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Exploding dice, and some unexpected results for Savage Worlds

We roll an $n$-sided die, adding to the result a second roll, and a third, and so on, as many times as $n$ is rolled. So, for example, the outcome of an exploding six-sided die might be 3, or 6+2=8, or 6+6+1=13, and so on.

Over many such outcomes, what is the average number of times the die will be rolled? What is the average value of an exploding die?

The probability that the die will be rolled exactly once is the probability of rolling less than $n$: $P(1) = (n-1)/n$. The probability that there will be exactly two rolls is the probability of a roll of $n$ followed by a roll of less than $n$: $P(2) = 1/n \cdot (n-1)/n = (n-1)/n^2$. Similarly, the probability that there will be exactly three rolls is $P(3) = (n-1)/n^3$, and, in general, it is evident that $P(k) = (n-1)/n^k$ for all $k$.

The mean number of rolls is therefore $$\sum_{k=1}^\infty k P(k) = (n-1) \sum_{k=1}^\infty \frac{k}{n^k}.$$ To solve the sum, first let $x = 1/n$, so that we can express the sum as $\displaystyle S = \sum_{k=1}^\infty k x^k$. Then, with a little algebra, $$\begin{array}{rll} S & = & x + 2x^2 + 3x^3 + ... & \\ xS & = & x^2 + 2x^3 + ... & \text{(multiply by x)} \\ (1-x)S & = & x - x^2 - x^3 - ... & \text{(subtract from the previous equation)} \\ x(1-x)S & = & x^2 - x^3 - ... & \text{(repeat...)} \\ (1-x)^2 S & = & x. & \end{array}$$

Hence \begin{align*} S & = \frac{x}{(1-x)^2} \\ & = \frac{1/n}{((n-1)/n)^2} \\ & = \frac{n}{(n-1)^2}. \end{align*}

Finally, the mean is equal to $(n-1) S$, or $$\text{average number of rolls }=\frac{n}{n-1}$$.

Now we can find the average value of an exploding roll: this is simply the mean of a single roll of the die, $(n+1)/2$, multiplied by the mean number of rolls, $n/(n-1)$; i.e., $$\text{average value }=\frac{n(n+1)}{2(n-1)}$$.

Here is a table showing the average number of rolls, and the average total value obtained, for each of the common dice:

Die Average
d4 1.3333334 3.3333333
d6 1.2 4.2
d8 1.1428572 5.142857
d10 1.1111112 6.111111
d12 1.0909091 7.090909
d20 1.0526316 11.052631

As we would expect, the average roll increases as the number of sides increases, though the likelihood of re-rolling decreases. So, while a smaller die is more likely to explode, the explosion won't contribute as much as it would for a larger die, and the average result is smaller.

In some roleplaying game systems, such as Savage Worlds, a player will attempt to meet or exceed a specified target number with an exploding roll. Based on the averages, we would assume that using a die with more sides would make it easier to succeed, and for most target numbers we would be correct. There are certain cases, however, where the assumption doesn't hold.

Consider, for example, a target number of 6, and the probabilities of meeting or exceeding that value with an exploding d4 and d6.

Clearly, the probability of succeeding with a d6 is 1/6; this is simply the probability of rolling a 6. In order to succeed with a d4, however, one must roll a 4 (probability 1/4) followed by a 2, 3, or 4 (probability 3/4); the probability of success is thus 3/16, which is over two percent better than the d6.

Similarly, with a d6 and a d8 against a target number of 8, we find the respective probabilities of success are 5/36 and 1/8, and again the smaller die has an advantage.

In fact, when any of the standard dice is compared with the next larger die, except in the case of the d12 and the d20 where the gap is large enough to offset the effect, the smaller die will have an advantage if the target number is equal to the number of sides on the larger die. In systems where a larger die corresponds to a higher level of skill, therefore, the probabilities won't always support the assumption that a higher skill should correspond to a greater likelihood of success.

Here is a graph of the probabilities of rolling x or higher for exploding versions of the standard dice. Note that the d4 curve overtakes that of d6 when x = 6, the d6 curve overtakes that of the d8 when x = 8, and so on; only the d12 remains at all times below the next larger die (the d20).

Using zero-based dice ($n$-sided dice numbered zero to $n-1$) gets rid of the horizontal steps in the graph and fixes the problem:

Note that for zero-based exploding dice, the average total result will be equal to half the number of sides, i.e., slightly less than for normally-numbered dice. For example, an exploding 0-9 die will average 10/2 = 5 instead of about 6.111. Adding 1 to the final result would make the average results very close to those of the normal dice while preserving the shapes of the probability curves.