Understanding probability of being dealt a two pair just need help wrapping my head around this:
I understand that the sample space consists of C(52,5) elements.
And that the logic behind calculating the probability is that we first find the product of 
choosing 2 ranks from the 13 available and 1 rank from the 11 remaining. Then find the product of this with the square of choosing 2 suits from 4 (for the pairs),and with choosing 1 from 4 (for the remaining card).
i.e. C(13,2)C(11,1)C(4,1)[C(4,2)^2]
But I don't understand (logically) why this doesn't provide the same result (off by a factor of two)
as if we first choose 1 rank from 13, then 1 from 12, then the rest is the same.
i.e. C(13,1)C(12,1)C(11,1)C(4,1)[C(4,2)^2]
Why make a special case of the pairs?
Why not even choose 3 ranks from 13 initially - as oppose to choosing 2 then 1?
i.e. C(13,3)C(4,1)[C(4,2)^2]
 A: Think of it this way: we have to paint a car using three colors, we can chose out of three colors, the way it works is the three colors are mixed and then used to paint the car. Clearly  all three colors are the same and the order in which we chose them don't matter. So there is only 1 way to choose(or not choose;)).
Another option is that 1 of the colors is chosen as the main color and then the other 2 are mixed and are used to paint the interior. Then there are 3 ways because the only choice you have to make is which of the colors is the main color.
Finally there is a third option in which 1 color is used for the middle of the car, one for the back and one for  the front. Then there are 6 options because every permutation of the colors changes the way the car is painted.
Your example is like the second example. You need to chose the non-pair color first because it plays a different role. On the other hand the 2 pair ranks play exactly the same role. So the order in which they are chosen is irrelevant. So it is the same as chosing 2 ranks from the remaining 12.
A: In your second computation, you can pick your two ranks in two ways. If you divide $C(13,1)C(12,1)$ by $2$, though, you get $C(13,2)$.
Similarly, if you just pick three ranks, how do you know which ranks have pairs? You have to choose two from the three ranks for pairs, yielding:
$$C(13,3)C(3,2) = C(13,2)C(11,1)$$
which is the same as the original formula.
