Need help understanding poker hand probability/combinatorics.

I have some trouble understanding a difference when calculating the probability for a full house versus two pairs.

For two pairs the number of possible hands is:

$$\binom{13}{2} \binom{4}{2} \binom{4}{2} \binom{11}{1}\binom{4}{1}$$

Which makes sense for me as we first choose 2 values out of 13 possible, then 2 suits out of 4 possible twice to get the two pairs, then we draw the last card.

However, when applyng this logic to a full house, I encounter a problem.

For a full house the number of possible hands is:

$$\binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2}$$

Here we first choose 1 value out of 13 possible and 3 suits of 4 possible to get the three of a kind. Then we choose 1 value out of 12 possible and 2 suits out of 4 possible to get the pair.

Why wouldn't we choose 2 values out of 13 possible, then 3 suits out of 4 possible to get the three of a kind then 2 suits out of 4 possible to get the pair like this?

$$\binom{13}{2} \binom{4}{3} \binom{4}{2}$$

Or the other way around, apply the full house method to the two pairs:

$$\binom{13}{1} \binom{4}{2} \binom{12}{1} \binom{4}{2} \binom{11}{1}\binom{4}{1}$$

I hope someone can help me understand the inconsistency in these two calculations.

• Because for the full house choosing, say, $K,2$ is not the same as choosing $2,K$ (as $KKK22\neq KK222$). – lulu Apr 6 '16 at 13:34
• That makes sense, thanks. So I guess as long as there is symmetry between the draws the two-pair method is to be used? – NinjaNymo Apr 6 '16 at 13:40
• Exactly. The binomial symbol disregards the order...which is fine so long as there is symmetry. – lulu Apr 6 '16 at 13:42

You can't apply the two pair method to the full house because the 2 "ranks" out of 13 that you choose are not symmetric like a two pair.

Consider a two pair with 5's and 4's

5-5-4-4-X

If you switch the 5's and the 4's the hand would stay the same

However, for a full house you choose one "rank" to be the 3 of a kind and one rank to be the pair

5-5-5-4-4 is different from 4-4-4-5-5

By your "two pair method" applied to a full house you don't distinguish the above two hands so you undercount.

Now, you could apply the "full house method" to two pairs, but with a few changes.

If you apply this method to two pairs, 5-5-4-4-X would be distinguished from 4-4-5-5-X, which is not desired. (If 5 is chosen from the first selection then 4, and vice-versa)

So every two pair will be counted twice, so you need to divide by two

"Two pair method" for two pairs: $\binom {13}{2}\binom {4}{2}\binom {4}{2}\binom {11}{1}\binom {4}{1}$

This is just a factor of two away from the "full house method" for two pairs: $\binom {13}{1}\binom {4}{2}\binom {12}{1}\binom {4}{2}\binom{11}{1}\binom {4}{1}$, because:

$\binom{13}{2} = \frac{1}{2} * \binom{13}{1} \binom{12}{1}$

Hope this helps!

Note: (Another way to think about the two pair) For the last card in the two pair, you don't need to do $\binom{11}{1}\binom{4}{1}$ explicitly. As 5's and 4's are ruled out, you can choose any one of the 52-8 = 44 remaining cards (which is the same as $\binom{11}{1}\binom{4}{1}$)