Women's Performance in 2011 WSOP Main Event - A Deeper Dive

Poker Grump brought up an extremely interesting topic over the weekend (at least to me) where he attempted to analyze, from a statistical standpoint, whether the number of female cashers in this year's main event could be considered within the sample variance of a formal sampling test. Or whether, without any causality offered (smartly), some factor or factors contributed to their cash rate being significantly lower than their male counterparts.

Let me be clear up front. I am IN NO WAY bashing female poker players or women in general. I am looking at this solely because I find it intriguing from a math perspective and PG brought it up.

At first read, my hunch (and comment on the post) was that, statistically speaking, the sample size was sufficient to reject the null hypothesis, that the cash rate of the female players in this year's ME was within acceptable tolerance of variance from the expected value of the statistic. Upon further thought and investigation, I decided that underlying distribution used in his analysis (the binomial) was inappropriate for the question at hand (although directionally informative) and wanted to try to take the question further.

At it's root, the binomial distribution models performing some kind of experiment N times where the probability of a successful trial is p. It comes with several assumptions, however, where a few of the important ones include independence of outcome between experiments and a constant potential success rate. Essentially, it is the same as sampling with replacement from a population of size N where a success will occur with rate p.

The original analysis was done from the women's perspective, i.e. what percent of the time, if cash rates are truly the same, would there be 13 or fewer female cashers out of the 242 entrants (5.4% "sampled") when the expected value is 24 (10.1%). I decided to use the same distribution to perform the test from the men's perspective and found that, using this methodology, over 15% of the time would more men cash than there are places in the money. This clearly indicates a deficiency in the test.

So, what would the appropriate distribution be? The true answer to that question is way above my pay grade, as the odds of success for each individual in the field varies wildly, but if we assume for the sake of argument that each entry in the field has the same chance of cashing, I think that the problem is modeled nicely using the hypergeometric distribution. This distribution is frequently taught in statistics classes using examples such as 'there are a red and b blue marbles in a box; what is the probablity that if we choose c of them, that d of our choices will be red.'

So, in our example, we have 6,623 red marbles and 242 blue marbles and want to know what the distribution of blue marbles would be if we pull 693 samples from the box. Using the hypergeometric, we find that the expected mean is the same as the binomial, 24, and should occur almost 9% of the time. The distribution will return 13 or fewer only 0.562% of the time, which is very close (and even more rare) than the PG's binomial.

This was not meant to be a knock on Grump in any way. I believe he was trying to get a quick and dirty answer (and still pretty accurate), and succeeded. Being who I am, and having a little more than enough statistics knowledge to be dangerous, I attempted to see what a more rigorous approach to the problem would lead to.

What did it lead to?



Blue marbles performed less than red marbles in the 2011 main event.

And there go another 5 minutes of your life you're not getting back...