Stud Versus Hold'em: Part I

Brian Alspach

Poker Digest Vol. 3, No. 17, August 11 - 24, 2000

I recently had a discussion with a veteran poker player concerning the similarities and differences between seven-card stud and hold'em. The discussion was wide ranging but there was one opinion he expressed which has led to this series of articles. He stated that seven-card stud is a game of two-pair while hold'em is a game of straights and flushes. We debated that statement as I argued there were many contexts in which his statement was incorrect. Nevertheless, I would like to see whether or not there is any mathematical basis for his statement.

Put on your seatbelts because the mathematics involved in this discussion is going to be more sophisticated than usual. What I propose to do is work towards the probabilities of a flush winning in seven-card stud and in hold'em. We shall then do the same for straights. This is not an easy task. One source of difficulty is the variance in the answer depending on the texture of the game. It's patently obvious that higher limit games are going to reduce the incidence of straights and flushes because most players in these games do not stick around with flush and straight draws nearly as often as lower limit players. Thus, we are going to make assumptions about the game and try to extrapolate from there.

I have had my own experiences relating to the preceding paragraph. I shall never forget my first no fold'em hold'em experience. It happened some years ago at the Tulalip Casino in Marysville, Washington. I had been playing hold'em for about two years at the $4-$8 level in Vancouver, British Columbia and had become accustomed to the texture of that game. I sat down at a $4-$8 table at Tulalip and watched in amazement as hand after hand saw either nine or 10 players (if I folded, there were only nine) calling, raising, capping the bet, and building enormous pots. The number of hands won by straights and flushes was truly astounding. Needless to say, I was hammered and left the table shell-shocked. I managed to split one pot and that was it. On the other hand, where I usually play now, the biggest game is $10-$20 and it is far removed from no fold'em hold'em. Many hands are won without a showdown, and those which do have a showdown frequently are won by a single pair or even ace high.

Another source of difficulty is the actual computation itself. This is going to be one of those cases in which we use inclusion-exclusion. This is the method we introduced and developed in the ``I'm In...No, I'm Out'' series in Poker Digest, Vol. 2, Nos. 14-19. Let's undertake a brief review of this technique.

We start with an example. Suppose you are given 500 people and are given the following information: 240 of the people have brown eyes, 210 of the people have brown hair, and 120 people have both brown eyes and brown hair. The question is how many people have neither brown eyes nor brown hair? To obtain the answer we subtract 240 from 500 to remove the people with brown eyes. We then subtract another 210 to remove the people with brown hair. This leaves us with 50. Is this the correct answer? Of course not! We see that the 120 people with both brown eyes and brown hair have been subtracted one too many times. So we add 120 to obtain 170 people with neither brown eyes nor brown hair.

The example above is typical of inclusion-exclusion arguments. We have a set of objects which is easy to count (all the people) and we have a list of attributes the objects may possess (brown hair or brown eyes in the example). For any sublist of attributes, it's relatively easy to count the number of objects having that sublist of attributes and possibly more of the attributes, but it's difficult to count the number of objects having exactly that sublist of attributes. (In the example we are given the number of people with brown eyes, but some of them also have brown hair while others don't, so we are not given the number of people having only brown eyes.) We want to know how many objects have none of the attributes. The answer is obtained by subtracting the number of objects having one of the attributes, adding the number of objects having two of the attributes, subtracting the number having three of the attributes and so on.

Let's return to the poker problem. In order to be able to derive some reliable numbers to work with, we are going to do our counting in the following contexts. For hold'em we are going to work with a 10-handed game in which all players stay until the end. For seven-card stud we are going to work with a seven-handed game in which all players stay until the end. These are the numbers we use to form the basis of our discussion.

We deal with flushes first because they are easier to count. What we want are the probabilities that someone has a flush in hold'em or seven-card stud, respectively (remember the contexts we are assuming from the previous paragraph). We shall see that the determination of this probability for hold'em is easier. In fact, we use different methods for the two games. For seven-card stud, instead of trying to calculate directly the probability of one or more players having a flush, which would involve too many cases, we determine the probability nobody has a flush and subtract that from one. Let us see that we are precisely in the land of inclusion-exclusion.

We assume there are seven hands of seven cards each dealt. The objects are then all possible deals. An object (that is, a deal) has attribute i if the i-th player is dealt a flush. We are interested in the number of objects having none of the attributes, that is, the number of deals in which no player is dealt a flush. This is the classical setting for inclusion-exclusion, and we will pursue it further in future parts of this series.

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