Brian Alspach

Poker Digest Vol. 3, No. 26, December 15 - 28, 2000

Let's look at the probabilities involved in bad beat jackpots for seven-card stud. Calculating these probabilities for seven-card stud presents problems not encountered with similar calculations for hold'em or Omaha. The main source of these new problems is the fact that there no longer is a board and players do not share any cards in forming their hands. In some ways, this is a help.

For example, if one player has a full house with aces-full, then no one else can have a full house which beats it. The only way the player can lose is if another player has four-of-a-kind or a straight flush. This is, in fact, a move in the direction of simplifying the computations.

For both hold'em and Omaha, we calculated probabilities of certain kinds of semi-deals occurring. We shall discuss the relationship between the semi-deals and real games in some detail in the concuding article for this series in the next issue. The semi-deals we must use for seven-card stud are constrained in a different way. This problem has occurred before. In order to calculate accurate numbers, we must work with seven-card hands. However, since a deck of cards contains only 52 cards, we are constrained to work with seven hands.

The game normally is dealt to eight players.

Thus, what we are calling a semi-deal for seven-card stud is a choice of 49 cards from the 52 partitioned into seven hands of seven cards each. To count the total number of semi-deals, we first count that there are C(52,49) = 22,100 possible choices for 49 cards from 52. We then multiply by how many ways we can partition any choice of forty-nine cards into seven hands of seven cards each. The latter number is obtained by dividing 49! by 7! to the eighth power and it is huge. The product is even bigger and there is little point served in writing it out here. Nevertheless, it is the number we use to determine all the probabilities for seven-card stud probabilities.

Dealing with quads (four-of-a-kind) in seven-card stud is straightforward because all four cards of the given rank must be in the player's hand, and the player can't possibly have a better hand. Let's see how to determine the probability that two or more players have quads in seven-card stud. There are C(13,2) = 78 ways of choosing two ranks from thirteen. Since eight cards are now determined, there are C(44,3) = 13,244 ways to complete one of the hands and C(41,3) = 10,660ways to complete the other set of quads to a hand. There are C(38,35) = 8,436 choices for the 35 cards comprising the remaining five hands.

You then count the number of ways of partitioning 35 cards into five hands of seven cards each and multiply all of this together to obtain the number of semi-deals for which two or more players have quads.

At this stage we have not counted the number of semi-deals exactly because those semi-deals, rare as they may be, having three or more players with quads have been counted too many times. The solution to this is to count the number of semi-deals with three or more players, four or more players, and so on, having quads. You then apply inclusion-exclusion to your numbers to come up with the exact value. I have done so and the exact probability of the occurrence of a semi-deal in which two or more players have four-of-a-kind is .0000754661 which is about one-half the probability of two players having quads in Omaha. Interesting, eh!

The preceding discussion for quads provides a hint as to where difficulties arise in doing these calculations for seven-card stud. Even though we were able to obtain an exact probability for two or more hands with quads, we had to employ inclusion-exclusion. The reason we could get the exact value is that the number of individual calculations having to be performed is fairly small. One of the dangers present when inclusion-exclusion must be used is that the number of individual calculations may grow too large making exact calculations tedious beyond bound. However, even in the latter bad situation, normally one can get a very accurate result -- say to six or seven significant digits -- by working with the first few values. That is precisely what happens when we look at straight flushes.

In the case of two or more players getting straight flushes, we employ inclusion-exclusion, but it gets messy. The reason it gets messy is that a player may have a straight flush of length five, six or seven and we have to handle this in some way. The way I chose to handle it was to flag a straight flush by the largest card appearing in the straight flush. If it happens to be something like a six, then when we are completing the hand to seven cards, we cannot include the seven of the same suit in the hand. On the other hand, if it is an ace or if there is a second straight flush in the same suit which uses the card of next highest rank, then we have no restriction on how we complete the hand. This means subcases begin to proliferate and we obtain only an approximation.

The probability of the occurrence of a semi-deal in which two or more players have straight flushes is .0000024144 with accuracy to five significant figures. At the time this article was due, I did not have time to work out any further numbers for seven-card stud. In the last article of this series, I intend to fill in considerably more detail for seven-card stud, provide several numbers missing for hold'em, and discuss what these numbers mean in relation to real games.

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