Normal distribution with dice

Question

I'm wondering how to control the normal distribution that comes from summing dice rolls only using different numbers of dice, different combination of types of dice (d4, d6, d8, d10, d12, d20) and simple math (+,-,•,/ and perhaps ^)?

I know that if you only use one type of die the mean value and the standard deviation increases linearly with the number of dice with a constant that depends on the number of sides on the dice.

I also realize that I can shift the distribution with plus and minus. But I don't know if there is a equally simple way to control the standard deviation.

In summary: how do I control the standard deviation of a normal probability distribution only using different combination of dice and simple math?

Answer 1

Here is a blogpost that gives you an overview of the distributions of summed dice as the number of dice increases. In short, as the number increases, it becomes increasingly well modelled by the normal distribution.

However, there is a small gap between the analytic solution that we get for the probability distribution of dice and the normal distribution. My intuition is that this will not be important - because rolling the vast number of dice that would allow statistical tests to detect this gap is probably not something you will do in this game.

If you are happy using d6s then a roll of $n$d6 is relatively well approximated by

$ \mu = \frac{6n+1}{2}$, $ \sigma = 0.175\sqrt n $

when $n$ gets sufficiently large. I imagine I would use $n=4$ or 5.

Answer 2

I think you are using words that you don't understand. The sum of multiple dice does not follow a normal distribution. See my explanation for the sum of two $n$-sided dice case in my answer here.

Supposing you remove the word "normal," you can see how the mean and standard deviation interact by the linearity of expected value.

The expected value of a discrete random variable, $X$, is given as $E[X] = \sum\limits_{x\in\mathcal{M}} x\cdot Pr(X=x)$ (where $\mathcal{M}$ is the sample space)

The expected value operator is linear. I.e., $E[\alpha_1 X_1 + \alpha_2 X_2] = \alpha_1 E[X_1] + \alpha_2 E[X_2]$

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The mean of a random variable $X$ is defined to be $E[X]$

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Let $X$ be a random variable described by the result of a die throw. If $X$ is a fair $n$-sided die with sides labeled $1,2,\dots,n$, then the mean, $\mu = E[X]$ is equal to $\frac{n+1}{2}$.

The variance of $X$ in this case is $\frac{n^2-1}{12}$.

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Let $X$ be a random variable described by the result of a die throw. If $X$ is a fair $n$-sided die with sides labeled $1,2,\dots,n$, then the expected value of $\frac{1}{X}$ is $H_n\cdot \frac{1}{n}$ where $H_n$ is the $n^{th}$ Harmonic Number

Suppose our random variable, $Y$, is a linear combination of random variables. Let $X_1$ be the random variable described by the result of an $a_1$-sided die, $X_2$ be the random variable described by the result of an $a_2$-sided die, ..., $X_k$ be the random variable described by the result of an $a_k$-sided die.

Let $Y=\alpha_1 X_1 + \alpha_2 X_2 + \dots \alpha_k X_k$

Then $E[Y] = E[\alpha_1 X_1 + \dots + \alpha_k X_k] = \alpha_1 E[X_1] + \dots + \alpha_k E[X_k] = \alpha_1\frac{a_1+1}{2} + \dots + \alpha_k\frac{a_k+1}{2}$

Note, that for the case where $X_1, X_2,\dots, X_k$ all describe the result of rolling an $n$-sided die, you get $E[X_1+X_2+\dots+X_k] = E[X_1]+\dots+E[X_k] = \frac{n+1}{2}+\dots + \frac{n+1}{2} = k\cdot \frac{n+1}{2}$, matching your intuition.

When you use multiplication, $E[XY] = E[X]E[Y]-Cov(X,Y)$, however noting that when $X$ and $Y$ are independent (which they will be when $X$ and $Y$ are for different dice [not necessarily different types of dice]) it implies that $Cov(X,Y)=0$ and that $E[X,Y] = E[X]E[Y]$

This gives us a way to calculate the mean of any combination of results of dice and $+,-,\times$. Noting that division is just multiplication by the multiplicative inverse of the other, we get that $E[X/Y] = E[X\cdot \frac{1}{Y}] = E[X]E[\frac{1}{Y}]$ and the term on the right will rely on the $n^{th}$ harmonic number.

I see no easy way of giving a general rule to use when using exponentiation of one random variable to the power of another apart from referring you to calculate it directly using the definition of expected value. For finding the expected value of a random variable to an integer power, I refer you to the definition of Moments and to Generalized Harmonic Numbers as well as for the specific case of $E[X^2]$ being the $n^{th}$ Square Pyramidal Number divided by $n$.

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For standard deviation, $\sigma$, of a random variable $X$ note that $\sigma^2=Var[X] = E[(X-E[X])^2] = E[X^2]- E[X]^2$

With all of this information, you should have the tools to calculate whatever you are looking for.

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Example

You have three fair dice, a d4, a d6, and a d8. Let $x,y,z$ represent the results of throwing the d4, d6, and d8 respectively. Suppose that you are playing a game where you get to attack and deal $z+xy-2$ damage. What is the expected amount of damage and what is the standard deviation?

The average amount of damage is: $E[Z+XY-2] = E[Z]+E[X]E[Y] - 2 = \frac{8+1}{2} + \frac{4+1}{2}\cdot\frac{6+1}{2} - 2 =4.5 + 3.5\cdot 2.5 - 2 = 11.25$

The variance is: $Var[Z+XY-2] = E[(Z+XY-2)^2] - (E[Z+XY-2])^2 = E[Z^2 + X^2Y^2 + 4 + 2XYZ - 4Z -4XY] - 11.25^2$

$= E[Z^2] +E[X^2]E[Y^2]+4+2E[X]E[Y]E[Z] - 4E[Z] - 4E[X]E[Y] - 11.25^2$

$=\frac{(8+1)(16+1)}{6} + \frac{(4+1)(8+1)}{6}\frac{(6+1)(12+1)}{6}+4+2\cdot 2.5\cdot 3.5\cdot 4.5 - 4\cdot 4.5 - 4\cdot 2.5\cdot 3.5 - 11.25^2$

$=42.4375$

The standard deviation then is $\sqrt{42.4375}\approx 6.51$.