Brent@GMATPrepNow wrote:Not all cultures play cards.

Say it ain't so!

(On a more serious note, if this question were on the GMAT, the testwriters would probably just tell you the composition of a deck of cards. They don't seem to mind assuming all cultures roll six-sided dice, however!)

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.