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When trying to find how to simulate rolling a variable amount of dice with a variable but unique number of sides, I read that the mean is $\dfrac{sides+1}{2}$, and that the standard deviation is $\sqrt{\dfrac{quantity\times(sides^2-1)}{12}}$.

I doubt that the $12$ comes from the formula because it seems strongly linked with the examples of using two six-sided dice.

Is the formula for the standard deviation correct? If not what is it?

Olivier Grégoire
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  • I think it would be more helpful if you described your rolls in words as I don't think that many people in this community are programmers – John K Apr 21 '17 at 16:05
  • @TheBro21 Is "how do I compute the standard deviation of rolling 100 times a 6-sided die" better? – Olivier Grégoire Apr 21 '17 at 16:08
  • I don't think there's too big of a problem with your title. Instead, replace your code with something more non-programmer understandable – John K Apr 21 '17 at 16:10
  • This is probably just my inexperience talking, but what quantity does "quantity" refer to here? Is it just the number of dice rolled? – Newbyte Sep 27 '23 at 09:24

3 Answers3

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Note that

$$Var[X] = E[X^2] - E[X]^2 = \sum_{k=0}^n k^2 \cdot P(X=k) - \left [ \sum_{k=0}^n k \cdot P(X=k) \right ]^2$$

For a single $s$-sided die, that implies:

$$Var[X] = \frac{1}{s}\left(1^2 + 2^2 + 3^2 + ... s^2\right) - \left(\frac{1}{s}(1 + 2 + 3 + ... + s)\right)^2$$

This simplifies to

$$Var[X] = \frac{1}{s} \cdot \frac{s(s+1)(2s+1)}{6} - \left(\frac{1}{s} \cdot\frac{s(s+1)}{2}\right)^2$$

or

$$Var[X] = \frac{s^2 - 1}{12}$$

So the $12$ is just part of the equation. It comes from the fact that the sum of squares equation has denominator $6$, and the sum of consecutive integers equation has denominator $2$ (which gets squared to $4$). And $\text{lcm}(6, 4) = 12$.

Marcus Andrews
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The formula is correct. The 12 comes from $$\sum_{k=1}^n \frac1{n} \left(k - \frac{n+1}2\right)^2 = \frac1{12} (n^2-1) $$ Where $\frac{n+1}2$ is the mean and k goes over the possible outcomes (result of a roll can be from 1 to number of faces, $n$), each with probability $\frac1{n}$. This formula is the definition of variance for one single roll.

dioid
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$X$ is a random variable that represents our $n$ sided die.

Frequence distibution $f(x) = \begin {cases} \frac 1n & x\in \mathbb N, 1\le x \le n\\ 0 & \text {otherwise} \end {cases}$

The mean of a n sided die $E[X] =\frac 1n \sum_\limits{i=1}^n i = \frac 12 (n+1)$

the variance
$E[X^2] - E[X]^2 = \frac 1n \sum_\limits{i=1}^n i^2 - \frac 14 (n+1)^2\\ \frac 16 (2n+1)(n+1) - \frac 14 (n+1)^2\\ \frac 1{12} (n^2-1)$

the $\frac 1{12}$ is intrinsic to the math

user317176
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