Enumerating Omaha Hands: Part I

Brian Alspach

Poker Digest, Vol. 5, No. 14, June 28 - July 11, 2002

I am going to emphasize Omaha for a while because there appears to be less information available about this game. For example, I could not find any information on the basic problem of determining the numbers of hands of various types. The information for Omaha may be available, but my search came up empty. So this is the first problem I am going to consider.

Most people who have played much poker have seen a list of the possible 2,598,960 five-card poker hands. The list has appeared in books (usually correctly, but not always!) and is posted at several web sites. Similarly, the list of the possible 133,784,560 seven-card hands is readily available, although you will have to work a little harder to find the list.

First, we have to decide what we mean by ``all different possible Omaha hands.'' We use the classification of seven-card stud hands as an analogy. What we do for seven-card stud is take all possible subsets of seven cards -- there are $C(52,7) = 133,784,560$ -- and then use the best possible subset of five cards from the seven as the hand. So in Omaha, this means we should take all possible configurations of a five-card board and a player hand of four cards, from which we choose the best possible five-card hand under the restriction that we must use precisely three cards from the board and two cards from the player's four cards.

This means the number of possible configurations is the product of the number of five-card boards and the number of ways of choosing four cards for the player from 47. This product is $C(52,5)C(47,4) =
463,563,500,400$, which is a big number.

There is a second way to view the possible number of distinct hands. First, choose nine cards from 52, and then choose five of the nine cards for the board, with the remaining four cards going to the player. If you multiply $C(52,9)$ and $C(9,5)$, you will find that you get the same number: 463,563,500,400.

The reason I mention two different ways to obtain the number of possible Omaha hands is that they suggest two different strategies for enumerating all the Omaha hands. The first way of computing the number suggests the strategy of looking at each of the possible 2,598,960 boards, and then running over all possible four-card combinations to go with particular boards. You could do it in the reverse order and look at all possible four-card player combinations, and then run through all possible boards for each combination.

The second way of calculating the number suggests a different strategy: Examine each of the possible subsets of nine cards chosen from 52, and look at the 126 ways of breaking a set of nine cards into a subset of four cards for the player and a subset of five cards to form the board.

I am going to employ the latter strategy for the simple reason that there is an intermediate step in the enumeration at which we can check the calculations for correctness. It always is nice in a messy calulation if there is a point partway through the calculation where correctness may be verified.

I now want to whet your appetite for the calculation upon which we are embarking by looking at the special case of straight flushes. Not only does this give a hint of what needs to be done for enumerating Omaha high hands, it also answers a particular question sent to Poker Digest by Dick Williams.

In order for a player to have a straight flush, the set of nine cards must contain five, six, seven, eight or nine suited cards of consecutive ranks. There are 24 9-sets with nine suited cards of consecutive ranks since the smallest card may be any ace through six. There are 28 ways of choosing eight suited cards of consecutive ranks. We complete each of them to a 9-set by choosing one more card. For the eight beginning with an ace or a seven, we may choose any remaining card except the single card making nine suited cards of consecutive ranks (we've already counted them). Thus, we may choose any single card from 43. For the other 20 choices of eight suited cards of consecutive ranks, we may choose any single card from 42. Altogether, we obtain $(8\cdot 43)+(20\cdot 42)=
1,184$ 9-sets with exactly eight suited cards of consecutive ranks.

In a similar way, there are 29,240 9-sets with exactly seven suited cards of consecutive ranks, 484,352 9-sets with exactly six suited cards of consecutive ranks, and 6,073,320 9-sets with exactly five suited cards of consecutive ranks.

For each of the 9-sets just described, we now have to determine the number of the 126 partitions giving the player a straight flush. Sometimes the answer is less interesting than the logic used in obtaining it. I now am going to show you a nice method. First, let's see that for a 9-set composed of nine suited cards of consecutive ranks, no matter how you partition the 9-set into a player hand and a board, the player has a straight flush.

For purposes of illustration, let's assume the 9-set is the 3-4-5-6-7-8-9-10-J of spades. Now choose any subset of four of the cards to go into the player's hand. Pretend you have a sliding window that lets you look at five consecutive cards and see which of them are in the player's hand. In the left-most position, the window let's you look at the 3-4-5-6-7 of spades. If the player has exactly two of those cards in her hand, she has a straight flush.

Now suppose you slide the window two positions to the right so that you are viewing 5-6-7-8-9 of spades. Again, if she has exactly two of those cards in her hand, she has a straight flush. Thus, in order to prove that she has a straight flush no matter which of the four cards are in her hand, we need to prove that there is a position for which the sliding window views exactly two cards in the player's hand. How do we do this?

The key to completing the proof is to observe how the view from the window can change as you slide it one position left or right. It is easy to see that the number of cards in the player's hand you view upon sliding the window one position diminishes by one, stays the same, or increases by one. That is, it can never change by more than one.

Now start the window all the way to the left viewing 3-4-5-6-7. If you see exactly two cards in the player's hand you are done because the player has a straight flush. Suppose you see only one card in the player's hand from this view. This means the right-most view of 7-8-9-10-J has either three or four cards in the player's hand. So in sliding the window from the left-most position to the right-most position, you go from seeing one card to seeing three or four. However, we know that as we slide the window one position at a time from left to right, it can change by at most one for each position. Since we have gone from seeing one to seeing three or four, we know there must be an intermediate position for which exactly two cards are visible. That is, she has a straight flush.

Nice argument, eh? Thus, for nine consecutive suited cards, 126 partitions give the player a straight flush. Similar arguments give 120 partitions for eight consecutive suited cards, 108 partitions for seven consecutive suited cards, 90 partitions for six consecutive suited cards, and 60 partitions for five consecutive suited cards.

We then multiply the various numbers of partitions by the numbers of 9-sets and sum to obtain 411,293,904 straight flush Omaha hands. Dividing by the total number of Omaha hands given above, we obtain a probability of .0008872 that a player ends up with a straight flush. This is approximately once every 1,125 hands.

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