This is known as a hypergeometric distribution.
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Would this just be a 4/52 x 3/52 x 2/51 for the first?
It would be approximately 4 times that since you basically get 4C3 attempts. This estimate is only too high by 4/(52C4).
The chance of exactly 3 deuces (and not all 4) is: 4*48 / 52C4 ≈ 1 in 1410
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And in 5 card omaha what would be the odds of getting three 2's with two of them being spades and clubs?
There are two choices of suit for the third deuce, and 48C2 choices of non-deuces, so it's 2(48C2)/(52C5) ≈ 1 in 1152
