Standard Deviation for sums of fair dice given the number of dice, and the number of sides on each die $n$ 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?
 A: 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
$$\sigma=\sqrt{n\cdot\left(\frac{(x+1)\cdot(2x+1)}{6}-\left(\frac{x+1}{2}\right)^2\right)}$$
