Also, maybe I'm missing something, but I'm getting a different answer from Rich. (If I do it his way, the actual simplification I get is 166/663, but that doesn't seem to be right either.)
Assuming this is without replacement (i.e. both cards are drawn at once):
Prob(Black,Black) = (26/52) * (25/51) = 25/102
Prob(Ace,Ace) = (4/52) * (3/51) = 1/221
But we've overcounted the pair of black aces, as they appear in each of the above. So we subtract
Prob(Black Ace,Black Ace) = (1/26) * (1/51)
This gives us 25/102 + 1/221 - 1/1326 = 55/221. (Close, but just a hair off.)
Where I differ from the original post is the notion that there are 12 ways to pull four aces or 2 ways to pull two black aces. That assumes the order matters, but I'm not sure it does here: there's only one pair of black aces (Ace of ♠, Ace of ♣), for instance. Likewise, the # of pairs of black cards is (26 choose 2), so the probability of two black cards is (26 choose 2)/(52 choose 2).
If this is done WITH replacement, then we have
Prob(Black,Black) = (26/52)(26/52) = 1/4
Prob(Ace,Ace) = (4/52)(4/52) = 1/169
Prob(Black Ace, Black Ace) = (2/52)(2/52) = 1/676
So 1/4 + 1/169 - 1/676 = 43/169.
