# Exploding dice

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.