Probability of getting two pair in poker I was looking at this website http://www.cwu.edu/~glasbys/POKER.HTM and I read the explanation for how to calculate the probability of getting a full house.  To me, the logic basically looked like you figure out the number of possible ranks and multiply by the number of ways to choose the cards from that given rank.
In other words, for a full house $P=$
$$\frac{{13\choose1}{4\choose3}{12\choose1}{4\choose2}}{52\choose5}$$
Following this logic, I tried to calculate the probability of getting two pair.  My (incorrect) logic was that there are 13 possible ranks for the first pair and $4\choose2$ ways to choose two cards from that rank, 12 possible ranks for the second pair and $4\choose2$ ways to choose two cards from that rank, and 11 possible ranks for last card and $4\choose1$ ways to choose a card from that rank.  
So I tried $P=$
$$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$
Obviously my solution was incorrect.  I read explanation and the correct answer is $P=$
$$\frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$
I'm still a bit fuzzy on where I went wrong though.  Can anyone help me understand this problem a little better?  Thank you very much for your help.

When making a tree, you have 13 choices for the first kind, 6 ways to get it, 12 for the second kind, 6 ways to do that, and 44 cards that are not the first two kinds:
13x6x12x6x44
The problem with this reasoning is that you will have 10s and 4s on one branch and on another you will have 4s and then 10s--the SAME hand.  In other words, you have doubled your hands.
So logic says, divide 13x12 by 2 to take away all the double answers that will show up on your tree.
So....it is 
(13x12/2)x6x6x44
yielding 123552.
For the probability, you need to still divide that by:  52C5
But this verifies WHY your second answer is correct and the first one is wrong:
13C2 x 4C2 X4C2 x 44C1 / 52C5
