in omaha, need following prob. solved, need help. pls help math gurus...?

Question

Have 52 variables. nCx=270725 pls break down into poss. combos& prove. Ex. if 4 out of 52 variables are equivalent, how many times will the same var.'s happen 4 x in a row? answer is 52x3x2x1/4! or 52x3x2x1/4x3x2x1, or 13. BUT how is calculation figured when the 1ST instance can be any 1 var., & 1 of the following 3 must be equiv.(if only 3 var.'s remain), etc.? Pls prove the following:
given x=52, a=1st var, b=2nd var, c=3rd var...therefore, xabc/4!=13, where a,b,&c are certain values not equiv. to x or a or b or c...so when x=a(3x) or b=c(3x) or x=b(3x) or x=c(3x), for ex., how does one solve that?
At any rate, pls prove so that all solutions and show work(to the extent possible on yahoo! text fields), so that, when added together, equal 270,725, that will help me out immensely. TY in advance!

Thanks, yes, what are the number of hands that would be four of a kind (13), no. of hands that would be trips (Ex. Q-Q-Q-3) no. of hands that would be 2 pair (ex. 5-5-7-7) no. of hands that would be 1 pair (ex.2-2-Q-J) and no. of hands that would be No pair, (ex.3-4-J-A). When all of these numbers are added together it should equal 270,725. I keep getting close to this number but not exactly on it, so I'm wondering where my math is flawed, basically. Thanks.

Answer 1

I'm not quite sure what you are asking.
Are you asking what the probability is of being dealt a pair in Omaha hold'em? (270725 is the total number of possible hands in that game, and 13 is the number of hands with 4 of a kind, obviously) Please describe the problem in terms of the game.

OK (in the following, C(a,b) is the combinitorial function-combiniations of b things out of a things)
3 of a kind:
There are 13 ranks. Then, there are 4 cards of each rank. Then, there are 48 cards remaining to choose from. Therefore, there are 13*4*48 = 2496 possible hands here.

1 pair, other two different: 13 ranks. Out of each rank we must choose two, so multiply by C(4,2) = 6. Then there are 48 cards to choose from, then 44 cards not identical to the first three to choose from. But we must divide by 2 here since each combination of these last two would be represented twice (i.e., Q♦,K ♣ and K ♣,Q♦ would both be represented if we didnt divide by 2)
So the total here is 13 * 6 * 48 * 44 / 2 = 82368 hands with ONLY one pair.

2 pairs: Out of 13 ranks we choose 2, or C(13,2) = 78. Then for each of these two ranks, there are 4 cards (suits) from which we choose two [C(4,2)=6 twice]. So, 78 * 6 * 6 = 2808 hands with 2 pair.

All different: 52 cards, then 48 cards, then 44 cards, then 40 cards to choose from, then we divide by 4! = 24 to account for the multiple arrangements of 4 cards.
Therefore, 52 * 48 * 44 * 40 / 24 = 183, 040 hands.

So we have 13 + 2496 + 82368 + 2808 + 183,040 which does equal 270,725.

Hope that helps.

p.s. We can see by this that the probability of getting a pair or better is 87672 / 183040 ≈ 32.4%
(13 + 2496 + 82368 + 2808 = 87672 pair or better hands. Of course in Omaha, connector hands are great as well, i.e., 9-10-J-Q)