re: PT stats, how many hands needed to see patterns?

[MicroBob]

I think you're within 1% of your true VPIP in WAY less than 10k hands.

For most of my opponents, they hit a fairly reasonable VPIP after only 200 hands with them and it very rarely deviates by more than a couple points after that from my limited observation.

If I have 1000 hands on someone then the VPIP I have on them is pretty much going to be their true VPIP.

Most of the time you don't have to worry about it.
If a player is 18/12 after 200 hands then he is probably going to be 18/12 after 1000 hands and 18/12 after 10,000 hands (unless you get some hands with him with many more or fewer players than the first 200 hds).

This is a stat that seems to converge extremely quickly.

[uDevil]

[ QUOTE ]
I've had PT since last year, and was curious to know how many hands I need in my database before I can start seeing how my stats compare with a perfectly random distribution? Specifically, I have 5000 hands in my database on FTP and while my pocket eights have won every single time of the 12 times I've had them, my kings have won 65% of the time. That seems odd.

[/ QUOTE ]

Although I'm sure people mean well, there's very little useful information in this thread. The number of hands required to have a given level of confidence depends on the variance associated with the statistic you are interested in. A number like 100,000 means nothing without knowing what the confidence level is.

You may want to do some reading about confidence intervals.

You can find web apps that will do calculations for you. Here is one.

[creamfillin]

uDevil, I've always wondered about these concepts but never bothered to research them. I guess we're all a bunch of little kids blurting out big numbers.

[LionelHutz00]

some stats have higher standard deviations than others. vpip has a relatively low standard deviation, so your stats for vpip will converge to their true values relatively quickly. winrate has a high standard deviation so it takes many many many hands for that stat to converge to its true value.

[Sciolist]

Does anyone know how to calculate the standard deviation of PT stats? It calculates your win rate SD for you, so it's way easier to know your confidence interval. However, I don't know of a way to calculate the SD of VPIP, for example. I'm not maths expert, and I'm even a bit shaky on how to calculate SD in the first place, let alone creating a query to do it from the PT SQL DB.

I made a post about this on the probability forum a while back and got absolutely no replies.

[uDevil]

[ QUOTE ]
Does anyone know how to calculate the standard deviation of PT stats? It calculates your win rate SD for you, so it's way easier to know your confidence interval. However, I don't know of a way to calculate the SD of VPIP, for example. I'm not maths expert, and I'm even a bit shaky on how to calculate SD in the first place, let alone creating a query to do it from the PT SQL DB.

I made a post about this on the probability forum a while back and got absolutely no replies.

[/ QUOTE ]

You do need to write an SQL query. Which means you have to understand how the PT data tables are set up. I learned this myself by poking around a little using Microsoft Access. I queried columns for my player ID and the vol_put_money_in_pot column then took 10,000 records and copied them to Excel, where I just used the stdev and average functions to calculate the standard deviation and mean for my VP$P.

I got a mean VP$P of 25.2 (this is for SH LHE games) and a standard deviation of 43.4/hand. So after 1000 hands, I could have 95% confidence that my VP$P would be between 25.2+(1.96)43.4/sqrt(1000) and 25.2-(1.96)43.4/sqrt(1000), i.e. between 23.2 and 27.7.

For the OP, you'll be dealt a specific pocket pair one time in 221. The standard deviation (see binomial distribution ) for the number of times you'll be dealt a specific pocket pair (per hand) should be sqrt((1/221)*(1-1/221))=.0671. In 5000 hands, you can have 95% confidence that you will be dealt between 5000(1/221-1.96*(.0671)/sqrt(5000)) and 5000(1/221+1.96*(.0671)/sqrt(5000)) pairs of 8's, for instance. That is, between 13.3 and 31.9 pairs of 8's.

I'm not a mathematician and I make a lot of mistakes, so if these numbers are important to you, you should check them.