what are the odds of flopping 2pair (only) in omaha with 4 random cards?

** WARNING: Rough approximation **

If your hand is unpaired, then basically you need to hit two of your 4 cards.

Chance of pairing a selected card is a little less than 3*3/48 = 75%, say 1/6 approx.

So odds of missing all 4 cards is about 5/6 ^ 4 ... close to 50/50. The inverse of this is also 50/50.

I look forward to reading how much I'm off by.

Well one of us (or both) is way off.

Assuming hero has no pair, the probability of exactly 2 pairs on the flop by pairing 2 of the 4 hole cards is

C(4,2)*3*3*36 / C(48,3) = 11.2%

C(4,2) is the number of ways of selecting 2 of the 4 hole card ranks
3*3 = number of combos of 1 each of the selected ranks that pair
36 is the number of non-matching cards for the 3rd flop card
C(48,3) is the number of possible flops

The chance of hitting at least one of your 4 unpaired cards is

1 - C(36,3)/C(48,3) ≈ 58.7%

but that's not what he asked.

You can also make 2-pair by pairing 3 of your hole cards since only 2 count.

[C(4,2)*3*3*36 + C(4,3)*3*3*3] / C(48,3) ≈ 11.9%

ignoring straights and flushes.

12/48 x 9/47 x 36/46 x3

Add 12/48 x 9/47 x 6/46 if you want to include 3 pair

I'm thinking of outlawing unneeded combination symbols on this website.

I would typically give that solution if I were talking to a baby. When you're an adult, you use adult techniques. Your method is needlessly redundant as it divides by 48, 47, and 46 twice. Then you have to multiply by 3 because you're using permutations. I generally add verbiage to explain that the unpaired card can appear in any of 3 positions. That makes it harder to understand for people who already know combinations, which as you say, should be any 8-year old with an IQ over 89. And if they don't know it, they need to learn because you can't always use fractions without creating a hell of a mess. Techniques on this forum should be judged by their general usefulness, not as if this is the only problem you're ever going to solve in your whole life.

I would like to outlaw using x for multiply, like we're back in first grade using flash cards. x is a variable. It's like a guy I used to work for said about Macintosh computers. When they're designed for idiots, you can't help but feel like an idiot when you sit in front of one. They use the term "folders" instead of directories, and they have little pictures of folders and filing cabinets, lol.

My technique is better when it is directed at the person who asked the question in the first place.

I always give both solutions to people I'm not sure about. They often have some some passing familiarity with combinations, and the solution with combinations is often much more elegant. In this case I was talking to statmanhal who already gave half the solution with combinations.

I always try combinations as my first approach for then I don’t have to worry about dealing with orderings. If you review previous postings of solutions to these types of problems, often wrong solutions are a result of people screwing up the ordering factors.

You could call the law, "unnecessary display of horsepower". That's something you can actually be fined for in some municipalities. It's given to people with hotrods that smoke their tires and make a lot of noise. A buddy of mine used to get tickets for that in his souped up Camaro.

I disagree that combinations are unneeded or should be outlawed. If anything, we should make a sticky about combinations, and make everyone read it before participating on this forum. If you want to go very far in probability, you need to understand combinations, so we might as well start people off right. You can give someone a fish by showing them an easy technique for a specific problem, or you can teach someone to fish by showing them methods that are much more generally useful, more concise, and in many cases, less error prone.

I'd still show the fraction method when it applies because it's an important starting point, and you can make people immediately feel that they can understand something. But it only takes you so far before you run into a problem where you have to multiply by some combination anyway, or there are just too many terms and the fractions become cumbersome. Probability is an area where it's easy to convince yourself that the right answer is wrong, and one of the most common problems is screwing up order issues with permutations. Lots of people would forget to multiply by 3. Combinations avoid that. It's not hard to understand what they represent, and you can evaluate them directly in google.

Heehaww has been doing inclusion-exclusion problems using the double factorial because he's been reading Brian Alspach. I've generally done them by writing out long strings of combinations because nobody knows what a double factorial is, most calculators won't compute them, even R doesn't do it unless you provide your own or download a special package (but Excel does), and while combinations are longer, the terms follow a simple pattern and are not complicated. But the advantage of using the double factorial is that you don't have to worry about who gets what hands which really simplifies things. You only have to worry about how the cards are paired off. With my way, I'm considering permutations of hands, even though I'm using combinations for the cards within a hand. So using combinations is to the double factorial what your fractions are to using combinations. Like combinations, the double factorial takes some getting used to, but the rewards are worth it.

While being able to see simple solutions to particular problems is important, it's also important to develop an arsenal of techniques that you can rely on. If a given problem is the only one you were ever going to do in your whole life, then a particular method or a clever solution might be less complex. But over a lifetime of problems, using a smaller number of more powerful techniques is less complex than a lot of different ad hoc techniques which can be more error prone, especially if you aren't yet aware of potential subtleties. Was this the reason your boss didn't want you teaching your method to the other actuaries?

On the other hand, you don't want to become so complacent with higher level techniques that you lose the ability to think and see simple things, like knowing that if you have AA, the probability that one of 9 opponents has AA is just 9/1225 like you once lambasted Brian Alspach for doing.

I don't really want to outlaw combinations. I mean how else would I realize that with eight girlfriends I could have 219 different, interesting, evenings?

For the less combinatorially inclined among you, DS is such a player that 2 girls at a time isn't interesting enough for him!

Yeah but as you probably know, the multi-factorials are only in the special case of when you want groups of 2 (ie preflop hands in hold'em).

If they were Omaha hands, I'd be using long strings of choose-3 combos (square pyramidal numbers I believe). That wouldn't count order of groups just like !!'s don't for hold'em hands. Others might use strings of choose-4's and but then they'd have to divide away the orders.

Or he's limiting himself to a 6-some or less while including the evening spent with himself.


"I've moved beyond threesomes, and I'm now into foursomes."

-House M.D.

Say you want to count the ways to partition 36 cards into 9 foursomes. You could do

35!!*17!!/3^9

≈ 3.88035036597428e+23

That's 35!! ways to partition them into pairs, times 17!! ways to partition the pairs into foursomes, divided by 3^9 because each foursome could have been made from 3 different initial pairings. That's the same as

C(36,4)*C(32,4)*C(28,4)*C(24,4)*C(20,4)*C(16,4)*C( 12,4)*C(8,4)*
C(4,4) / 9!

≈ 3.88035036597428e+23

Interesting, this might be the best way. Were you thinking that when you did it, or did you just use double-factorials for their own sake and it just happened to be a good method?

For that matter you can use single-factorials, though it's more stuff: 36! / (4!^9) / 9!

There's another way of writing/understanding that: (36 P 27) / (4!^9)
We start with 9 arbitrary cards being in different hands. Then we arrange the other 27 cards around them (linearly). This counts order of cards within each hand, so we divide by 4! nine times.

The choose-3 way I was talking about is the most direct (but perhaps not as preferable as yours), since it never counts unwanted orders to begin with and thus requires no denominator:

C(35,3) * C(31,3) * ... * C(3,3) = same answer

(Same reasoning as the double-factorial method for 2-card hands.)

Edit -- oh cool, that's the same as:

35! / 32!!!! / (3!⁹)

I thought just now how I would do it with the double factorial, and the first thing that occurred to me was doing the pairs, and then pairing the pairs.


That's the standard way. It's more basic than the double factorial.

For those that don't know:

(2k-1)!! = P(2k,k)/2^k

= (2k-1)(2k-3)*(2k-5)*...*1.

LOL 32!!!!. Let's use that all the time now. David will be hanging himself from the rafters.

Introducing double-permutations (DS will love these even more), a generalization of double-factorials.

n PP r = n(n-2)(n-4)...(n-2r+2)

35 PP 9 = 35*33*31*...*19

= (35 PP 9)(17!!)² / 3!⁹

Should we use double-perms or quadruple-factorials?

Of course you realize that the only combination you have to do is C(8,2) = 8*7/2 = 28. Then you can just do 256-1-8-28 = 219. There are 256 total combinations including the 0 combination because each woman is either in or out, no pun intended, so 2^8. Of course in your excited state, you may not be able to do this simple math.

So what's the easiest way to show that all the combinations of x is two to the xth?

That was it. Each thing is in the combination or not. So if you want to exclude women 1,3, and 5, that's NYNYNYYY. 2^8 possible strings for 2^8 combinations. That's why there are 2^x subsets of a set with x elements. That's true even when x is infinite.

Another way is to note we want the sum of the coefficients of (n+1)^x which is (n+1)^x with n=1.