Hi guys - I've done a bit of searching around this, but was hoping one of you kind folks could help me out in case an answer is readily known/calculable.

Situation is:

Online microstakes on a site with a relatively reg-heavy pool. Zoom format.

We have 25 hands on a villain and his VPIP is 10. The typical player in this pool runs about 22/18. We haven't observed any hands this player's been in (due to Zoom) and there's no other info in terms of fishy screenname/avatar etc.

My questions are (and I do have intuitive answers to these, but interested in the maths to back it up):

1) Are we better assuming this player's VPIP is "really" 10 (or, say, 8-12 if that's easier), or that it's 22 (i.e. the median for this player pool)? Can we calculate the probabilities of these two outcomes?

2) If the answer is 22 (as I suspect...) how many hands would we need before we'd be better off playing against villain as if he's a genuine VPIP 10 player. (Again, backed up with maths if possible)

3) How do we go about doing the maths to work this out? I'm guessing we need to do some Bayesian stuff given there's significant prior information we have on the population?

Stats become more relevant as the sample goes bigger. Card distribution is random and may give false stats on a villain. There is no specific number of hands, just the more the better.

In 50 hands I have had a vpip of 0 and another 50 it might be 45. I am a vpip 19 player.

I have posted several versions of the following, which uses binomial distribution confidence interval theory to develop a required sample size. It has not been challenged as far as I know, yet I haven’t seen any great acceptance either. Use or ignore as you see fit.
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For a binomial measure such as VPIP or 3-bet percentage, to be C% confident that the sample mean is within d percentage points of the true value when the true value is believed to be P, the required sample size is

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

where Zc is the standardized normal deviate corresponding to a C% confidence interval. For 80%, Z = 1.28; 90%, Z=1.645; 95%, Z = 1.96.

You have to make a guess on P; N is maximized with P = 0.50.

Example: Suppose limited data on a villain indicates a VPIP of 20%. How large a sample is required so you can be 95% confident that the sample VPIP is within 3% of the true value.

N = (1.96^2)*0.2*0.8 / 0.03^2 = 683

For 90% confidence for a 5% deviation, it reduces to 105.

I have something similar that I've used for a long time but we seem to differ slightly. Mine is

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

Which are the same except that I get an N that is 1/4 the size of yours. I don't remember where I first saw or read this and actually, although I'm pretty sure my *code* is right, I'd have to go look at it to see if I've transcribed it right.

I think we may have the same formula. It depends how we define my Zc and your Z_A/2. I showed the values I use; do they agree with yours? It look like your symbology of _A/2 is for a 2-sided interval.

Well, unfortunately, although I have a value in my code, I didn't specify what it corresponds to. I probably thought I wouldn't forget. The value I have is 0.82245 which is half of the z-score for a 90% confidence interval. I'm not sure if I'm wrong here or not. I suppose what I should is make up a simulation and see. I wrote this probably 8 years ago and haven't used it in at least 5.

This question can be answered by the "bayesian estimator". You obviously have to take into account the population and adjust your observation.

it will be somewhere in the range of 11-21 and will depend on the standard deviation of your population.

The first step is to construct the prior distribution from the population tendencies (the simplest prior distribution possible is probably the beta distribution). The beta distribution is based on two parameters (alpha,beta) which can be derived from the mean and the standard deviation.

Estimator = (alpha+X)/(alpha+beta+n)

x = number of observations
n = total sample

I'm guessing here but in the notation z_A/2 I think you divided z_A by 2 when it was meant to denote using the Z score for a two-sided confidence interval. I can't think of a reason to halve a Z score.

The bayesian approach suggested above is a valid alternative to the classical confidence interval approach. I would suspect a relatively large standard deviation indicating some heavy nits and some crazy maniacs.