Cracking Aces: Part I

Brian Alspach

Poker Digest Vol. 3, No. 7, March 24 - April 6, 2000

Bib Ladder appeared out of nowhere the other day and asked me the following question, ``This morning, professor, I read on the Internet there is a 31.1 percent chance of winning with aces against nine random hands played to the end of the hand. It was reported this was found using simulation. What does it mean and should I pay any attention to it?''

``That's a good question, Bib. Let's take a seat at that booth in the corner and I'll explain it as best I can.''

We sat in the booth, ordered two cups of coffee, and I gave him the following explanation. Suppose one wanted to know the true probability pof a player holding a pocket pair of aces in hold'em winning the hand against nine other players who stay all the way to the end.

Theoretically it's easy to do. You set aside a pair of aces and examine all possible ways of dealing hands to the remaining nine players and spreading five cards to form the board. You then tally those for which the pocket aces win and divide by the total number of possible ways. Let's take a look at how theory and practicality part ways.

We care only what the other nine hands are and not who gets them. This is what we have called semi-deals in the past and considering the number of semi-deals, instead of deals, substantially reduces the computation. The first step is to choose the 18 cards to be distributed to the remaining players. This can be done in C(50,18) ways (remember, we have removed two aces). Now we have to partition the 18 cards into nine hands of two cards each. This can be done in 17!! ways. (Recall that $17!! = 17\cdot 15\cdot 13
\cdots 3\cdot 1$ -- a large number.) Let's see how we obtain 17!!

First we choose two cards from the 18 in C(18,2) ways. Next we choose two cards from the remaining 16 in C(16,2) ways. Continuing in this manner, we obtain $C(18,2)C(16,2)C(14,2)\cdots C(2,2)$ ways of chopping the 18 cards into nine hands of two cards each. However, when choosing two cards at a time, we could have gotten any particular given partition in 9! different orders so we must divide the preceding product by 9! because that is how many times we are counting the same semi-deal. If one writes out the product, one gets a fraction in which the numerator is 18! and the denominator is 29 times 9! Evaluating the expression leaves 17!!

As far as the board is concerned, since we are assuming all players stay to the end, it doesn't matter in which order the five cards appear. Thus, we are interested only in the cards comprising the board. There are C(32,5) boards because we are choosing five cards from the remaining 32.

So the total number of possibilities we need to examine is simply expressible as 17!!C(50,18)C(32,5). The naive might say just let a computer examine all the possibilities. Our intuition isn't capable of appreciating how rapidly these numbers grow. If we had a computer which was able to examine 328,185 possibilities per second, it would take approximately 12,000,000,000,000 years to examine all the possibilities. Hooking up 100,000 computers to work in parallel or increasing the speed of the computer by a factor of ten, a hundred or a thousand still leaves the problem intractable. The point is, given our current state of computational knowledge, a brute force examination of all possibilities is out of the question. This means we have to come up with other methods to attack the problem of determining p or approximating it closely.

One clever and elementary approach is to randomly sample a large number of the possibilities and see what proportion of the random sample leads to pocket aces winning. Call this proportion p'. This is what people do although it is stretching it to call it simulation, a term I reserve for more sophisticated analysis of models.

Anytime one uses random sampling, one must be concerned with the accuracy of the results obtained. There may be problems with the randomization process, for example. Over the years there have been some monumental mistakes because of the samples not being truly random. There are standard statistical measurements of accuracy for random sampling.

If you use 1,000,000 as the number of random samples, then we have 99 percent confidence the true probability p lies between .310 and .312. What the latter means is that 99 percent of the time we obtain a value for p' by sampling 1,000,000 random outcomes, the true value of p will lie between p' - .0012 and p' + .0012.

Therefore, we have considerable confidence that .311 is very close to the true value of p. Accordingly, in using this figure for p when discussing how to play pocket aces in various scenarios, we can be assured any problems we encounter with the conclusions we reach come from fuzziness introduced by other assumptions and not the assumption that p = .311.

We continue in the next article with a discussion of the relevance of the .311 figure to actually playing a pair of pocket aces.

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