What is the probability of obtaining a hand with TWO PAIRS in a standard 5 card game of poker? What is the intuition behind, 'What is the probability of obtaining a hand with TWO PAIRS in a standard 5 card game of poker?'



I know the solution is, 

$$\frac{\binom{13}{2}\binom{4}{2}\binom{4}{2}\cdot\binom{11}{1}\binom{4}{1}}{\binom{52}{5}} = 0.047539$$

I understand the reasoning behind $\binom{11}{1}\binom{4}{1}$. That is, the number of ways a single card (After the two pairs have been selected) can be selected is $11\cdot4$. 

However, the $\binom{13}{2}\binom{4}{2}\binom{4}{2}$ part confuses me a little. Why doesn't $\frac{\binom{13}{1}\binom{4}{1}\binom{12}{1}\binom{4}{1}\cdot\binom{11}{1}\binom{4}{1}}{\binom{52}{5}}$ work? I've read that it counts solutions twice, but I don't understand how? How is that $\binom{13}{2}$ - 78 cards can be selected amongst $\binom{4}{2}\binom{4}{2}$ - 36 suits?
First pick the ranks of the first pair and second pair. There are 13 choices, then choose which two you want of those of each rank. This gives $\binom{13}{2}\binom{4}{2}\binom{4}{2}$ in total.

$\binom{13}{1}\binom{4}{1}\binom{12}{1}\binom{4}{1}$ doesn't work because choosing one rank and then choose another separately implies that the pairs aren't the same type in the sense that the order is not interchangeable. For example, it is saying that $AAKKQ$ is not the same as   $KKAAQ$. Also, you are choosing two per rank, not one, thus it should be $\binom{4}{2}$.
The key to correctly applying the combinatorics is to acknowledge how many kinds of MULTIPLICITY - four-suit ("quad"), three-suit ("triple" or "set"), two-suit ("pair" ), or one-suit ("singleton") - exist in a hand AND how many times EACH one occurs. 

For example, in a five-card full-house hand, TWO multiplicities exist, once each, a triple and a pair. Thus TWO ranks are selected from the thirteen available using $\binom{13}{2} = 78$. Then, each selected rank (say, K and Q) can be used in EITHER multiplicity to make DISTINCT full-house hands - KKK full of QQ, or vice versa - which is SECOND choice, $\binom{2}{1}=2$. 

But, in a six-card full house of the form AAABBB of ranks A and B, both of which are triples (and, of course, the higher of which is strategically played as the rank "full of" the lower) the second choice would not exist, so the "$\binom{2}{1}=2$" multiplication would not be performed. 

Likewise in a six-card three-pair scenario $\binom{13}{3} = 286$ would indeed select the three ranks of each pair, but because they are ALL pairs (the same multiplicity) no second choice under "$\binom{3}{1}=3$" or "$\binom{3}{2}=3$" could possibly attach. 

But a seven-card THREE-pair hand with ONE singleton would require $\binom{13}{4}\binom{4}{3} = 715*4 = 2860$ ways to fully assign ALL ranks to ALL possible variants. 

One final example: ten-card four-of-a-kind of the form AAAABBBCCD: it contains four multiplicities - a quad, a triple, a pair, and a singleton, once each. So $\binom{13}{4}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1} = 17160$ ways to conceive of distributing all selectable ranks through all multiplicities. 

If it had been a quad of the form AAAABBCCDD, the result would be only $\binom{13}{4}\binom{4}{1}\binom{3}{3} = 2860$ distinct possible rank-multiplicity assignments. This explains why our (intuitive) experience is that triples and quads with singletons, rather than with multiple pairs, are the most likely.