An Omaha Question

Brian Alspach

Poker Digest Vol. 4, No. 19, September 7 - 20, 2001

A recent message on the newsgroup,, included the following question about Omaha. Players A and B are heads up before the flop. The flop comes with three hearts and player A has no hearts. Player A wants to know the probability that player B has a heart flush.

The question was answered quickly and accurately on RGP under the assumption player B holds a random hand. The answer is determined in the following manner. There are 45 unseen cards from player A's viewpoint. The total number of possible hands for player B is C(45,4) = 148,995. Player B has a heart flush when holding either two, three or four hearts. The 45 unseen cards are composed of 10 hearts and 35 non-hearts. There are C(10,4) = 210 ways B can hold four hearts, there are C(10,3)C(35,1) = 4,200 ways B can hold three hearts, and there are C(10,2)C(35,2) = 26,775 ways B can hold two hearts. Summing gives us 31,185 hands giving player B a heart flush. Dividing by 148,995 yields a probability of .209 or roughly 4-to-1 odds against player B having a flush.

The above answer led to several comments about the answer possibly being meaningless because one is not really dealing with a random hand in player B's possession. This, of course, is a valid point. In most cases, heads-up play before the flop indicates at least one raise and only two players deciding to stick around to see the flop. It does not make sense to treat the hands as random. In spite of this, I believe we still can use some mathematical analysis on the hand.

If we know something about player B's tendencies; for example, suppose we know there is a reasonable chance B is double-suited with two cards in each suit, then what is the probability B has a heart flush? The first rough approximation we might consider is to say that any two suits are equally likely. There are six combinations of two suits out of four, and hearts are in three of the combinations. This gives a probability of .5 that B has a heart flush given our assumption B has two suits of two cards each. Of course, there is a problem with what we just did and all readers are ready to jump on it. I know, I know, we have ignored the fact there are three hearts on board.

The reason we made the preceding rough approximation is to indicate that by making a reasonable assumption about the shape of a hand B would have been willing to take to battle heads up, we get a number considerably different than the random assumption produces. Thus, we see the value of A being able to make some kind of projection about the shape of the hand held by B. Now, under the same assumption about B's hand, let's see what happens when we take into account the cards A can see. For now let's assume A also has two cards from each of two different suits. This means player A knows the suit distribution of the 45 unseen cards is 13-11-11-10 and is assuming B has a suit distribution of 2-2. So the total number of possible hands for B is not C(45,3); instead, it is the number of four-card hands with suit distribution 2-2, which can be formed from 45 cards with suit distribution 13-11-11-10. How do we calculate this?

If player B has two cards from the 13 card suit and two cards from hearts, then this can be done in C(13,2)C(10,2) = 3,510 ways. Do the same for each possible combination of two suits and sum the six products. Altogether we obtain 20,065 possible hands for B of which 8,460 produce a heart flush. This gives a probability of .422 that player B has a flush.

We have seen that a reasonable assumption on the kinds of hands B would be playing gives a substantially different probability of B holding a flush. The astute reader now is going to complain that the assumption we made about B's hands dealt only with suit distribution and not at all with rank distribution. This is a very important point as well and we should now stand back and see what is going on here.

If player B is making decisions on which hands to play according to the ranks of the cards as well as the suits of the cards, the important feature to remember about the interplay between ranks and suits is that there are the same numbers of various ranks in each suit. In other words, if player B is reducing hands to be played because of rank criteria, the reduction essentially will be proportionally balanced over all suits. This implies that rank considerations will not have much effect on the .422 just derived. It certainly is the case that the ranks of the hearts on board, as well as the ranks of the four cards in player A's hand, have some effect on the probability, but the vast majority of the doubling of the probability when moving from an assumption of randomness for B's hand to an assumption about suit distribution arises because of the 2-2 assumption.

Thus, it is certainly reasonable for player A to assign a probability of .4 that player B has flopped a heart flush. Player A can perform the same calculation (not at the table!) as her own suit distribution varies. The probability will vary somewhat, but it will stay considerably larger than .21.

Another interesting variation on the above question is to ask what is the probability someone has flopped a heart flush given that all ten players have seen the flop and player A again has no hearts? This question is the subject of my column in the next issue of Poker Digest.

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