Hi,

In no limit texas hold'em online people often use a heads up display (HUD) which tracks specific player stats such as voluntarily put $ in pot % (VPIP) and pre flop raise % (PFR).

I'm curious as to what the probability of running certain stats over a specific number of hands is if said person's true stats over an infinite sample are given.

My question:

If over 18 hands someone has the stats 80/50 and their true stats are 28/22, what is the frequency at which this player gets these stats over 18 hands, or what's the standard deviation from the real stats.

Please provide the work so I can follow it and be able to understand how to calculate this on my own. If my wording/question is incorrect or misworded please just ask for clarification or make whatever assumptions are necessary.

Thank you

There are several different, yet related, ways to address your question.

To answer your question most directly, you can use the Binomial Distribution to calculate the probability of observing your opponent's stats given his underlying true probability. Consider VPIP. Say his true prob is P and you observe a frequency percentage of W over N hands.

Then you can simply tally the probability of a binomial distribution with probability P would have W or more "successes" in N trials. This is given by:

Sum[H goes from Round(W*N) to N] C(N,H)*(P^H)*((1-P)^(N-H)) where C(X,Y) is the choose function giving the number of combinations in which you can choose Y items out of X items without replacement and when the order does not matter.

In your example above, P=.28, W=.80, and N=18. By plugging the numbers into a calculator or computer program, you will see that the probability of observing .80 (say 14 or more "successes" out of 18 trials) is very low if the true underlying prob is really .28.

Since VPIP and PFR tend to move together (though not necessarily), it is probably prudent to test these separately, or maybe just VPIP. Note that this entire subject is rather complex and it is not easy to address it comprehensively in short posts in a forum like this.

Anyway, I hope that helps.

Edit: In cases in which the observed frequency is much lower than the true prob, you would take the sum in the other "direction". That is, rather than at least W*N, you would calc the prob of at most W*N (sum would go from 0 to Round(W*N in this case).

Okay, thanks. The answers I get seem to be counter-intuitive to me based on my experiences.

For example, using a binomial calculator @ http://stattrek.com/online-calculator/binomial.aspx

I've plugged in:

probability of success = 0.28 (assuming my vpip is 28%)
number of trials = 18 (number of hands played)
number of successes = 0 (meaning my vpip would be 0, haven't played a hand in 18 hands)

If I'm understanding this it says the probability of this occurring is 0.27%, which seems very low. And for my original question in the OP it was an answer 10^-6, incredibly low. Is there some other factor I am not considering here?

1) You're assuming a fixed frequency of 28% (eg: the player might choose to play very differently when a loose maniac is at the table compared to playing vs a table full of super-nits, etc).

2) You're not accounting for the "context" of the 18 hands (eg: a true 28/22 player may play the 18 hands very differently if the first hand were played UTG as opposed to in the cut-off, or [related the point (1)] may play the 18 hands very differently if there is a maniac at the table raising every hand, etc).

3) You're not accounting for serial correlation (eg: the player might have been caught bluffing several times recently and deliberately tightened up, or the player might have just gone on monkey tilt after getting his aces cracked, etc).

Juk

All true, yes. I cannot quantify those, however.

There was a HUD that tried to address (1):

http://forumserver.twoplustwo.com/16...r-more-368151/

http://web.archive.org/web/201005221...euth-HUD.shtml

(sadly the project is long since dead now and most likely with the current "anti-HUD" sentiment today; it would be insta-banned if anybody tried to create anything similar now...).

This thread has some discussion about how to address (1) and (2) too:

http://forumserver.twoplustwo.com/25...ating-1516954/

This thesis outlines a different idea of how to deal with (1) and (2), and also whosnext's point about how "VPIP and PFR tend to move together" (see section 4):

http://www.ai.rug.nl/~mwiering/Tom_v...eij_Thesis.pdf

(3) is much harder to deal with though (even with huge amounts of data to aid you) and is probably one of the main reasons why poker AI research has moved away from exploitative/model-based methods and towards game-theoretic/pseudo-optimal methods.

Juk

Using a binomial distribution assumes random deviation from the mean. The stats in question measure player decisions, not a random process.

For these you need the actual historical distribution and standard deviation to make predictions, not just the actual mean. You then can find the likelihood of a particular sample.

good point


Can you explain the basic difference between these for me? Thanks, sorry if its a basic question.

Have a read of the first part of both of these:

http://poker.cs.ualberta.ca/publicat...vidson.msc.pdf
http://poker.cs.ualberta.ca/publications/IJCAI03.pdf

Juk

thanks for the link juk, ill give them a read

What would be a reasonable std deviation for someone with loose-ish stats, say 28/22? If the standard deviation is known, what would it be measured in? If it is known how do we use it to answer the question in OP?

This issue was recently covered in this very forum. See the link below.

http://forumserver.twoplustwo.com/25...-size-1574339/

After reading that thread (it is a short thread), feel free to ask further questions.

Man I'm dumb. I totally forgot about Z. It's all coming back to me, thanks!

In OP, specifically looking at PFR being 50% and expected value to be 22% and solving for sample N needed to be 95% confident in value we would need a value of N equal to:

N = (Zc^2)*P*(1-P) / d^2

N = sample = ?
Zc = (95% confident) = 1.645
P = (known value) = 0.22
d = 5% = 0.05

N = (1.645^2)*(0.22)*(1-0.22)/(0.05^2)

Above equation should be that you need N hands to be 95% confident that the player has a 22% PFR. Solving the above equation we get:

N = (2.706)*(0.22)*(0.78)/(0.0025)
N = 0.464/0.0025
N = 185

However, I don't think this really answers my original question in OP (although it is useful) because in this equation we are NOT taking into account the current statistics we have. The OP in above thread had same issue, but I'm not sure I follow it 100%. Can we go through this specific example (my OP) with your other equation:

((P)(N)+S)/(N+T) and whatever you are trying to convey in post #6?

I will break up my replies into two posts to respond separately to the two different questions.

In statmanhal's "classical" approach, the "d" is the observed difference between the assumed true underlying probability value (P) and the observed probability frequency (W).

So the required number of samples does indeed depend upon your sample observations. Of course, you need fewer observations if the sample frequency is far away from P to be 95% confident that the sample is not from that binomial process than if the sample frequency is quite close to P.

Secondly, I cannot do better than quote my post from the other thread.

This "Bayesian" approach differs from the "Classical" approach in that it is not focused upon testing the hypothesis that your observed frequency (W) did not come from some assumed-to-be-true binomial process with underlying probability (P).

Instead the Bayesian approach asks the question: given that I believed prior to observing new data {frequency W} that the underlying process was binomial with probability P, what is my new belief about the underlying binomial process {what is the new probability P'?}.

So if you observe a little data with W near P, you would not move your new estimate of P (call it P') much at all (P' would be very close to P).

But if you observe a lot of data with W far away from P, you would rightly move your new estimate of P (P') significantly (P' would be far away from P).

Bayesians talk about their "prior" belief about P and then, after observing new data, updating their belief about P with a "posterior" distribution.

The formula above, ((P)(N)+S)/(N+T), is a way to update your belief about the underlying process.

For example, if prior P=.22 in a prior sample of 1,000 (N=1,000), and you observed a new sample of 28 "successes" out of 100 new sample observations, then under the assumptions described in the previous thread's post, your "posterior" updated expectation for the binomial probability is given by:

P' = [(.22)*(1000) + 28] / [1000 + 100] = (220+28)/1100 = 248/1100 = .2254545

You will readily see that the more samples that the W is based upon, the more you will adjust your updated frequency estimate. Of course, if N=T, then your updated estimate is simply the average of your prior belief and your new sample frequency. And if your sample is quite small relative to the number of your prior observations, you will not adjust your estimate very much based upon the new sample data.

One "challenge" with this specific approach is that you have to have a good sense of what N is, the number of previous observations from which the prior P sprang. Of course, this reflects how confident you are in the prior P.

This is where the previous thread recast the Bayesian approach listed above to one dealing with an estimate of the standard deviation (either from prior observations or as a measure of your "confidence" in your prior estimate) around your prior estimate of the underlying binomial frequency.

Under this specific approach of using a standard deviation, new formulas were posted to derive the updated "posterior" estimate of the binomial frequency.