# An Omaha Question Revisited

Brian Alspach

Poker Digest Vol. 4, No. 20, September 21 - October 4, 2001

Previously, I've discussed the following question which was posted on RGP. 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 answer of .209 was found quickly and accurately on RGP under the assumption player B holds a random hand. We then explored whether or not the answer is different under other assumptions about player B's hand. What we found is a probability of .422 that player B has a heart flush under the assumption player B is playing a hand with two suited cards from each of two suits, and player A also has two suited cards from two different suits. The essential point is that an assuption about the types of hands B would play gives a result far removed from an assumption of two random cards. Finally, we mentioned a variation on the above question: 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? Let's discuss this variation.

We make the assumption the nine players are holding random hands. We use the method of inclusion-exclusion for this problem. Since the solution is fairly brief, this problem gives us the opportunity to see how inclusion-exclusion works. A reader who works through the details below is going to have a good idea how to handle inclusion-exclusion.

Let P(i) denote the probability that player i has a heart flush. The total number of possible hands for player i is C(45,4) = 148,995 since we are choosing four cards from 45 unseen cards. Player i has a heart flush if her hand has two, three or four hearts. There are 10 unseen hearts. This means there are C(10,4) = 210 ways she can hold four hearts, 35C(10,3) = 4,200 ways she can hold three hearts, and C(10,2)C(35,2) = 26,775 ways she can hold two hearts. Summing these numbers yields 31,185 hands giving her a heart flush. Dividing by 148,995 gives a probability of .2093 that player i has a heart flush.

For those of you who know something about inclusion-exclusion or remember something about it from earlier articles, we add and subtract various terms which alternately overestimate and underestimate the correct answer. As we proceed to deeper and deeper levels of the estimation, the approximation gets closer and closer to the correct answer. The first term in our estimation is obtained by multiplying P(i) by nine because any of nine players may have a heart flush. Upon carrying out this product we obtain 1.8837 which we denote by A. Wow! Clearly, this is miles from the correct answer because a probability cannot exceed 1. Why is A so big? The trouble is caused by overcounting many deals. Any deal which has exactly one player with a heart flush has been counted exactly once, whereas, any deal with exactly two players holding heart flushes has been counted twice in A, deals with three players holding heart flushes have been counted three times, and so on. We can see that a large error has accumulated.

We now want to compensate for the overcounting of deals with multiple flushes. Let P(i,j) be the probability that both players i and jhave heart flushes. The total number of hands they may have is the product of C(45,4) and C(41,4) which is 15,088,723,650. Note that we do not divide by two because i and j label the hands. Things begin to get messier now because there are nine combinations of choices of hearts they can have giving both of them flushes. As an example, if i has four hearts and j has three hearts, there are C(10,4) = 210 choices for i's hand and 35C(6,3) = 700 choices for j's hand. Multiplying the two gives 147,000 choices for their hands in this subcase. Counting the total number of ways both can have flushes yields 503,845,650. Dividing by the total number of hands above gives a probability of .0334 that both i and j have flushes. There are 36 combinations of pairs of the nine players. This means the second term, denoted B, in our approximation is thirty-six multiplied by the preceding probability giving us B = 1.2021. We then have A - B = .6816. At least it is less than 1 so that we know our current estimate is within the allowed range of a probability.

What does the value A - B measure? Any deal with exactly one player having a heart flush contributes once to A and contributes nothing to B. Thus, it contributes once to A - B. Any deal with exactly two players having heart flushes contributes twice to A and once to B. Since B is being subtracted from A, a deal with exactly two players having heart flushes contributes once to A - B. Thus, deals with exactly one or two players having heart flushes have been counted precisely once in A - B.

Where does the error in A - B originate? Well, if a deal has exactly three players with heart flushes, then it is counted three times in Aand three times in B. Since B is subtracted from A, such a deal has not been included at all. Deals with four or five players having heart flushes are actually counted negatively in A - B. Thus, A - Bis undercounting heart flush deals.

We next compensate for deals with exactly three players having heart flushes. Let P(i,j,k) denote the probability that players i,j,k have heart flushes. The total number of hands they can have is the product of C(45,4), C(41,4) and C(37,4). There are 23 combinations of hearts which can give all of them heart flushes. Working through all the combinations produces 3,564,739,158,750 ways all three can have heart flushes. Dividing by the total number of hands they can have yields a value of .003577 for P(i,j,k). There are C(9,3) = 84 ways of choosing three players from nine so that the next term, denoted C, is obtained by multiplying the probability P(i,j,k) by 84 to obtain C = .30048.

Let's examine A - B + C = .98208 which is the next level of approximation. Using the same analysis as before, any deals with exactly one, two or three players having flushes contribute exactly once to A - B + C. Deals with four or five players having flushes are being overcounted so that the approximation A - B + C is too big. However, this approximation is going to be close to the exact value because the probability of four players having heart flushes is small since there are only ten hearts available. The probability of five players having heart flushes is very small for the same reason. Of course, it is impossible for six players to have heart flushes. Thus, this is a problem for which we can get an exact answer easily.

We carry out the same process for four and five players having heart flushes obtaining D = .02491 and E = .0003071, respectively. The value A - B + C - D + F = .95748 is the exact probability that one or more players have a heart flush after a flop of three hearts given that one player has no hearts amongst her four cards and nine other random Omaha hands have been dealt. This may be surprising to some readers (as it was to me), but some results stray from our intuition.

The previous probability actually is useful. In the previous article we saw that an assumption about the suit structure of a single opponent's hand produces a dramatic increase in the probability of that opponent having flopped a flush. Thus, if many opponents in your Omaha game have a tendency to stay in when they have flush possibilities, and if a flop has come with three cards in the same suit of which you have none in your hand, there is a very high probability someone has flopped a flush when there are many callers because the probability already is very high for nine random hands.

Home | Publications | Preprints | MITACS | Poker Digest | Poker Computations | Feedback
website by the Centre for Systems Science
last updated 26 October 2001