Since im a dummy figuring these things out,id like you to help me out with the calculations below:

So,say 4 players(3 opponents and me) see a turn in an Omaha game.The board reads 39K9 r,i have a pair of Ks and i want to know the chances of my hand being good(not counting sets,overpairs or other Kings that have me outkicked,just the possibillity of someone having a 9).
So far i have seen 4 cards in my hand and 4 cards on the board,so there are 44 unknown cards to me.
My opponents hold 12 cards combined,so the possibillity of them holding a 9 is 44/12= a little less than 4 to 1?

Are these assumptions any good? :/

First of all, if it were 48/12, that would be 3 to 1, not 4 to 1, and 36/12 would be 2 to 1, and 44/12 would be 2.7 to 1. But this isn't how you calculate the answer. There are 2 nines, so the probability that at least one is in your opponent's 12 cards is 1 minus the probability that both nines are still in the deck which is 44-12 = 32 cards. So

1 - 32/44 * 31/43 ≈ 47.6%

That's 1.1 to 1 against your opponents having a 9. The 32/44 is the probability that the first 9 is in the deck, and 31/43 is the probability that the second nine is in the deck given that the first nine is in the deck.

If you know combinations, you can do

1 - C(42,12)/C(44,12) ≈ 47.6%

C(44,12) is the total number of ways your opponents can have 12 cards, and C(42,12) is the number of ways to have them without a 9. You can compute those in google with say "42 choose 12". That's how I first did the problem, then added the other method to please Sklansky.

lol,was pretty sure i would screw that up!thanx man!!