Let's say my winrate is W BB/100 hands, and my std deviation is S BB/100 hands.

How many hands until I'm 95%, 99%, or 99.9% certain I'm a winning poker player?

I used to know how to figure this stuff out quite well, but it's been a while...

(Edit: subject asks for "hours", but hands is fine - I'll assume something like 30 hands per hour... Also, I do realize that this requires the bad assumption that my skill level and game difficulty remain static)

Assume you have sample data that indicates you have a positive win rate of W/100 with a standard deviation of S/100. To determine what sample size you need to assure with a specified confidence that you are a winning player, you can use the following formula:

N =100*( Zc*S / W)^2

where Zc is the normal distribution factor corresponding to a one-sided C% confidence interval: Z80 = 0.84, Z90 = 1.28, Z95 = 1.645. Note that N increases with C and S, and decreases with W.

Example: You have an estimated win rate of 5 per 100 hands with a corresponding standard deviation of 75/100. To be 80% confident you are a winning player,

N = 100*(0.84*75 / 5)^2 = 15,876.

For C=90%, N = 37,000 (approx.) and N = 61,000 for 95% confidence.

The above assumes your win rate is distributed normally (valid by the Central Limit Theorem) and that your win rate and standard deviation are relatively constant over the sample period. If your win rate is not constant but is increasing, then the sample size result can be considered to be conservative.- i.e., it is more than you need.

Ok so 5BBs/100 with sd of 75, that's 15,876 hands to be 80% sure I'm a winning player. That's over 500 hours if I play live at 30/hands per hour, which would be over 3 months of play playing 40 hours/week every week. (Note - I have a normal full time job, not planning on changing to poker, just trying to put this into perspective.).

Am I correct in assuming that this also means that for those values, I would be down 20% of the time for any given 3 month stretch?

Fwiw, my winrate *seems* to be much much higher than this (playing live 2/5), and my SD is probably a lot higher too (I'm very often up or down 100BBs from one hour to the next - so if my SD is like 100BBs/hour or 100BBs/30 hands, then that should be like 100*sqrt(3.33)/100 hands right? Or ~180/100 hands. Maybe that's an overestimate though). But I've only recently started logging my sessions - and it's only 12 sessions for 102 total hours, so probably much too soon to try to start calculating these stats.

How many sessions would it take to have a good idea of my SD/hour? (I saw another post here, i think by BruceZ where he showed how to calculate SD per hour given only session totals - but I don't know what the standard error in this SD calculation would be).

No.

80% confidence as used here theoretically means the following: If you played 16,000 hands a number of times with the stated win rate and standard deviation and calculated the lower 80% confidence limit on win rate each time, then 80% of such limits would be greater than zero. Many people equate the confidence level to win probability when it actually relates to the limit calculation. Some say that’s a distinction without a difference.

Regarding sample size for standard deviation, I seem to recall several threads/posts related to that. Try the 2p2 search function if someone doesn’t respond here.

Thanks for your responses... but not sure I'm quite following you here.

Stated differently -- if my "true" winrate and SD were the same as my sample winrate and SD, would I be correct in stating that for any 3 month stretch, I'd have a 20% chance of being down in money? (Using the example values)... I do understand the distinction between our sample statistics vs our "true" statistics, but want to make sure I'm correct about this equivalence if we assume the sample statistics match our actual winrate and SD. I'm guessing we could use some clever Bayesian model to get more precise about all of this...

Here's my interpretation (it's been a long time since I was in the classroom for this stuff). You are 80% confident you are a winning player. The other 20% you don't have confidence - maybe you are, maybe you aren't - you need more data.

Yeah my understanding is that if we collected 100 players with these stats, we'd expect that about 80 of them are in fact winning players, and 20 are actually losing players running good. (But I don't think this is actually quite true, because we're making assumptions about the prior distribution of winrates across all players - I.e., that any winrate is as likely as any other).

Look at it this way.

Suppose you were to make a confidence interval estimate such as “Based on the sample data of N hands, I am 80% confident that my win rate is greater than Xbb/100.” Now if your sample win rate was 5bb/100 and the sample standard deviation was 75bb/100, then you would need 15,876 hands to have X =0, signifying a winning player. Hope that helps.

Heeded also, thanks to you

Really to make this a math question is incorrect. You cann't know you are a winner by numbers, you have to know what your opponents do wrong and how much gain you make from them, Can you even cause any player you know of to fold in error? well if you can do that more then the rake and the blinds and then more then the bills you are a winner, those math answers are trying to measure the chance of making a profit if you have been but as I said if you dont exactly know how you are a winner then it is running good.