from which the other card is drawn. There are $\binom{4}{1}$ ways to choose the suit of that card. $$\binom{13}{1}\binom{4}{4}\binom{12}{1}\binom{4}{1}$$ Your answer is incorrect because you forgot to account for the rank and suit of the other card.
Straight flush: There are $\binom{10}{1}$ ways to choose the low card and $\binom{4}{1}$ ways to choose the suit. $$\binom{10}{1}\binom{4}{1}$$ so your answer is correct.