Bad-Beat Jackpots: An Example

Brian Alspach

Poker Digest Vol. 4, No. 2, January 12 - 25, 2001

I've had several inquiries about how inclusion-exclusion and approximations show up in the bad-beat computations, and what they mean. I want to answer the questions by looking at a simple example which should provide considerable clarification.

Let's consider the following question. Suppose we deal three hands of two cards each and ask for the probability that at least one player receives a pair. This is a straightforward question, but providing an answer already illustrates the principles involved in obtaining the probabilities we did in the previous five articles.

We don't care which of the three players receive a pair; we are interested in semi-deals rather than deals.

As a first step, let's determine the total number of semi-deals. There are C(52,6) = 20,358,520 ways of choosing six cards from 52 to be dealt to the three players. Any set of six cards can be partitioned into three hands of two cards each in $5!! = 5\cdot 3 = 15$ ways. To get the total number of semi-deals, we take the product of the two preceding numbers and obtain 305,377,800 semi-deals. In the rest of this article, we denote the number of semi-deals by A(0).

As a next step, we count occurrences of pairs in semi-deals. There are 13 choices for the rank of the pair, six choices for the pair of that rank, C(50,4) = 230,300 ways to choose four cards from the remaining fifty, and three ways to partition these four cards into two hands of two cards each. Taking the product of these four numbers gives us 53,890,200 which we shall denote by A(1).

The next step is to count the occurrences of two pairs in semi-deals. We must split this into two subcases since the two pairs may have the same rank or different ranks. If they have the same rank, there are 13 choices for the rank, three choices for the two pairs of that rank, and C(48,2) = 1,128 choices for the remaining hand. The product, which is 43,992, counts the number of occurrences of two pairs of the same rank in semi-deals.

If the two pairs have different ranks, there are C(13,2) = 78 choices for the ranks of the pairs, there are six choices for each pair, and there still are 1,128 choices for the remaining hand. The product is 3,167,424 which when added to 43,992 yields 3,211,416. We denote the latter number by A(2).

Finally, the occurrences of three pairs in semi-deals also has two subcases: Either all the pairs have different ranks, or two of them have the same rank. In the former case, there are C(13,3) = 286choices for the three ranks and six choices for each of the pairs. The product gives us 61,776 occurrences of three pairs of distinct ranks. If there are two pairs of the same rank, then there are 13 choices for the rank occurring twice, 12 choices for the other rank, three choices for the two pairs of the same rank, and six choices for the pair of the remaining rank. Taking the product of all these numbers gives us 2,808 occurrences of three pairs, where two of them have the same rank. Denote the sum 64,584 by A(3).

What does A(3) actually count? It's the exact number of semi-deals with each of the three hands being a pair. This is easy to see because each such semi-deal is counted only one time in the calculation of A(3).

What A(2) is counting is much more subtle and I am not going to labor the point. The key to understanding this article is the realization that any semi-deal with three pairs is counted three times in the computation for A(2) because it contributes one for each subset of two of the three pairs in the semi-deal. Therefore, if we want to know how many semi-deals have exactly two pairs in them, we must remove three times A(3) from A(2). Doing so gives us 3,017,664 and we use B(2) to denote this number.

Similarly, if we want to know how many semi-deals have exactly one pair in them, we must remove the semi-deals which have been counted too many times in A(1). Semi-deals with three pairs are counted three times in the computation of A(1), and semi-deals with exactly two pairs are counted twice in A(1). Thus, we must remove three times A(3) and two times B(2) from A(1) to get the number of semi-deals with exactly one pair. Doing this arithmetic gives us 47,661,120 semi-deals with exactly one pair and we denote this number by B(1).

Now we can answer the original question. The number of semi-deals with at least one pair is B(1) + B(2) + A(3) = 50,743,368. To get the probability we divide by A(0). The preceding, of course, means there are 254,634,432 semi-deals without any pair by subtracting 50,743,368 from A(0) as the latter number is the total number of semi-deals.

Now look at the expression A(0)-A(1)+A(2)-A(3) and substitute the numbers above. You will find you obtain 254,634,432 which is the number of semi-deals without any pair. This is not an accident and tells us why this is an instance of inclusion-exclusion. Namely, in the preceding expression, semi-deals with one or more pairs are included, then excluded, and so on, leading to them contributing zero to the sum.

The above illustrates why inclusion-exclusion enters these kinds of computations. Now let me say a little about approximations.

The example above needed only four ``levels'' of computation, that is, as we calculate A(0), A(1), A(2) and A(3) we are adding more and more conditions to obtain successive numbers. We can think of them as levels because we have to work backwards to calculate the exact numbers of objects at the previous level. When the semi-deals involve 10 hands, such as in Omaha and hold'em, then working through all 11 levels simply becomes too complicated in most instances. However, we can use approximations which increase in accuracy as you use more levels.

For example, if we use A(0) - A(1) as an approximation for the number of semi-deals without a pair for the problem discussed above, we obtain 251,487,600 as compared to the exact answer of 254,634,432. This is an error of about 1.25 quarter percent. On the other hand, if we use A(0) - A(1) + A(2), we obtain 254,699,016 which now is within 1-in-40,000 of the correct answer. This is pretty accurate. The approximations I used for the bad beat jackpots were even more accurate.

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