Jens Nockert

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.