In the previous post I showed how, when rolling a set of identical dice, we can calculate what sum is most common by multiplying the expected value of one die, or by finding the middle value between the lowest and highest possible sums. I was curious if the same techniques work when rolling dice that have different numbers of sides. It seems to, but with a complication.

Rather than rolling two six-sided dice, let's consider rolling one six-sided die and one ten-sided die. The expected values are 3.5 and 5.5, respectively, which sum to 9. Nine is also the middle value between the highest and lowest sums possible (2 and 16). And, working out the table of sums, it turns out 9 is indeed a most common sum when rolling these two dice.

But, unlike when rolling identical dice, more than one sum may be most common. When rolling six- and ten-sided dice, sums of 7, 8, 9, 10, and 11 all share equal likelihood. To see why, look at the sum table:

+ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |

2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |

4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |

5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |

The unequal dimensions of the table create more than one longest diagonal, giving the sums on those diagonals equal probability.

Something similar happens when rolling more than two dice, though it's tougher to visualize. For six-, ten-, and twenty-sided dice, the most common sums are all the integers from 15 to 22.

However, there do still exist scenarios where one sum can be more common than all others. It doesn't seem to happen at all when rolling only two dice, but when rolling three it appears to happen when exactly one die has an odd number of sides and the die with the most sides has fewer sides than the sum of the sides of the other two dice. To illustrate, here are some example dice combinations that have a single most common sum:

- 6, 7, 8
- 6, 7, 10
- 6, 7, 12
- 6, 8, 9
- 6, 8, 11
- 6, 8, 13
- 6, 9, 10
- 6, 9, 12
- 6, 9, 14
- 6, 10, 11
- 6, 10, 13
- 6, 10, 15
- 6, 11, 12
- 6, 11, 14
- 6, 11, 16
- 6, 12, 13
- 6, 12, 15
- 6, 12, 17

With four dice, it seems exactly two dice have to be odd, and the die with the most sides has to have fewer sides than the sum of the sides of the other three dice:

- 6, 7, 8, 9
- 6, 7, 8, 11
- 6, 7, 8, 13
- 6, 7, 8, 15
- 6, 7, 8, 17
- 6, 7, 8, 19
- 6, 7, 9, 10
- 6, 7, 9, 12
- 6, 7, 9, 14
- 6, 7, 9, 16
- 6, 7, 9, 18
- 6, 7, 9, 20
- 6, 7, 10, 11
- 6, 7, 10, 13
- 6, 7, 10, 15
- 6, 7, 10, 17
- 6, 7, 10, 19
- 6, 7, 10, 21

Or at least, that's how it looks. I haven't probed the math to see if those constraints are inherent or simply an artifact of the combinations I happened to generate. If you have any insight on this behavior, feel free to email me.