Still Counting

Brian Alspach

Poker Digest Vol. 1, No. 2, August 14 - August 28, 1998

I'm back ... still counting ... and will further demonstrate poker situations where mathematics plays a role. My goals are threefold: (1)to convince some people who think they are mathematically inept that probably they are not; (2) to provide easy-to-understand explanations for some of the mathematical aspects of poker; and (3) now and then to stir up a little controversy. My material will vary from elementary to sophisticated; much will be old and some will be new. It is very likely, however, that the old material will be new to some readers.

In the first article of this series, I was showing my friend Bib Ladder some elementary aspects of counting via counting three-card poker hands brought about by a question he had regarding whether or not a three-card straight should beat a three-card flush. Our conversation continues.

``Bib, I have indicated to you the principles of knowing when to add, knowing when to multiply, and avoiding duplicate counting. None involve a formula, however, when we were counting three-card flushes, we implicitly used a formula that is so important we must make it part of our counting arsenal. In counting poker hands, the situation of counting how many ways a certain number of objects can be chosen from a given set of objects occurs over and over again. Let me derive a useful formula for this which, as I say, we will use all the time.

Let's suppose we want to know how many ways we can choose r distinct objects from n objects. We just extend what you did before and try not to panic because we are working with variable values. The first object can be any of n, the next can be any of n-1, the next any of n-2, and so on until we have chosen r objects. So the last object chosen, that is, the r-th object, can be any of n-r+1 objects. Since we are choosing them successively, the connective here is ``and'' meaning that we multiply. This gives us $n(n-1)(n-2)\cdots(n-r+1)$ choices. However, a fixed selection of r objects could have been chosen in any of $r(r-1)(r-2)\cdots 2\cdot 1$ ways forcing us to divide by this number. Let me always use the notation C(n,r) for the number of ways of choosing r distinct objects from n objects. We have just seen that

\begin{displaymath}C(n,r)=\frac{n(n-1)\cdots (n-r+1)}{r(r-1)\cdots 2\cdot 1}.\end{displaymath}

That formula gives us $13\cdot 12\cdot 11/3\cdot 2\cdot 1=286$ for the number of three-card flushes of a given suit and this is the number you obtained earlier. Henceforth, I shall use C(n,r) freely when talking about counting with you.

Previously we determined that there are 48 three-card straight flushes. Now let's count the number of three-of-a-kind hands. First, the rank of the cards may be any of 13 ranks. Once the rank is determined, there are four cards of any given rank so that there are four ways of choosing three of them because any one of the cards may be excluded to form the triple. We now multiply because we first chose the rank and then the triple. The connective word is ``and''. So there are $4\cdot 13=52$ three-of-a-kind hands. Does that make sense?''

``Yeah, professor, it makes sense and is actually easy. So far they've all been easy.''

``So, Bib, not only do we see why a three-card straight flush beats three-of-a-kind in three-card monte, but we can see that it barely beats it.

Counting pairs is next. There are 13 ranks for the pair. Once the rank is chosen, you want to count the number of ways of choosing two cards from the four of that rank. We have just seen that this is C(4,2) which easily is evaluated as six. The non-pair card can be any of the remaining 48. Thus, there are $6\cdot 13\cdot 48=3,744$ three-card hands with a single pair.

To determine the number of no-pair hands, simply count all possible three-card hands and delete those that have already been counted in one of the previous categories. The number of three-card hands is $C(52,3)=52\cdot
51\cdot 50/3\cdot 2\cdot 1=22,100$. There are 5,660 hands counted above so that there are 16,440 no-pair hands in 3-card monte.

That's it, Bib. Not so bad, right?''

``Yeah, thanks professor. Maybe I'll have a little harder question next time.''

The table below summarizes the number of three-card poker hands. The ``Probability'' is the probability of having the hand dealt to you.

hand number Probability
straight flush 48 .0022
3-of-a-kind 52 .0024
straight 720 .0326
flush 1,096 .0496
pair 3,744 .1694
no pair 16,440 .7439

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