Statistics term, defined by the square root of the variance (which is the average squared distance from the mean).
This may seem complicated but it really isn't, for example say you have a data set of 1, 3, 5, 7, 9. The mean is 5, squares of the distance are 4^2 + 2^2 + 0 + 2^2 + 4^2 = 32 + 8 = 40. divide by 5 and you get 8 as the variance, sqrt(8) is the standard deviation. Although actually I think for some reason you divide by n-1 so you'd divide by 4, get 10 and sqrt(10), but that's irrelevant for large data sets anyway (if you've played 50k hands, does it matter that you divide by 49999 or 50k?).
Anyway, what does this mean? Well basically the relevance of this is in a normal set of data. I'm sure most have seen the bell curve. Basically a large data set that is determined randomly will have an approximately normal distribution.
Anyway, if we assume a normal distribution (which we can't, since in a normal distribution any amount of loss or win would be possible although extremely improbable while in NLHE if you leave before you get too deep you cannot win 1k big blinds in a hand) we can then approximate the probability of any given result in any given amount of time.
First off to apply this you have to learn how to determine the standard deviation of a sample when given standard deviation/100 hands. The way this works is that when you combine data sets, you add the variances, which means that to get standard deviation you square the standard deviations, add together, then take the square root again. For example, say you have an 80 bb/100 standard deviation yet you have 1k hands.
You would end up with this expression:
sqrt((80^2)*10) ~= 253 BB/1k hands
ok but now we get to how we can use this. I can post one later, but look on the internet for a z-table (think that's what it's called). Basically it lists the amount of area under a bell curve for each standard deviation. As a rule of thumb, 1 standard deviation either way encapsulates 68%, 2 is 95%, and 3 is 98%. So let's say that in these 1k hands you have won 253 BBs, meaning you are up the exact standard deviation in that amount of time (and also winning at an amazing 25.3 BB/100).
The probability of you actually being a losing player is the amount that is less than 1 standard deviation. If the 1 covers 68%, 32% is outside it and half of that, 16% is below that. So if you are winning at 25.3 BB/100 after 10 hands with an 80 BB/100 standard deviation, you can say with 84% certainty you are a winning player. Now try the numbers yourselves with bigger samples and more reasonable winrates. You'll understand why it's so easy for people to have upswings and downswings lasting 10k hands, but how if you play hundreds of thousands of hands and have a good winrate, you can be pretty certain you are a winning player. Also we can use this with other numbers as well, not just 0. For example say you've been running at 6 PTBB/100 for the last 20k hands. You can figure out the probability that you are at least a 5 PTBB/100 winner.
I think I've pretty much covered everything, the one thing I'm interested in actually though is the normal distribution. I don't think we can treat hands as if they follow the normal distribution simply because we will have a lot of small losses and a few big wins if we are good. With a normal distribution the odds
of winning an amount x units above our mean (winrate) and an amount x units below our mean are identical. There's a test you can use to test how normal a data set is and I think I may run it on my data. I'll let you know how it goes.