# How many ways are there to get a full house?

I've seen many other ways to do this but I want to understand why my way doesn't work.

So the full house looks like this: XXX YY, where Xs are cards of one rank and Ys cards of another.

These can be picked the following way:

1. The first X: One can pick any of the 52 cards.
2. The second X: There are 3 other Xs to pick, so one can pick any of the 3.
3. The third X: There are 2 other Xs to pick, so one can pick any of the 2.
4. The first Y: The rank can't be the same as that of X, so there are $$52-4=48$$ options.
5. The second Y: There are 3 other Ys to pick, so one can pick any of the 3.

Therefore number of full houses would be $$52 \cdot 3 \cdot 2 \cdot 48 \cdot 3=4,4928 \cdot 10^4$$, which is not the correct answer.

So where does it go wrong?

• You are counting the ways to draw a full house in order, but of course that isn't the usual definition.
– lulu
May 27, 2021 at 16:04
• the actual result is $13 \cdot 4 \cdot 12 \cdot 6 = 3\,744$ May 27, 2021 at 16:06
• Mention the fact that "full house" is a "poker" game card configuration. It looks maybe so evident to you, but it' always good in an international site like this one to add this kind of information. May 27, 2021 at 19:17

A simple way is the following:

1. Pick the rank of the three-of-a-kind: there are $$\binom{13}{1}=13$$ ways of doing so.
2. Pick the three suits that will go into the three-of-a-kind; there are $$\binom{4}{3}=4$$ ways of doing so (equivalently: pick the card from that rank that is not being picked, which can be done in $$\binom{4}{1}=4$$ ways).
3. Pick the rank of the two-of-a-kind: there are $$\binom{12}{1}=12$$ ways of doing so (because one rank has already been chosen for the three-of-a-kind).
4. Pick the two suits that will be in the two-of-a-kind: there are $$\binom{4}{2}=\frac{(4)(3)}{2}=6$$ ways of doing so.

Total: $$13\times 4\times 12\times 6$$.

Your errors: you are counting the order in which the cards are chosen; so for example, you count separately the case of aces of clubs, spades, and hearts, followed by the two of hears and two of diamonds; and the case of the aces of clubs, spades, and hears, followed by the two of diamonds and hearts; and the case of the aces of clubs, hearts, and spades, followed by the two of hearts and diamonds, etc. These are all the same hand, but you are counting them separately.

How much did you overcount? For each choice of three-of-a-kind, you counted it six times, as many times as you have to list the three cards in order. And for each choice of two-of-a-kind, you counted it twice, once for each ordering of the two cards. In total, you counted each hand twelve times, so you need to divide by $$12$$ in order to compensate for the overcounting.

• thanks! Can you also explain, why doesn't it work if I first choose the two ranks, then the cards: $\binom{13}{2}\cdot\binom{4}{3}\cdot\binom{4}{2}$ May 31, 2021 at 7:12
• @mathslover: Because the two ranks are not interchangeable: it matters which one corresponds to the three-of-a-kind and which one corresponds to the pair. If you just choose two ranks, you aren't saying how you are deciding which one is which. May 31, 2021 at 10:53

You could have selected the three X cards in a different order; to compensate for this, divide by $$3!$$. You could also have picked the two Y cards in a different order; to compensate for this, divide by $$2!$$.