Poker is played with a regular deck of cards which contains 52 cards. In the deck of cards there are 4 suits (♠♡♢♣), each in 13 denominations. A hand of poker consists of 5 cards.
I need help with the following problem.
How many poker hands have a pair? Meaning, 2 cards in one denomination and 3 cards of other values, the other 3 cards can't be of the same value!
Now they solved it like this, using binomial notation $C(n,k) = \binom{n}{k}$:
- Pick 1 of 13 denominations $C(13,1)$
- Pick 2 of 4 colors $C(4,2)$
- Now pick 3 of 12 denominations $C(12,3)$
- Now each of the three values can be combined with 1 of 4 suits $C(4,1)^3 = 4^3$
You get the following $C(13,1) \cdot C(4,2) \cdot C(12,3) \cdot 4^3 = 1098240$ which is correct.
But why don't they write it like this:
- Pick 4 of 13 denominations $C(13,4)$
- Pick 2 of 4 suits for one of the value above $C(4,2)$
- Now each of the three other values can be combined with 1 of 4 suits $C(4,1)^3 = 4^3$.
You get the following $C(13,4) \cdot C(4,2) \cdot 4^3 = 274560$ (Why is this wrong!?)
Note! $274560\cdot 4=1098240$ (if this might help your explanation for me)
Best Answer
From the $4$ denominations that we have chosen, we must still choose the one in which we will have a pair, or equivalently the three in which we will have singles.
That's because for example $\spadesuit$K, $\heartsuit$K, $\diamondsuit$$2$, $\spadesuit$$3$, $\clubsuit$J is different from $\spadesuit$$2$, $\heartsuit$$2$, $\diamondsuit$K, $\spadesuit$$3$, $\clubsuit$J.