Using a Bayes’ Estimator to Provide a Best Estimate of a Random Villains’ Opening Frequency Given a Small Sample   Hi guys, I still don’t have access to PokerStars so I’ve been working on theory instead. Chad asked an interesting question that got stuck in my head until I worked out my solution. Here’s the question:   “What maths would we use to find  the distribution of likeilhoods of a specific outcome given a finite sample, for ex say someone raised 4/5 buttons and we are looking to find the likelihood of their pfr being greater than 70% how would we go about this (or just point me in thew direction of the right theory and i'l give it a shot). Thinking in the context of making significant adjustments in the face of small samples.”   So first off, a disclaimer: I’m not Mers. This is the first time I’m doing this kind of problem. I do have some experience with this kind of stuff since I have a degree in Physics and Math and I took a grad level stat class, but I’m not a stats wiz. I might still be doing stupid stuff, and I might be flat out wrong. But here is my best take on it, along with the research I did to get to my conclusions. It’s pretty long but the results are pretty simple.   So the idea of finding a distribution and estimating a parameter (in this case opening frequency) is a problem in statistics called estimators. Finding “best” estimators is an interesting subject and has a lot of depth of it. That being said, I’m doing a cursory job and jump to the chase and avoid much of the underlying theory. Read more about estimators here: http://en.wikipedia.org/wiki/Estimator   So as we start the problem we immediately reach a cross roads because there are two main approaches to the problem. The first is the direct method--all we know is that someone raised 4/5 buttons. We focus in and care only about did they raise or do something else. The data is in the form of {1,1,0,1,1} where 1 = raised, 0 = didn’t. This means that the sample represents a Bernoulli distribution. So here we could use the Bernoulli distribution http://en.wikipedia.org/wiki/Bernoulli_distribution and calculate Chad’s question. But I didn’t do that. I’m working for myself and so I’m going to answer the question in an expanded way that makes more intuitive sense for me and will provide a more accurate estimate.   The problem is that I don’t think that the way we have formed the problem is accurate, and I don’t think solving it will give practical information. Chad wants to use the information to make decisions like can we make a significant adjustment like 3b bluffing. For me the question becomes interesting when we immerse the problem in reality--we aren’t playing against an unknown coin. We are playing against a player within a sample population. I want to use more advanced math and give an accurate estimate of the following question: I am playing against a random $30 ST player, the effective stacks are between 20 and 25 bb, and this player just opened 4/5 of his buttons. What is his most likely opening frequency, o? Hint: it’s not 80% as it was in Chad’s question. Why not? Because the average $30 player has an established average frequency and we have to use that to get his most likely opening frequency. In his article about 3betting Mers specifically stated that his estimate of mean opening frequency is 55%. We want to use that kind of info.   So here’s where we get deeper into the math. We want to use prior information to make our decision more accurate. This kind of statistics is called Bayesian statistics. http://en.wikipedia.org/wiki/Bayesian_statistics. The basic idea is that behind the parameter o (the opening frequency) there is a prior probability distribution, we use this together with our random sample (the 5 hands in which he had an o of 4/5) to create a posterior distribution of what we think his opening frequency looks like. I want to use the best possible Bayes’ estimator to do this. http://en.wikipedia.org/wiki/Bayes_estimator   To make life easier I’m going to use a common trick in Bayesian statistics--I’m going to use a convenient type of prior distribution. For each random sample distribution there’s a “conjugate” distribution that is particularly easy to work with called the conjugate prior. http://en.wikipedia.org/wiki/Conjugate_prior. For the Bernoulli distribution this is the Beta Distribution. http://en.wikipedia.org/wiki/Beta_distribution. It’s kind of ugly distribution itself, and it might not be the most accurate estimate of the actual prior, but it means we can black box our math a lot and all of the formulas we’ll need in the end end up pretty and exact.   So let’s figure out this prior distribution. The Beta distribution has two parameters (called hyperparameters) alpha and beta (called the shape and the range). One way to do Bayesian stats is just to guesstimate the parameters using reasonable assumptions (like using Mers’ 55% mean for instance). I’m sure this works fine for someone who has an intuition for the Beta Distribution, but after playing around with the two hyperparameters I realized that I don’t. And I want this to be relevant to $30 ST games in particular, and I don’t intuitively know what that looks like either. So I’m going to use actual data to do this--just some small set for everyone to get an idea of the results we can get using this method.  I used a small random sample of villains from my database of FullTilt superturbos, which unfortunately was found through Poker CoPilot which has really terrible filtering so it’s not filtered for any blind level or stake. The only filter I used was that I used the villains against whom I had the most hands so that at least each data point wouldn’t be skewed by small sample size. I supplemented this data with some data from Mr. Bambocha’s HEM which also isn’t necessarily filtered for effective stacks. This won’t be super accurate, but if there’s interest then together as part of the FastTrack forum we can always make this more. For now I just wanted something that was reasonable and so here is the data I used: {0.6; 0.55; 0.69; 0.62; 0.23; 0.65; 0.23; 0.27; 0.41; 0.6; 0.34; 0.63; 0.08; 0.64; 0.0; 0.54; 0.8; 0.65; 0.38; 0.38; 0.2; 0.56; 0.55; 0.56; .59; .53; .61; .54; .43; .7; .57; .77; .51; .46; .69; .54; .57; .15; .36; .62; .58; .24; .58; .70}   I then plugged these values in to this calculator http://www.wessa.net/rwasp_fitdistrbeta.wasp to get the best Beta distribution fit for the data. This gives me the Beta distribution hyperparameters alpha and beta, called shape1 and shape2 on that website, that best correlate with the data set. And now we have our prior distribution--it’s defined entirely by being a Beta distribution and having alpha = shape1 = 2.48 and beta = 2.68.  I graphed it using Mathematica’s free webapp and it looks like this: http://www.wolframalpha.com/input/?i=beta+distribution+2.48%2C+2.68 So now we can use it to solve our initial problem.   Using the following site http://www.ds.unifi.it/VL/VL_EN/point/point4.html about Bayes’ Estimators, I look at the part about the Bernoulli distribution. Using their notation, we have a = 2.48, b = 2.68, Xn = number of opens = 4, n = number of hands = 5. I now use their formulas to get the answers I want. First I use their formula right after bullet 1. The posterior distribution is a beta distribution (that’s the magic of a conjugate prior, the posterior distribution is of the same form) given by a’ = a+Xn = 2.48 + 4 = 6.48’ and b’ = b + (n-Xn) = 2.68 + (5-4) = 2.68+1 = 3.68’. Again I plotted it and it looks like this: http://www.wolframalpha.com/input/?i=beta+distribution+6.48%2C+3.68. This represents our now population distribution--we are not dealing with a random anymore, but instead a random who then proceeded to open 4/5 buttons. The posterior distribution represents that population sample. This is the qualitative part of the answer--like how likely is it over 65%.    But I want what is the directly useful result: the answer to MY problem--what is the best estimate of villain’s opening frequency given a prior. This is given by the Bayes’ estimator after bullet 3. -- Un = (Xn + a)/ (n + a + b). Plugging in the variables, we get that Un = (4 + 2.48)/(5+2.48 + 2.68) = .638 = 63.8% = o. We have it!   So I know this was a wall of text with math involved so CLIFFS: Given a population sample we can use the data to compute a prior distribution in the form of a beta distribution with the hyperparameters a and b. In this case I used a small sample and got a = 2.48 and b = 2.68. Then we use our random sample of data from the current villain in the form of total number of opens Xn and total number of hands n. In Chad’s case that was Xn = 4 and n = 5. Lastly this lets us compute the most likely opening frequency o%  = (Xn + a)/ (n + a + b) = 63.8%.   Some things that I know can be improved and worked on further: 1. Improving the random villain data set. The more we put in this the more accurate our estimator will be and therefore the more useful. This could be made very accurate by having it filtered by blind levels and stake. As it is, I have some data points but there should be many more and it would really be best to make sure they are accurately filtered by stake and effective stacks. This is really key to making this method accurate and useful in game--essentially once we have accurate alpha and beta numbers we will be able to always have an accurate idea of villain’s ranges. 2. Working out more uses for the posterior distribution we calculate--for example you can use it later on in a match as a better prior to calculate off of. It also contains in it more information like variance and likelihoods of other opening frequencies besides the most likely one. 3. Using this kind of approach to dealing with non binary ranges. By incorporating limping, folding, and creating a distinction between open shove and minraise frequencies we can try to make our estimator more powerful, nuanced, and perhaps more accurate. It will involve generalizing our conjugate prior and random sample distributions from the univariate Beta and Bernoulli to their respective multivariate analogs Dirichlet: http://en.wikipedia.org/wiki/Dirichlet_distribution and Categorical http://en.wikipedia.org/wiki/Categorical_distribution.