I've found many *many* pages on how to figure combos but after extensive web searches and poring over the omaha probabilities wikipedia page, I can't seem to get a definitive answer on:

A) If I hold any double-suited hand, what are the FLOP COMBOS that are a flush draw in one of the two suits, (four to a flush iow) excluding the chance of flopping the flush? So I'm not concerned about the combos where I get the flush, or the board is a rainbow, nor any weird flops like all 1 rank/trips on the board. If I backdoor my second suit that's just gravy, I still want to include such flops

It seems that the combos on this should be:

22C1*10C1*(46 cards left less the 9 flush outs, so 37) / 6(permus of 3 cards)

Yet this yields 1356.6 repeating. Unless my understanding of poker/probability is badly skewed, you can't have 2/3 of a combination.

B)If I hold any single-suited hand <same as above>

11*10*37/6 yields 678.3 repeating (???)

So that leaves my math at fault...what am I doing wrong? Should I be naming actual suits so that my hand is --- and the board is --//? That seems untoward and runs counter to the way the wiki builds binomial coefficients but for the life of me it's not 'clicking.'

If someone could explain where I'm going astray and how to correctly build these binomial coefficients on my own I'd be eternally grateful, I've crunched numbers for going on 12 hours now with no joy.

Also, if you could show me an example w/ a wrap/double-wrap draw and two pair drawing to a boat, and how to calculate the combos where the draw is completed on the turn and draw is completed on the river, that would be super-awesome!

Thank you, 2+2ers and I look forward to reading your replies!

PS: there IS a separate wiki page on rank draws in omaha, but again I can't suss out how to apply the principles used there to my own calculations.

Double Suited:

There are 48 unknown cards from which 3 are flopped. For a flush draw for one of the two suits, there are 11 remaining of that suit, from which the flop has to have 2 and the third card has to be from the remaining 37 cards of the other 3 other suits. Therefore, for a draw from either suit,

Pr(flush draw given double suited) =

2*C(11,2)*C(37,1)/C(48,3) = 23.53%

Single Suited

This is just a bit more complicated for the draw can come from the suited suit or either of the other two suits. The equation is

Pr(flush draw given single suited) =[C(11,2)*C(37,1) + 2* C(11,3)]/C(48,3) =13.67%

In Omaha you must use exactly 2 of your four cards. You don't have a draw if there is a monotone board and you have a card of that suit. So the probability of a FD is just half of the double-suited case.

Yup, you can tell I'm not an Omaha player. Thanks for the correction.

Thank you for the responses guys, I appreciate that.

I'm still a bit confused though.

OK so the "2*" represents that I have two suits from which to draw in my hand, i.e. double-suited? And then the "C(11,2)" represents 11 cards (of either suit) and getting 2 of those on the flop...the "*C(37,1)" I'm okay with...but what about the divisor "C(48,3)"? Shouldn't I divide by the number of permutations for any 3 cards? i.e. 6 permutations? In other words, the dividend is my number of combinations INCLUDING permutations of the same cards, so given Q-T-5 flop, I can achieve that in six different orders, so the divisor should be six, right?

Not that I'm questioning your calculations, I trust yours over mine any day. I'm just trying to understand why it is this way so I can extrapolate and start building my own equations for other situations. Teach a man to fish, and all that.

Also, if the divisor is C(48,3) on the flop, does that mean it's C(48,4) and C(48,5) on the turn and river, respectively? Thanks again, 2+2ers, looking forward to hearing back from you!

His dividend did not include permutations, only combinations (combinations don't count order, only the possible selections). But you're correct that if order were counted in the dividend, it would also have to be counted in the divisor. In this problem, order doesn't matter. That can be expressed either by counting every possible order (permutations ÷ permutations), or by counting no order (combos ÷ combos).

Combos are just permutations with the order divided out.
C(n,r) = (# permutations) / r!

So his C(11,2) already has a 2! divisor. And then multiplying by a combo from a different set altogether (the set of 37 from a different suit) doesn't count order.

The permutation way would be: (11 P 2)*37 *3 / (48 P 3) = 11*10*37*3 / (48*47*46)
The *3 is to count the 3 orders not yet counted. The 11P2 counted order of the flush cards relative to each other (a 2! is accounted for). But then like I said, the multiplication by 37 does not yet count order of where the non-flush card can go relative to the flush cards, so the *3 achieves that. Altogether, 2!*3 = 6 = 3! so the numerator has counted all orders.
Yep, C(48,4) is the divisor for flop+turn (before seeing the flop), and C(48,5) is the divisor for the whole board.

Ok, awesome...I think it's making more sense now. Based on my current understanding the formula for when holding a double-suited hand of ending up w/ a four flush by the river (flush not completed) is:

2*C(11,2)*C(37,3) / C(48,5) = 2*55*7770 / 1712304 = 49.9% of ending w/ a four-flush in one of the suits...now suppose I wanted to take this a step further and determine:

A) all the boards where I went to the river and never made even a four-flush, flopped one of my suit or less
B) got my four-flush in one suit and completed a flush in the second hand-suit, i.e. my hand: --- board on 5th St.: ----
C) 1 - (.499 - the B) edge cases + the A) cases) = my percentage of getting a flush or straight flush, right?

Thanks again for the help, it's starting to come together for me, finally.

A)
i) Flop none of suit = C(26,5)
(not selecting from 37 because can't have either suit)
ii) Flop one of suit = 22*C(26,4)

Total = 394680

You might be wondering why it can't be done in one step: C(26,4) * 44
Doing that is equivalent to: 2*11*C(26,4) + C(26,4)*C(22,1)
That second term is supposed to represent getting 5 blanks, but instead of C(26,5), it draws from the same set twice separately. Drawing from the same set multiple times counts order, in this case 5 orders (the separately-chosen 5th card can be in any of 5 places). Therefore, C(26,4)*44 counts the case of C(26,5) five times when it should only be counted once. So if we start with C(26,4)*44, we must subtract 4*C(26,5) to arrive at the correct answer of 394680. That's an example of inclusion-exclusion.

B) 2*C(11,2)*C(11,3)

C) I think not. But it looks like I've interpreted (A) differently than you, especially considering your .499 number which does count the (B) cases like you said. For (A) you seem to be allowing for cases like 1 of a suit and 4 of the other.

I can recalculate (A) if you like, but I can answer (C) without math.

.499 = all the ways for 2 of one suit and 3 or fewer of the other, but plus some extra: it double-counts the ways you can have 2 four-flushes. (See why?)
.499-B = ssxxx where xxx is not ccc. Still with the extra ssccx.
.499-B+(your A) = cssxx + ssxxx + ssccx
1-that = ccccc + ddddd + hhhhh + ...(lots of stuff, not all flushes)

So the answer is no. Even without your extra ssccx, the final result would still have non-flush cases. You forgot that there are 2 other suits that can ruin the flush.

Edit: of course, the easier way to check is to just directly calculate the ways to make a flush.
2*[C(11,3)*C(37,2) + C(11,4)*36 + C(11,5)] / C(48,5)

Okay, I get all that, but I want to make sure I'm doing the rest of the calculations right. Thank you for your continued assistance with this, heehaww.

So if I might back away from the turn/river calculations for a minute, and explore the flop some more...I've made a little table, could you please point out any mistakes I may have made? For these calculations I've eschewed the double-suited hand as I see it as kind of a 'special case,' and with your insight I'm pretty sure I can make determinations on that type of hand accurately. Here is the table, using clubs as the only *important* suit and x, y, and z as the unimportant ones:

BOARD:
CCCC CCCx CCxy Cxyz/CDHS
==================================

CCC
C(9,3) C(10,3) C(11,3) C(12,3)
84 120 165 220

CCx
C(9,2)*39 C(10,2)*38 C(11,2)*37 C(12,2)*36
1404 1710 2035 2376

^^^Pretty confident about these calcs... vvv these, not so much.

Cxy
9*39*26 10*[2(12*13)+13^2] 11*[12^2+2(12*13)] 12^3*4(suits)
9126 4810 5016 6912

xyz
13*13*13 12*13*13 12*12*13 12^3*4
2197 2028 1872 6912

xxy
3*C(13,2)*26 [C(12,2)+2*C(13,2)]*25 [C(13,2)+2*C(12,2)]*24
4*C(12,2)*38
6084/5550/5040/10032

xxx
3*C(13,3) C(12,3)+2*C(13,3) 2*C(12,3)+C(13,3) 4*C(12,3)
858 792 726 880

I realize that's a lot of work and the table didn't quite come out 'right,' but if you could offer your insight one more time I'd be greatly indebted to you.

PS only after looking it over have I realized I forgot the Cxx cases...I'll crunch those numbers tomorrow, it's getting awfully late.

They're correct.

Divide by 2 because 39*26 counts order of whether x or y comes first. Also could have said 9*C(3,2)*13^2
Fyp. Times C(3,2) not C(4,3) because the club is forced. The other 2 suits out of 3 can vary.
Just 12^3. We need 3 non-club suits out of 3, so C(3,3)=1.
2*13*C(12,2) + 2*C(13,2)*12 + 2!*C(13,2)*13 = 5616
2!*C(12,2)*12 + 2*C(12,2)*13 + 2*12*C(13,2) = 5172
(3P2)*12*C(12,2) = 4752
Fyp

Anything I didn't quote, you got right. You're getting the hang of it. Made the same mistake a few times of multiplying by / choosing from 4 when you didn't need to. For Cxy you got the harder ones right. For xxy you forgot some cases (which is why you had one addition instead of 2), and I'm not sure where your factors of 25 and 24 came from. I get that 26=2*13 but why were you removing cards?

Now the question is whether I slipped up anywhere :P

Safe money says you didn't slip up, as after adding the combos in my columns and including a row for 'xxy' flops, each column came to 17296!

Not sure where I was coming up with 25/24, it was late and I just wasn't reading my table very well.

Thanks again to all of you for your insight, and hopefully now that I've got a decent grasp of this I can pay it forward to some other new 2+2er in the near future.

Special thanks to you, Heehaww, for being so patient while I bumbled my way through the math.

Ok I have to renege on that, but this is just a last-minute thing...

When I extrapolate to the turn, I am coming up w/ this formula for figuring out the Cxxy turns on a CCCx hand: (for this example assuming the 'x' in my hand is a diamond)

10{[C(12,2)*13]+[C(12,2)*13]+[C(13,2)*12]+[C(13,2)*13]+[C(13,2)*12]+[C(13,2)*13]}

metaphorically: 10 remaining clubs * {ddh+dds+hhd+hhs+ssd+ssh cases}

and which simplifies down to 10{2[(13*C(12,2))+(25*C(13,2))]}.

however, this seems inefficient...is there some sort of 'pre-processing' I can do so that I don't have 6 different terms/cases to add? Or is that it, and that's as good as it gets? Thanks again 2+2ers!

I enjoy this stuff, and also I'm kinda paying forward myself.

I'd group together the dds & ddh, the dss & dhh, the ssd & dds. You can tell beforehand that each of those couples has the same number of combos, so it's 3 terms each with a factor of 2. I'd pull the 2 out and have 3 terms to add.

10*2*[13*C(13,2) + C(12,2)*13 + 12*C(13,2)]

And maybe I'd notice beforehand that two of the terms will have a C(13,2) which I'd combine to 25 as you did in your simplified form. So the preprocessing takes you directly to your simplified form. Beyond that, I don't think it gets much better.

Bummer, I thought maybe there was some Unified Theory of Omaha or something that could help with all the minutiae.

Nonetheless, it's great that you all are willing to help me out w/ these calculations and I'm indebted to you folks. Sometime in the near future I'll have all these numbers/combos/odds put together in a spreadsheet and I'll be sure to post it here so I can spread the wealth as it were.

Thanks again for all the help!