Brian Alspach

Poker Digest Vol. 3, No. 25, December 1 - 14, 2000

In this article we are going to look at the probabilities involved in bad-beat jackpots for Omaha high. Calculating these probabilities for Omaha presents a few problems which do not arise in similar calculations for hold'em.

Let's take a look back at the situation for hold'em. In the preceding article of this series, we saw that the probability of the occurrence of a semi-deal in which aces-full-of-10s or better loses is .00004803. Let's use that probability as a benchmark.

One of the differences for Omaha is that players have four cards in their hands. It is apparent that this is going to complicate Omaha calculations just because the number of possibilities is going to increase.

On the other hand, the rules of Omaha stipulate that the player must use two cards from her hand to form a hand. Thus, the problem of discarding some straight flush hands, which occurs in hold'em because sometimes a player's best straight flush uses only one card in that game, does not arise in Omaha because a player either can make a straight flush or cannot make a straight flush using two cards from her hand. So dealing with straight flushes is a little simpler in Omaha.

We still have to decide what to do about quads with regard to qualifying. We are going to make the assumption that quads do not qualify unless a player has a pair in her hand. Speaking of quads, just as we did in the first article in this series, we can fairly easily work out the probability that two different players have quads. The reason it is not too hard to do is because the board must contain two pair and we know the completion to quads forces the other two cards to go to different hands. We find the probability is .0001439 upon doing this calculation. Comparing that to the benchmark above, we see that two players getting quads in Omaha already is considerably more likely than aces-full-of-10s or better losing in hold'em.

And this hasn't accounted for the other ways quads can lose. What this tells us is that there is not much point in lowering the minimum qualifying threshold for Omaha below some kind of quads losing. Therefore, that is what we have done for this article and this makes the computation for Omaha much easier than the work we did before for hold'em. By not having to deal with the probabilities involved with full houses losing in Omaha, a lot of work is saved.

The basic outline for these calculations (full details are on my web site) is as follows. First, we want to know the total number of semi-deals in Omaha. There are C(52,5) ways of choosing the board. From the remaining 47 cards, we choose 40 to be dealt to the ten players and this can be done in C(47,40) ways. The trickiest part is working out the number of ways we can partition the 40 cards into 10 hands of four cards each.

First choose four cards to go to one hand, then choose four card from the 36 cards left, and so on until 10 hands have been chosen. Multiply all the preceding numbers and at the end you must divide by 10! because the same partition can be chosen in that many different orders. The final number has 42 digits!

The next step is to break the problem into smaller pieces. For example, we can determine the number of semi-deals for which two different players have straight flushes. Here you have to work out the number of different boards which allow this possibility. For example, if the board has exactly three suited cards, then they must have successive ranks in order for two players to be able to have straight flushes, and the smallest rank must be between three and 10.

This already hints rather strongly that there are many cases and subcases. Nevertheless, it is not such a terribly long project and in the following table is the final information. The column headed Minimum Qualifiers tells us the minimum qualifying hand. The probability is then the probability of the occurrence of a semi-deal for which two or more players have hands at least as strong as the minimum qualifying hand. Furthermore, a hand with quads does not qualify unless there is a pair of that rank in the player's hand.

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