Variance and expected values by dices, how does addition work?

Question

I have read through some stuff and I am confused now.

If we have a fair die and we just roll once, the expected value is going to be 3,5 and the variance is 2,916.

Well, it is easy to count by one die, but if we have 8 dice (6 sides) and we are interested in the sum of the roll, then the calculation can get really messy if we start to write up the chances of every possible outcome.

Luckily, the expected value is easy to count, it stays in the middle, so in this case, it's 28.

But the variance seems to be more difficult, it seems to me because the chances of the outcomes may vary who knows how.

Is there a formula you know about to count the variance in this situation, or do I have to write up all possibilities and count the probabilities?

Thanks

Answer 1

If random variables are independent, then the variance of their sum is the sum of their variances. That's actually the main reason why standard deviation is used as a measure of dispersion rather than mean absolute deviation or something else.

(A point of confusion comes when people computed variances using Bessel's correction, dividing by $n-1$ rather than by $n$, where $n$ is the sample size. But Bessel's correction is not appropriate here, and accordingly I see that you have not used it.)

Answer 2

Let the dice be labelled $1,2,3,\dots,8$. Let random variable $X_i$ be the number showing on die $i$. Then our sum $Y$ is equal to $X_1+\cdots+X_8$.

The variance of a sum of independent random variables is the sum of the variances. Thus $Y$ has variance $\text{Var}(X_1)+\cdots +\text{Var}(X_8)$. The variances of the $X_i$ are all equal, and you know them, so you know the variance of $Y$.

Remark: The distribution of $Y$ can be found, but it is a quite messy and lengthy business. In principle we could find this distribution, and use it to compute the variance of $Y$. Luckily, there is a much easier way to do it!