Is it possible to simulate a d20 roll using smaller dice

combinatoricsdiceprobability

I'm wondering if it's possible to simulate a d20 using smaller dice, and if so, how would one figure out how many you would need?

This question provides an algorithm to do so using 2 coins with p = 1/2 and p = 1/n, but is there an algorithm that would work for multiple like dice of the same probability? eg. can you find an x such that if you roll x 6 sided dice you are left with exactly 20 distinct outcomes?

Best Answer

If you aim for a fixed number $x$ of dice, I agree with the comment of Thomas Andrews that the product of the numbers of their sides has to be a multiple of $20$. However, if the number $x$ may vary, then this algorithm provides a solution:

  1. Roll a 6-sided die with outcome $\omega_1$. Set $a_1=0$, if $\omega_1\in\{1,2,3\}$, and $a_1=10$ otherwise.

  2. Roll a 6-sided die with outcome $\omega_2$. Set $a_2=0$, if $\omega_2\in\{1,2,3\}$, and $a_2=5$ otherwise.

  3. Roll a 6-sided die with outcome $\omega_3$. Set $a_3=\omega_3$, if $\omega_3\in\{1,2,3,4,5\}$. If $\omega_3 = 6$ repeat step 3.

Eventually, the algorithm will terminate and $a = a_1 + a_2 + a_3$ will be uniformly distributed in $\{1,\dots,20\}$.

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