# Probability of getting full house in five-card poker when cards in suite is missing

What is the probability of getting full house in five-card poker given the following cards are missing?

Missing cards: $$\heartsuit$$2, $$\heartsuit$$5, $$\heartsuit$$10, $$\heartsuit$$Jack, $$\heartsuit$$King

My attempt on trying to solve this was to first calculate the combinations of hands which gives full house and then remove the combinations of the missing cards. This is what I came up with:

$$\displaystyle \frac{{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}-{8\choose 5}{1\choose 1}}{{47\choose 5}}$$

But I'm not sure this is a valid method... Any tips on how to approach the problem?

This does seem like a valid strategy, but I don't see where the $${8\choose 5}{1\choose 1}$$ came from.

The full houses can be put in four categories: (1) both pair and three of a kind from one of the ranks with a missing card, (2) just pair, (3) just three of a kind, (4) neither

(1) We have 5*4 options for the ranks. For the pair, there is one card missing, leaving three cards as options. Since we're making a pair, we need to choose two out of those three to be in the pair, which is the same as choosing one out of the three to not be in the pair, so that's 3 options. For the three of a kind, since there are only three cards remaining, we have to take those three cards, giving 1 option for the three of a kind, for a total of 5*4*3=60.

(2) We have 5 options for the pair rank, 8 options for the three of a kind rank. There are 3 options for the pair, and 4 options for the three of a kind. 5*8*3*4=480.

(3) 8 options for pair, 5 for three of a kind. 6 options for pair, 1 for three of a kind. 8*5*6*1 = 240.

(4) 8*7 options for the ranks, 6 options for pair, 4 options for three of a kind. 8*7*6*4 = 1344.

Total: 2124.

• @Accumulation For example in the first case you say: "[...] 3 options for the pair and 1 option for the three of a kind [...]". What are you referring to? 3 options of what? We have already chosen ranks ${5\choose 1}{4\choose 1}$ – phx Oct 2 '18 at 15:52

It is a valid method but $${8 \choose 5}{1 \choose 1}$$ is not the number of hands to remove. The number of full houses that include the heart $$2$$ is $${3 \choose 2}{12 \choose 1}{4 \choose 2}$$ that include it as part of the three of a kind plus $${3 \choose 1}{12 \choose 1}{4 \choose 3}$$ that include it as part of the pair, for a total of $$360$$. The same applies for all the other cards, so we might think we should remove $$5\cdot 360=1800$$ hands. The problem is that we have twice removed the ones that include two of the missing cards.

The number of full houses that contain the $$2$$ of hearts in the triplet and the $$5$$ of hearts in the pair is $${3 \choose 2}{3 \choose 1}=9$$. There are $$20$$ ways to choose the two ranks to have cards missing from each, so we should add in $$180$$. This is an example of the principle of inclusion-exclusion.

We therefore remove $$1800-180=1620$$ full houses because we are missing the five cards.