Using confidence intervals to determine winrate with low sample size?

[BIG NIGE]

Does anyone know what the formula is for determining the minimum expected winrate for a specific level of confidence (eg, 90% or 95%), given a sample size and actual winrate? I learned this in statistics class a few years ago but have since forgotten it. Do I need to know the standard deviation? For example, suppose I have a sample size of 8, and an actual winrate of 135.2% ROI. With 90% confidence, what is my minimum expected winrate? This is for one-table, 9-player $22 SNGs on Pokerstars, so if SD is necessary to know then just use a typical one. Thanks to anyone who can help.

[TotalScience]

I'd say ballpark you have atleast 100% ROI.....though really, are you serious about 8 sample size or just using it for an example?

[BIG NIGE]

Serious. I have four wins, one 2nd, and three losses. And I'd like to find out what the formula is, so that I can calculate my minimum winrate with X% in the future.

[pzhon]

The typical standard deviation is about 1.7 buy-ins, or 170%. After n SNGs, the standard error (standard deviation of your observed ROI) is about 170%/sqrt(n). A rough 95% confidence interval for large n is the observed result +- 2 * standard error.

When you use a small sample, a normal approximation may not be accurate, particularly at the tails of the distribution. For n=8, you should expect the normal approximation to be poor, although the SNG distribution is not very wild, so this only causes a small problem.

A larger issue is whether you were planning to test your results after exactly 8 tournaments, or not. It is much more likely for your true ROI to be 2 standard deviations below your observed ROI at some point than at the end of a prespecified interval. If you stopped because the results look flattering, this introduces a bias.