# Exploding dice

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It is too warm to do any actual work in Sweden today, so I instead sat in my apartment, playing Classic Battletech with myself, and after a while; I thought that since I am an adult and can have any imaginary friends I like, that I wanted to incorporate some Mechwarrior (the accompanying role-playing game to Classic Battletech) characters to create a more interesting environment to lose time in.

One of the characteristics of the Mechwarrior game is that it uses 2d10 (two ten-sided dice) for almost anything, but with the special rule that any 10 is rolled again, adding to the total. For example, if I roll a 4 and a 7, my result is 11, but if I roll a 6 and a 10, I get to roll once more, maybe a result of 3, for a total of 19. You can continue rolling dice for a long while, in theory, an infinitely long while.

## But what is the mean value of such a roll?

We'll reduce the 2d10 case to a simple d10 (one ten-sided die), because the two dice are independent of each other. So, to calculate the mean, we just have to sum up each possible value, multiplied by its probability, simple.

$$\frac{1 + … + 9}{10} + \frac{11 + … + 19}{100} + \frac{21 + … + 29}{1000} + …$$

Except that sum is not really that simple, and not really that general, we can simplify it a bit more, each addend for example, if we are using $$n$$-sided dice, and we are in the $$i$$-th explosion.

$$\frac{k \times n \times (n - 1) + n(n 1) / 2}{n^(k + 1)} = \frac{(n - 1) \times (k + 1 / 2)}{n^k}$$

Then we can build the sum,

$$\sum_{k = 0}^{\infty}{\frac{(n - 1) \times (k + 1 / 2)}{n^k}}$$

and we can calculate some prefix sums for $$n$$ = 10, to some arbitrary precision,

$$i$$ $$\sum$$
0 4.50000
1 5.85000
2 6.07500
3 6.10650
4 6.11055
5 6.11105
6 6.11110

and we see that it seems to tend towards $$\frac{55}{9}$$, or $$\frac{10}{9}$$ times better than a normal die. In the general case, for $$n > 1$$,

$$\frac{n(n + 1)}{2(n - 1)}$$

or $$\frac{n}{n - 1}$$ times better than the regular, non-exploding die.

## Comparing to D20

Another common system, with similar qualities is D20, which uses a single d20 dice, but with the special property that a 20 always succeeds. Let us compare the two systems, based on which TN (target-number) that you need to succeed.

In both systems, the lowest roll possible, always fails.

TN d20 2d10
2 95.0% 99.0%
3 90.0% 99.0%
4 85.0% 97.0%
5 80.0% 94.0%
6 75.0% 90.0%
7 70.0% 85.0%
8 65.0% 79.0%
9 60.0% 72.0%
10 55.0% 64.0%
11 50.0% 55.0%
12 45.0% 47.0%
13 40.0% 39.8%
14 35.0% 33.4%
15 30.0% 27.8%
16 25.0% 23.0%
17 20.0% 19.0%
18 15.0% 15.8%
19 10.0% 13.4%
20 5.0% 11.8%
21 5.0% 10.0%
22 5.0% 8.4%
23 5.0% 7.0%
24 5.0% 5.7%
25 5.0% 4.6%
26 5.0% 3.7%
27 5.0% 3.0%
28 5.0% 2.4%
29 5.0% 2.0%
30 5.0% 1.7%

The most interesting parts about the table, is around TN 13 where the automatic success for a single exploding die, which means that the probabilities go a bit funny there, something to watch out for when designing a game with exploding die. The same thing happens higher up in the table as well, but there it doesn't make much difference.

## Conclusion

In the end, if you want your rolls to be average more often, fail less often, but with a quite exciting system for heroic success, use the exploding dice system. If you want a system that is easer to calculate the probabilities on, use a single-die system with automatic success.

I calculated everything by hand, so be a bit wary about my calculations. I only did some quick checking with numerics the result ended up similar.