Exploding dice

The Savage Worlds rule system for tabletop role-playing games (best known for its usage in Deadlands) includes a unique mechanic: exploding dice. If a player rolls the highest possible number on a die, it is said to explode, and they can roll it again and add together the rolls. A die may explode multiple times in succession, leading to unusually large scores — for example, a 4-sided die has a 1 in 16 chance of exploding twice and scoring at least 9.

Expected scores

The expected score on an $s$-sided exploding die is:

$$\mathrm{E}\left(D_s\right)=\frac{s(s+1)}{2(s-1)}$$

The derivation of this formula is given at the bottom of this page.

Using this formula we can calculate the expected scores on exploding dice (shown to two decimal places), and see how they differ from the expected scores on normal dice:

The target number paradox

The fact that dice with fewer sides are more likely to explode produces a counterintuitive effect I call the target number paradox: an ($s$ − 2)-sided die is more likely to succeed in meeting a target number of $s$ than an $s$-sided die (provided $s$ > 4). For example, a 4-side die has a 3 in 16 chance (about 19%) of scoring at least 6, while a 6-sided die only has a 1 in 6 chance (about 17%).

To prove this, we first observe that for an ($s$ − 2)-sided die to score at least $s$ it must first explode and then roll anything other than a 1. This occurs with a probability $p$ given by:

$$p=\left(\frac{1}{s-2}\right)\left(\frac{s-3}{s-2}\right)$$

We then subtract the probability, $q$, of an $s$-sided die scoring at least $s$:

$$p-q=\left(\frac{1}{s-2}\right)\left(\frac{s-3}{s-2}\right)-\frac{1}{s}$$

$$=\frac{s(s-3)}{s{(s-2)}^2}-\frac{{(s-2)}^2}{s{(s-2)}^2}$$

$$=\frac{s-4}{s{(s-2)}^2}$$

This value is positive — meaning the ($s$ − 2)-sided die has a higher chance of success than the $s$-sided die — whenever $s$ > 4.

Deriving the expected score formula

Let the random variable $D_s$ represent the final score rolled on an $s$-sided die. We want to determine the formula for the expected value $\mathrm{E}\left(D_s\right)$.

First we determine the formula for the expected value after a specific number of explosions. Let the random variable $D_{s,n}$ represent the score rolled on an $s$-sided die that has exploded exactly $n$ times. We can see that:

$$\mathrm{E}\left(D_{s,n}\right)=sn+\frac{s}{2}=s(n+\frac{1}{2})$$

Multiplying these values by the probability of a die exploding $n$ times and summing over the possible values of $n$ gives us the expected score formula:

$$\mathrm{E}\left(D_s\right)=\sum _{n=0}^{\infty }\frac{s-1}{s}{\left(\frac{1}{s}\right)}^n\mathrm{E}\left(D_{s,n}\right)$$

$$=\sum _{n=0}^{\infty }\frac{s-1}{s}{\left(\frac{1}{s}\right)}^{n}s(n+\frac{1}{2})$$

$$=(s-1)\sum _{n=0}^{\infty }{\left(\frac{1}{s}\right)}^n(n+\frac{1}{2})$$

$$=(s-1)(\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{s}\right)}^n+\sum _{n=0}^{\infty }n{\left(\frac{1}{s}\right)}^n)$$

The two summations in this expression sum a geometric series and its derivative. We replace the summations with the well-known formulas for their values:

$$=(s-1)(\frac{1}{2}\left(\frac{s}{s-1}\right)+\frac{1}{s}{\left(\frac{s}{s-1}\right)}^2)$$

$$=\frac{s}{2}+\frac{s}{s-1}$$

$$=\frac{s(s+1)}{2(s-1)}$$