I want to know what dice to roll to get a given probability distribution(mainly normal distributions but exponential distribution would also be helpful).

I want a function $f$ so that

$$N(\mu,\sigma)\approx f(n_2, n_4, n_6,n_8,n_{10},...,n_s) $$

where $n_s$ are the number of dice with $s$ sides. I know that the central limit theorem says that the sum of stochastic variables from the same distribution approximates a normal distribution. And the sum of stochastic variables from different normal distributions gives a new normal distribution with $\mu_{sum}=\mu_1+\mu_2$ and $\sigma_{sum}²=\sigma_1²+\sigma_2²$. And with some basic statistics from this answer it gives

$$\mu=\sum\limits_{i=1}^s n_i\frac{i+1}{2}$$ and $$\sigma²=\sum\limits_{i=1}^s n_i\frac{i²-1}{12}$$.

This is how far I've gotten, I don't know how to invert this. Like for example, what function would I use to choose dice to approximate $N(60,15)$? The mean value can allways be adjusted with just adding and subtracting, the main issue is the standard deviation.

For context: I'm thinking about an idea for a pen and paper roleplaying game system where the probabilities of success are given in terms of the distribution and then the method of obtaining a stochastic variable from that distribution is up to the players. The function $f$ would give one group of methods.

  • Suggestion: For a continuous uniform random variable $U$ with values in $[0,1]$, and for a cumulative distribution function $F$, the random variable $F^{-1}(U)$ has $F$ as CDF. You could approximate $U$ with a dice by dividing the outcome by $6$. – Julian Wergieluk May 15 '15 at 12:29

After investigation I've realized that I've made things unnecessarily complicated.

If $X$ is the sum of enough n s-sided dice rolls to that the central limit theorem can be used then $X\in N(\mu_d,\sigma_d²)$ which means that $X*k+m\in N(\mu_d*k+m,\sigma_d²*k²)$. Index $d$ indicates that the value is of the actual distribution of the dice and not the one it approximates. $\mu_d$ and $\sigma_d²$ are given by the equations in the question.

\begin{cases} \mu=k*n*\frac{s+1}{2}+m\\ \sigma²=k²*n²*\frac{s²-1}{12} \end{cases}

which gives

\begin{cases} k=\sqrt{\frac{12\sigma²}{n²(s²-1)}}\\ m=\mu-\sqrt{\frac{3\sigma²(s+1)}{(s-1)}} \end{cases}

So if the distribution to approximate is $N(60,15)$ I would roll for example $n=10$ dice with $s=10$ sides and insert $\sigma²=15$ and $\mu=60$ into the equations to obtain $k=0.4264$ and $m=52.58$. Then I would multiply the sum of the roll with $k$ and then add $m$.

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