Combinations that make a 2 pair

Question

I was reading from textbook. It says that for a full house there are 13 values x (4 3) suits x 12 remaining values x (4 3) suits so 13 x 4 x 12 x 6 = 3744. That makes sense to me. It also says for 2 pair there are (13 2) choices of values x (4 2) suits for first card x (4 2) suits for x 44 remaining cards that are not of the same value or suit as the first 4 cards, that also makes sense to me. Until I think, why can't I multiple 13 values x (4 2) suits x 12 remaining values x (4 2) suits x 44 remaining cards? It seems I am double counting/miscounting combinations but I don't understand why.

Answer 1

As you know, a full house consists of three cards of one rank combined with 2 cards of another rank.

So there are 4 operations that must be performed:

1) Selecting the rank of the first three cards. That would be ¹³C₁ which calculates to 13 (Note: Combinations are calculated using ⁿC_r = n!/(r!(n-r)!) where "n" is the total number of objects in the set and "r" is the number you are choosing from that set. In this case, you are choosing a rank, which can be A, 1, 2, . . . K, and you are choosing just 1 rank)

2) Selecting 3 of the same cards from that rank which would be ⁴C₃ (there are 4 of each rank and you are choosing 3). ⁴C₃ calculates as 4. Note that you have this step incorrectly denoted in one part of your question.

3) Selecting the rank of the second three cards. That would be ¹²C₁ which calculates to 12. (Note that there are only 12 possibilities of rank left because you have chosen one already for the first 3 cards.

4) Selecting 2 of the same cards from that second rank which would be ⁴C₂ which calculates as 6. Note that you have this step incorrectly denoted in one part of your question.

Therefore there are 13•4•12•6 or 3744 total possible full house hands.

Please note this only gives you the total number of possible full house hands and NOT the probability of getting a full house. To get that probability, you would need to divide by the total number of possible poker hands.