Simple probability / combinatorics question that I was having trouble getting exactly right-- most resources I have on the subject are for Hold 'Em (where you're only dealing with two cards) or for hand combinations rather than specific cards.

What are the odds of drawing two specific cards in a hand of five? (As in, say, in a 52-card deck, the odds that in five-card draw your first hand contains both the Ace of Spades and the King of Spades.)

Answers that are generally applicable to larger or smaller decks of cards would be really helpful, thanks.

edit: The formula I had in mind was something like ((2/52)*(1/51 + 1/50 + 1/49 + 1/48)), but I don't know if that's right.

Say that N is the number of cards in the deck. N = 52 for the standard deck. You draw a hand made of K<=N cards (K = 5 in this case obviously). So the odds that your hand contains two specific cards are (C(a,b) is the number of combinations):

C(N-2, K-2) / C(N, K)

which is ~ 0.007541478 in your case.

Thanks. I need to apply this to a game with a bigger deck, so refresh my memory:

C(N,K) is calculated as N! / (K!(N-K)!), is that right? So the above is

(50! / (3!*47!)) / (52! / (5!*47!))

?

I did it this way with N = 52 and K = 5 and got the same answer as you, so I think I did it right.

Yes, you're right, although most calculators have a built in function for C(N,K). In google search you can do "N choose K."