Probability question - Omaha hi/lo

Can anyone tell me how to calculate the probability of open cards drawing three lows (distinct) in Omaha hi/lo.

If the only requirement is that the lows be different ( q = may two of the three match?) and all players stay to the end, I’d guess fairly high, like 45%+. And it would only be that low because so often there IS no possible low.

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To put it other way, how many times a table draws low in Omaha hi/lo. All rules of low apply: three distinct lows, other two can be anything. Probability is about 60%. But how to calculate it.

That sounds about right, but I’m judging by past observation. No idea how it would be calculated. You’d need someone smart for that, LOL.

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As a Omaha training guy, the probility 50% either way. Getting the Hi as same as getting the low. It is less when you consider getting the 5,4,3,2,A

The way to calculate it is to find the possibility of getting at least 3 cards from A to 8 in a 5 cards draw out of 52 cards…
So if you say:
n = 3 is the number of cards you need to be below 9
X = 32 is the total number of cards that match this condition (A to 8) 4*8 for all suits.
Y = 52 is the total number of cards
Z = 5 is the total number of cards drawn

The formula would be:


In case you need to know how C is calculated:
C(m,r) = r!/m!(r-m)!
And ! is the factorial, for example 4! = 4x3x2x1
8!=8x7x6x5x4x3x2x1 etc…

This will give you the possibility of getting exactly 3 out of the 5 cards below 9. You can repeat the same with n=4 and n=5 to get the possibility of getting 4 low cards or all 5 cards low.

I didn’t do the math because it’s too long but I got the formula and the information from this website (watch the videos), and I think it should match the 60% or any other possibilities you might need to calculate.

I hope this is what you’re looking for :slight_smile:



Thanks for your excellent lead and taking time to search for a solution.

This is the way to do, but needs two corrections need to be made.


The formula for three cards draw is

32c3 x 20c2 / 52c5

First term is probability of drawing low card and the second term is probability of drawing non low card.

The term 32c3 is expanded as

(32x31X30 )/ (1x2x3)

To say first card is any of the 32 low cards, second is any of 31, and third is any of 30. (Divided by 1x2x3 is to eliminate positional differences, like A,2,3 is same as A,3,2). This is not exactly true, first card is any of 32, but the second is any of other lows excluding the three cards having the same number as the low drawn in the first position (in other words no duplicate), which will be 28. Likewise the third will be 24. Modified formula will be

(32x28x24) / (1x2x3)


Second term 20c2 which is expanded as (20x19)/(1x2) says any of 20 non low cards and then any of 19 non low cards. This is also need to be modified as this depends on at which position non low is drawn. First position its 20 non low cards, but in the second position, after a low is drawn, it is 20 non low cards plus other three cards of the same number drawn in the first position, which will be 23 and third drawn after two lows will be 26 so on. This can’t be generalized in one formula. Has to be computed case by case as to where the non low is drawn.

There are ten combinations for three card draw 5c3 = (5x4)/(1x2) = 10. Five combinations for four card draw (5c4) and one combination for five card draw.

Computing the probabilities of these sixteen combinations:

Three cards:

 (1)  L L L N N   32/52 x 28/51 x 24/50 x 29/49 x 28/49 =  0.056  
 (2)  L L N L N   32      28      26      24      28       0.050       
 (3)  L L N N L   32      28      26      25      24       0.045       
 (4)  L N L L N   32      23      28      24      28       0.044       
 (5)  L N L N L   32      23      28      25      24       0.040       
 (6)  L N N L L   32      23      22      28      24       0.035       
 (7)  N L L L N   20      32      28      24      28       0.039       
 (8)  N L L N L   20      32      28      25      24       0.034       
 (9)  N L N L L   20      32      22      28      24       0.030      
(10)  N N L L L   20      19      32      28      24       0.026       

Four cards:
(11)  L L L L N   32      28     24       20      32       0.044       
(12)  L L L N L   32      28     24       29      20       0.040       
(13)  L L N L L   32      28     26       24      20       0.036       
(14)  L N L L L   32      23     28       24      20       0.032       
(15)  N L L L L   20      32     28       24      20       0.028       

Five cards:   

(16)  L L L L L   32      28     24       20      16       0.022       

Total probability                                          0.601    

The probability is 60.09% if computed to four decimal places, which I think is exact.


You’re absolutely right. I missed the part where the lo cards have to be different. Well done!
So yes, now you have your answer :slight_smile:

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Yes, and thanks, unless someone comments otherwise, or may be give a easier way to do.

Like, four cards draw can be generalized instead of going through each combination. Low combinations will be(32x28x24x20)/(1x2x3x4). Non low will be one card whose mean out is 26. So that,

32c4 x 20c1 / 52c5 can be modfied to

(32x28x24x20) / (1x2x3x4) x 26c1 / 52c5 which is 0.179.

An easier way would be to calculate the probability of NOT getting 3 or more lo cards, and then subtracting the answer from 1. You would then have to calculate the probability of getting 0,1, or 2 lo cards out of 5, which is done the same way but is easier to calculate (for 0 lo cards and 1 lo card it’s much easier to calculate because you don’t have to go through the long exceptions).

Out of curiosity though, why do you need to do all these long calculations? Since the answers are already available.

That doesn’t quite work, because the three low cards must be distinct. A board reading AAAA2 won’t have any high cards, but there still won’t be any available low draw.


I have no idea why I keep ignoring the fact that the lo cards have to be distinct lol. Thanks Sam :blush:

I tried to find the calculation in the net and I couldn’t find one. Then I tried to solve it myself, but got stuck. That’s when I posted the message if anyone knows or leads to the derivation.

When I tried to solve myself I was getting more than 80%. Your indicated method led to some 75%. When I reconcile the two methods I found the mistake I was making. Then I was able to solve. I posted here as the response to my earlier message and as well as I thought its worth sharing.

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And job well done on figuring it out :+1:
Thanks for sharing :slight_smile:

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