Lows in Omaha 8

Fuel55 posted recently about studying some odds in preparing for an Omaha 8 tournament. One of the numbers posted was that the odds of a low draw appearing on the board is 62%. This was computed as follows:

The chance of an Omaha 8 board even making a low is only 62%. The math behind this is fairly simple. Only 32 of 52 cards contribute to a low hand (4 x 2, 3, 4, 5, 6, 7, and 8). The probability of any one card being low-worthy is 32/52. The second low card can't pair first (28/51) and the third low card can't pair the first or the second low card (24/50). The 3 low cards can come in positions 123 on the board or 124 or 125 or 134 or 135 or 145 or 234 or 235 or 245 or 345 (10 combinations). So the net chance of the board making a low is 32/52 * 28/51 * 24/50 * 10 = 62 %.

That's pretty close to a correct answer, but there's a slight error in the computation. The reason is that boards with more than one low combination are counted once for each low combination they contain. In other words, 62% is actually an overestimate.

It's a bit involved, and almost certainly tedious to compute the actual probability, but since I'm in the middle of the most ridiculous piece of card deadness I have ever had the misfortune to experience in my MTT's at the moment, I thought I might as well put the long pauses between folds to some use.

Breaking news No sooner do I type that then I get a free look at the flop with the hammer from the BB. I flop bottom two pair with about a million limpers in the pot. Check, and by the time it comes back to me there's a nice amount of cash in the pot. I push and it all holds. Up to over 3000 chips from 800.

Back to our story.

Let's use lower case letters to denote small cards (different letters meaning non pairs), and a capital X to mean any big card (we don't care if they pair). The following codes represent the types of board we're interested in: abcde, abcdd, abbcc, abccc, abcdX, abccX, abcXX. Now we'll count the number of boards of each type. It turns out to be a lot easier to do this if you don't worry about the order in which the cards appear (trust me, I'm a mathematician -- the calculations below may look bad but if you throw in order as well it gets worse). One piece of notation that I can't do without: is C(n,k), which indicates the number of way to choose k distinct elements from among n. For instance C(8,5) is the number of ways to choose 5 different small ranks, and of course the total number of possible boards is C(52, 5) = 2598960.

Long break here, I hit some hands and I'm too old and the wrong sex to multitask effectively. But now I'm back (99th of 672, a real kiss your sibling result). Did you miss me?

It's time to count the number of boards for each of the codes we defined above.

abcde We need to choose 5 ranks, and then choose one of four cards from each. That gives: C(8,5) × 4^5 = 57344.

abcdd Choose a rank for the pair (8), and two suits (6), choose the other three ranks C(7,3) and suits for those (4^3). That gives: 8 × 6 × C(7,3) × 4^3 = 107,520.

Those computations get old real fast, so let's just give the formulas and numbers for the rest. As we say in the math biz, leaving the details to the reader.

abbcc: C(8,2) × 6^2 × C(6,1) × 4 = 24,192
abccc: C(8,1) × 4 × C(7,2) × 4^2 = 10,752
abcdX: C(8,4) × 4^4 × 20 = 358,400
abccX: C(8,1) × 6 × C(7,2) × 4^2 = 322,560
abcXX: C(8,3) × 4^3 × C(20,2) = 680,960

All we need to do is to add these numbers up, and divide by the total number of boards. That gives:

1,561,728/2,598,960 = 0.6009

or as near 60% as who gives a.

Gosh, all that work to weed out that extra 2%. Ah well, at least 60% is easier to remember!

Warning from the mathematician general It is quite likely indeed that there is at least one minor error in the computations above.