$n$ 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?
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It makes no sense to talk about 'the standard deviation for a probability'. @Remy has answered correctly (& very clearly). In case you also need to compute an approximate probability using his Answ, I left a very brief note with his Answer. Pls ask a question in a comment if you need more. – BruceET Mar 11 '18 at 09:18
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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
$$\sigma=\sqrt{n\cdot\left(\frac{(x+1)\cdot(2x+1)}{6}-\left(\frac{x+1}{2}\right)^2\right)}$$
Remy
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Thanks, I wasn't quite sure what he meant by "standard deviation for the probability of the sums" but I'm pretty sure he did not mean to say that. – Remy Mar 11 '18 at 09:04
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Nice presentation (+1). The question mentions computing probability. Although it makes no sense to "find the standard deviation for a probability." one can sometimes use the mean $\mu = E(S) =E(\sum_i X_i)$ along with your $\sigma$ to approximate probabilities such as $P(S < 10)$. By the CLT $S$ is nearly normal for surprisingly small $n.$ For example, rolling a std die with $s=6$ sides $n=10$ times, $S$ is very nearly normal. Then use $Z = (S-\mu)/\sigma$ ~ NORM(0,1) to approx $P(S < 10) = P[Z < (10-\mu)/\sigma)] .$ In case finding probab is the goal and this is too brief, OP can ask quest. – BruceET Mar 11 '18 at 09:07
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If it makes it easier, I have a simplified version of the standard deviation equation: $$\sqrt{\frac{n(x^2-1)}{12}}$$
cjferes
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