So I know the de facto standard for winrates is 100,000 hands. But I'm curious if that is also the number of hands at which the Dealt Cards will also converge to their expected value. So the simpler version of the question is after how many hands would we expect different players to have all received the same number of pocket aces, pocket kings, AKs, etc. aka at what point can people not b---- about their cards?

The more complicated version, at what number of hands would the number of (all) AA, KK, QQ, JJ, TT, AKs, AQs, AJs, AKo, and AQo hands be within 99% of the expected of value with 95% confidence? And I'm guessing that if AKs is the least common one, once that has converged all the rest would have as well. (At least I think that's how it should be worded.)

Should someon complain if he's dealt the first hand and it isn't to his liking? No.

Should someone complain if he's dealt the 100000th hand (or however many it takes) and everything has evened out? No.

So why would you think there is any point in between where he's 'allowed' to complain about his luck in cards? I don't get it.

People complaining about cards was meant as an aside, the real question is the probabilities.

Off hand, I’m not sure how to frame and solve the question of ALL hands converging for different players. For example, if AA and AK are of interest, there is an obvious dependence for if one player gets AA it obviously reduces the probability another player is dealt AK.

I can do an individual hand. If P is the theoretical frequency that hand H occurs in a trial, then the number of trials to be C% confident that the observed frequency is within D% of the true value is given by:

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, 99% , Z=2.33. This is based on the Central Limit Theorem, which may not be the best approach for very low probabilities.

For example, the probability that AK is dealt to a player is 16/1326= 1.21%. Then the number of trials (deals) needed to be 95% confident that the sample mean is within 1% of this value is

N = (1.96^2)*(0.0121*0.9879)/(0.01^2) = 459

The reason this number may be lower than you may have thought, is that a probability standard deviation over 1 trial is P*(1-P), which is quite low for low occurrence probability. If, for example, you select a hand type or range that has about a 50% occurrence frequency, about 9,600 deals would be needed.

Ah cool, thank you, don't remember if this was covered in my probability class twenty years ago.

Yah that doesn't make sense, 95% of the time after 459 hands the player will have gotten 4 AK? That doesn't jive with the reality of streaks IMO. Although perhaps there's a confirmation bias in that we only notice when we've gone 500 hands without an AK.

So for suited AK, at 99% so 4/1326 = .3016% ... if I did the math right that's 163.27. That can't be right, it's taking fewer trials to converge. Heh or does that just mean that after 160 hands we're very likely to have gotten AKs 0 or 1 time?

Although maybe that's the wrong question...we want to be something like 95% confident that the combo that is the biggest outlier is still within 1% of the expected value, which would probably take a lot more.

Perhaps the best explanation is that...I have no idea what "confidence interval" actually means.

A confidence interval refers to the goodness of the sample calculation.

For a 95% C.I., if you were to obtain a sample from a population N times and compute a confidence interval for each sample, then 95% of those intervals would include the true mean. It is the interval that is the “random variable,” not the population mean.

The confidence interval does not assign a probability to the population mean (an incorrect interpretation that is often made) but to the sample interval that is calculated. The narrower the interval imparts more “confidence” in your mean estimate.