Standard Deviation for sums of fair dice given the number of dice, and the number of sides on each die

Question

$n$ fair die are rolled, and each dice has $x$ sides, with the numbers on the sides going from $1$ to $x$, and with each side having a different number from the other sides. how do I figure out the standard deviation for the probability of the sums of the numbers that the die land on?

Answer 1

I am assuming you mean the standard deviation of the sum of $n$ dice, each with $x$ sides.

Let the random variables $X_1,X_2,X_3,..,X_n$ denote the results on the first, second, third$,...,n^{th}$ dice. The $X_i$ are independent. Because of this, we can simply take

$$\begin{align*} Var(X_1+X_2+X_3+...+X_n) &=Var(X_1)+Var(X_2)+Var(X_3)+...+Var(X_n)\\\\ &=nVar(X_1) \end{align*}$$

since the $X_i$ are identically distributed.

To calculate the variance of $X_1$, we calculate $E(X_1^2)-E(X_1)^2$

We have $$\begin{align*} E(X_1^2) &=\frac{1}{x}\left(1^2+2^2+\cdots+x^2\right)\\\\ &=\frac{1}{x}\left(\frac{x\cdot(x+1)\cdot(2x+1)}{6}\right)\\\\ &=\frac{(x+1)\cdot(2x+1)}{6} \end{align*}$$

and

$$\begin{align*} E(X_1) &=\frac{1}{x}\left(1+2+\cdots+x\right)\\\\ &=\frac{1}{x}\left(\frac{x\cdot (x+1)}{2}\right)\\\\ &=\frac{x+1}{2} \end{align*}$$

Thus

$$\begin{align*} Var(X_1+...+X_n) &=nVar(X_1)\\\\ &=n\cdot\left(\frac{(x+1)\cdot(2x+1)}{6}-\left(\frac{x+1}{2}\right)^2\right) \end{align*}$$

Finally, we take the square root to obtain the standard deviation

$$\sigma=\sqrt{n\cdot\left(\frac{(x+1)\cdot(2x+1)}{6}-\left(\frac{x+1}{2}\right)^2\right)}$$

Answer 2

If it makes it easier, I have a simplified version of the standard deviation equation: $$\sqrt{\frac{n(x^2-1)}{12}}$$