Standard deviation and the classical approach to probabilities?

Question

So I was taking this statistics class, and apparently, to determine population standard deviation I have to know the population mean, the number of members in the population, and the value of each of them.

That's just fine for theoretical work, and supercomputers, but suppose I'm playing dungeons and dragons, and I want to know the standard deviation for a 3d6 roll. I can quickly see that the mean is 10.5, and that all values will range between 3 and 18. With a little calculation, I can determine that there would be 216 permutations, which would make up the population... but do I really have to subtract the mean from each of these 216 values, square the result, and add them together before dividing by the number of permutations? Since I know what every single number is going to be, isn't there some easier way to get the right number without pushing over a thousand separate buttons on my calculator?

Sorry James, I tried that. Problem is the answer changes depending on the probability. Since each value has a different probability I'd rather use some other way. Also, I'd prefer to obtain the population standard deviation, since the population can easily be known based on the type and number of dice being used.

Answer 1

I think you forgot about the Central Limit Theorem.

For rolling one die, the mean is 3.5 and by exact calculation the variance is 2.91666...

For N dice, the mean is N * 3.5, and the central limit theorem says that the distribution tends to Gaussian with a variance of N * 2.91666... as N tends to infinity.

So a first approximation for 3 dice is mean = 10.5, std dev = sqrt(8.75) = 2.96.

Answer 2

yes there is. by using the binomial distribution.
If you knoe N and the probability you can figure out all the rest.
N (you know what n is, i am not familiar with DD so excuse me)
P (you know what P is)
Q= (1-P)
N*P = Mu
X Bar (sample )
SD = (sq root of N*P*Q)

The Xbar is if you want to test the significance of a sample against Mu
X (bar)- MU/SD then check z to z table
please note though, the farther the probability moves away from .5 inthe binomial distribution the less reliable it becomes. However, I think this will work fine for your purposes. Good luck