Standard Deviation Question, How to do it?

I have been trying to work out the variances of Online Hold'em inspector's best profile, and have done this spreadsheet:

http://www.cyberloonies.com/Solid%20Profit&Loss.xls

If you unhide the rows you can see the working out. You can see that on average you make 1.85 bb/hr at a typical table (tested against AI) but for 1000 hands there's quite a lot of variance - you couldn't be very confident that you would make money by this stage.

My sample size is 300. The graph looks a little bit like a normal distribution but not very much.

My question is would I get a normal distribution of data if the sample size was bigger, or any there too many random variables involved to ever get a normal distribution?

[AaronBrown]

I can't promise you a Normal distribution, but this looks very much like one given the sample size. The standard deviation appears to be about 3.25. The jagged peaks are not unusual, you've sliced the data pretty thin. Nothing is exactly Normal, and there are some minor deviations in your data, but nothing that would invalidate most methods based on Normal theory.

[AaronBrown]

I'm not being clear. X^a is a function for converting finishes (X) into probabilities (X^a). Once you have a (I'll get to how you get it in a minute) the function tells you your probability of any given finish. If a = 0.5 then your probability of finishing in the top 10% (X = 0.1) is 0.1^0.5 = 0.32. Your probability of finishing in the top half (X = 0.5) is 0.5^0.5 = 0.71.

So a is a measure of your skill. a = 1 means your finish in the tournament is random, you're as likely to come in first as last. a > 1 means you have less than an average chance of winning. a = 2, for example, means your chance of finishing in the top half (X = 0.5) is 0.5^2 = 0.25. So only one time in four will you even be in the top half. a < 1 means you're more likely than average to finish in the top half.

To estimate a, I asked the question, what a would make your actual finishes (0.04, 0.15, 0.33 and 0.47) the most likely? It takes a little math, but the answer comes down to:

ln(a)/(1-a) = [ln(0.04) + ln(0.15) + ln(0.33) + ln(0.47)]/4

This isn't a formula for a, but you can find the a that solves it using Excel or other methods. It happens to be a = 0.2857. You can check that:

ln(0.2857)/(1 - 0.2857) = [ln(0.04) + ln(0.15) + ln(0.33) + ln(0.47)]/4

So that's how you use a and how you find a.

[VivaLaViking]

Thank you for your patience. I have a simple query about the precedence of operation on the relationship you quoted. Is this the same as you cited?


ln(a) . . [ln(0.04) + ln(0.15) + ln(0.33) + ln(0.47)]
----- = ----------------------------------------
1 - a . . . . . . . . . . . . . . 4

If so, I will try to isolate the varable, a

ln(a) . . ln(0.04 * 0.15 * 0.33 * 0.47)
----- = -------------------------------
1-a . . . . . . . . . 4

I am initialy tempted to raise the L.H.S.'s and R.H.S.'s numerator to e. It's been a while (lol)