# Stud Versus Hold'em: Part IV

Brian Alspach

Poker Digest Vol. 3, No. 20, September 22 - October 5, 2000

We are in the midst of comparing the probabilities of at least one player having a flush or straight in seven-card stud and hold'em. In Part I of this series, we laid the groundwork; in Part II, we determined the probability of one or more players having a flush when two cards are dealt to 10 players and five community cards are dealt; and in Part III we approximated the probability of one or more players having a flush when seven cards are dealt to each of seven players. In this article we examine the probability for one or more players having a straight when ten players each receive two cards and five community cards are dealt.

Determining the exact probability for a flush in Part II of Stud Versus Hold'em was a fairly straightforward affair. However, determining the exact probability for a straight is very nasty. So let's look at a few of the issues which contribute to this nastiness.

The first piece of information we require is the total number of deals possible. There are C(52,5) possible boards. For a fixed board, there are C(47,20) sets of twenty cards to be dealt to the ten players. Each set of twenty cards can be dealt in 20!/210 different ways to the ten players. The product of these three numbers is the total number of deals. Using some symbolic mathematical package (I use MAPLE) the product is easy to obtain.

What we want to do is to determine how many of the total number of deals result in at least one player having a straight. The approach we take is to examine the ranks of the cards in the board. If there are only two ranks in the board, that is, the board either is a full house or contains four-of-a-kind, then no one can make a straight (not that it matters much in this case). If there are three ranks on board, then the rank pattern is either x,x,y,y,z or x,x,x,y,z. In both cases there are 192 rank patterns which allow straights. We must perform a similar analysis for rank patterns x,x,y,z,w and x,y,z,w,u. The first issue we must deal with is accurately counting the rank patterns which allow straights.

Another complicating factor arises from the fact there are different ways to complete the same rank pattern to a straight. For example, the rank pattern completes to a straight if some player holds either A-5 or 5-7, while completes to a straight if some player holds A-2 or 2-6 or 6-8. This is an indication that there might be many subcases to be considered. In fact, there are a horrendous number of subcases and that opens the doors for errors.

A third complicating factor arises out of the fact an ace can be used in small straights and big straights. This introduces the possibility of double counting straights involving aces. We have to be careful with our bookkeeping.

Once we are examining a rank pattern which allows completion to a straight, we have to determine how many deals for that fixed board actually produce straights. Some of the rank patterns require a non-trivial application of inclusion-exclusion. In all the applications we have presented in previous articles, the coefficients for the terms in the alternating sum follow a fixed pattern. For the problem under discussion, the coefficients must be played with.

After working through the rather horrid details, we have found that the probability of at least one player having a straight is about .27. However, it is important to point out that we have not excluded deals which also include flushes. And that, dear readers, is the final complicating factor. Since we are interested in deals which lead to somebody having a straight or a flush, we want to eliminate the flushes from the straights.

In the next, and last, article of this series, we shall discuss the elimination of flushes, the corresponding probability for seven-card stud, and what it all means for real games.

Home | Publications | Preprints | MITACS | Poker Digest | Poker Computations | Feedback
website by the Centre for Systems Science
last updated 6 September 2001