Standard Deviation question

I've read that the SD equation is 1.1/square root of "N". "N" represents the number of hands. One standard deviation is 68.3% of the time. Doubling the standardard deviation is 95% of the time and 3 standard deviation accounts for over 99.7% of the time I believe. With that said, here is a question broken down to a extremely small sample to make my point.

Lets say, for mathematically purposes only, that you play just 4 hands and the the average bet is $10 per mand. Total wagered amount is $40. Also assume that there are no splits, doubles or blackjacks.

SD for the above mentioned play is 1.1/square root of 4 = .55 or 55%. 55% times the total number of hands which is 4 = 2.2 hands. 2.2 times the average bet of $10 = $22. $22 is the SD. With 1% advantage you would have $.40 ($40x1%). You would potentially make $22.40 and also can lose $21.60 2/3 of the time. 2SD is $44 and 3SD is 66.

Here is my question, how does 2 SD or 3 SD work out when the variance in the SD is greater than the total amount wagered. 2 SD is $44. That would mean you can win $44.40 and lose $43.60. These numbers are not possible considering only a total of $40 was wagered and the most you can win is $40 if you won 100% of the hands.

3 SD is $66 and that is far greater than $40 wagered. Is the math wrong? I know that as you play more and more hands, the percentage for the SD goes down ie: 100 hands played would be 1.1/10 = 11%. 10000 hands played would be 1SD of 1.1%. 1.1% of 10000 = 110 hands. 110x$10 = $1,110. Obviously the total amount wagered is vastly greater than this (10,000 x $10 = $100,000)

Any thoughts on the calculation based on the 4 hand excercise?

None of the kind of analysis that you've done is intended to be on a sample as small as four hands. In fact, it is normally considered that you need something like 21 data points (not certain of the number) to even begin to consider the kind of mean-variance analysis that you have posted.

So, the short answer is: don't try to do this for four hands!

You can get mean/variance analysis not matter how large the sample is. However, there are some specific numbers unique to the normal distribution which may not carry over. With a normal distribution, you are within one standard deviation about 65% of the time. You are within 2 or 3 about 95% or 98% of the time. However, these numbers are not good for non-normal distributions. You may have to use the estimate of being within n standard deviations (1/2)^n of the time which is good on every distribution. So your results will not be as strong.

Reading Thorp and Griffin books I was surprised to read much about end play and various card combinations as the deck is depleted. My thoughts were that the authors were trying to get a grasp on the tails of the distribution. Similar to trying to find convergence between classical and quantimum theory (continuous versus discrete). Anyway I am sure in the derevaton of a normal distribution many assumptions are made as one approaches an infinite set of occurances. Most likely for small subsets a different distribution should be considered or many of the terms used in the normal distribution should be considered that are not considered in the assumed infinite set.

The normal distribution or a similar distribution is used in the Black Scholes formula, this distribution although not ideal is used widely in option pricing but again there are issues with the tails so I am told.