You're dealt 4 cards from a regular 52card deck.

What is the probability that you get 2 different pairs?

Different means (as example) that 2 pairs of 5's don't count.

0.0190

Hmmm...I get 216/20825 = .010372...

How did you get your results?

I got 396/20825 by doing C(13,2)* C(12,2) divided by C(52,4)

Here's my reasoning. After 2 cards received:

got pair : prob 3/51 [case 1]
not pair : prob 48/51 [case 2]

[case 1]:
3rd card: 48/50 (you don't want the 2 that match your pair)
4th card: 3/49 (to match your 3rd)

[case 2]:
3rd card: 6/50 (to get a match with one of 1st two)
4th card: 3/49 (to match your 3rd)

SO:
[1] 3/51 * 48/50 * 3/49 + [2] 48/51 * 6/50 * 3/49 = 216/20825 = .010372148...

I ran a simulation (a million tries) which confirmed that.

Can you explain your C(13,2) and C(12,2)?

You are totally correct Denis. I was a tad bit hasty in my answer. It should be:
C(13,2)*C(4,2)*C(4,2)/C(52,4) where
C(13,2) is the number of possible differing pairs and C(4,2) is the number of possible pairs for the denomination.

Yepper!