# Probability regarding getting a full house

I am trying to calculate the probability of getting a full house on a standard 5-card deal.

I am comfortable with how combinations, permutations and the fundamental principle of counting, but I am not sure how this problem works.

I understand that we need to avoid situations such as having a 4 of a kind or a flush, so can someone help me out ?

I would deeply appreciate it if you could give me another example like this as well.

• Umm, what exactly is a full house? – Lord Soth Dec 4 '13 at 22:48

You'll need to count how many different combinations of cards will result in a full house.

There are $13 \times 12$ ways to choose the particular two values of the cards that will make up your hand (i.e. kings over eights).

For each of these combinations, there are $_{4} C_{3} = 4$ combinations for the three of a kind, and $_{4} C_{2} = 6$ combinations for the pair.

Overall, there are $_{52} C_{5} = 2598960$ card combinations.

Hence the probability is $$\frac{13 \times 12 \times 4 \times 6}{2598960} \approx 0.00144.$$

• Thank you. I don't know why I was thinking that full house can come with a flush. – hyg17 Dec 4 '13 at 23:24

Hint: You need to count the number of full houses. They do not overlap with flushes or four of a kinds. How many choices for the rank of the triplet? How many choices for which cards of that rank? How many choices for the rank of the pair? It can't be the same as the triplet. Then how many choices for the cards of the pair, given the rank?