*Fireball damage: its calculation and its distribution.*

The spell components are ready:

Fortunately, as a wizard I am familiar with the arcane rule of tens. Prestidigitation ensues and the hidden number is revealed:

Apropos of nothing, I timed myself on a few more rolls:

- 65 hit points: 22.2s
- 63 hit points: 19.6s
- 56 hit points: 28.3s
- 63 hit points: 28.3s
- 62 hit points: 24.9s

An average of 24.7s spent to get a number that was usually within a couple hit points of 63.

## A Shortcut?

When younger I was tempted to roll a single d6, multiply it by 18, and be done with it. That is, use a d6 × 18 roll in place of the 18d6. However, even then I couldn't escape the feeling that this was not an equivalent roll. I could see that the latter method always produces a multiple of 18, but I wondered if it wasn't close enough all the same.
Both methods cause 63 hit points of damage on average, so in that respect they

Plotting the probability mass distributions is a good way to show how different rolls are. Here the 18d6 distribution is in red and the d6 × 18 distribution is in blue:

The standard deviations are 7.2 and 30.7, a large difference.

According to Chebyshev's inequality, the chance of being more than

Chebyshev's inequality is a rough upper bound that works for any distribution for which we know the mean and standard deviation. If we know the distribution, we can do a summation or integration to get the exact probability; for the 18d6 distribution the exact probability is less than 5%.

What if we allow two rolls of a d6? I wrote some code which searches for an expression with the same mean and the nearest standard deviation. To keep things simple the first d6 always gets multiplied by a non-negative integer and the second d6 is taken as is. With those constraints here are the best approximations:

Here is how well the approximation works for the arch-mage; the 18d6 distribution is in red and the d6×3 + d6 + 49 in blue:

*are*the same. But if the mean is all we care about, why roll dice at all?Plotting the probability mass distributions is a good way to show how different rolls are. Here the 18d6 distribution is in red and the d6 × 18 distribution is in blue:

The standard deviations are 7.2 and 30.7, a large difference.

## Chebyshev's Inequality

At this point, I'm going digress in an attempt to make the standard deviation seem useful.According to Chebyshev's inequality, the chance of being more than

*k*standard deviations from the mean can never be greater than 1/*k*^{2}. Thus, in the case of the 18d6 distribution, the chance the value is less than 49 or more than 77 is no more than 25%. The same probability for the d6 × 18 distribution is 66⅔%.Chebyshev's inequality is a rough upper bound that works for any distribution for which we know the mean and standard deviation. If we know the distribution, we can do a summation or integration to get the exact probability; for the 18d6 distribution the exact probability is less than 5%.

## A Better Shortcut?

Back to rolling fireball damage.What if we allow two rolls of a d6? I wrote some code which searches for an expression with the same mean and the nearest standard deviation. To keep things simple the first d6 always gets multiplied by a non-negative integer and the second d6 is taken as is. With those constraints here are the best approximations:

```
```

```
1d6: d6 × 0 + d6 + 0
2d6: d6 × 1 + d6 + 0
3d6: d6 × 0 + d6 + 7
4d6: d6 × 1 + d6 + 7
5d6: d6 × 2 + d6 + 7
6d6: d6 × 1 + d6 + 14
7d6: d6 × 2 + d6 + 14
8d6: d6 × 3 + d6 + 14
9d6: d6 × 2 + d6 + 21
10d6: d6 × 3 + d6 + 21
11d6: d6 × 2 + d6 + 28
12d6: d6 × 3 + d6 + 28
13d6: d6 × 4 + d6 + 28
14d6: d6 × 3 + d6 + 35
15d6: d6 × 4 + d6 + 35
16d6: d6 × 3 + d6 + 42
17d6: d6 × 4 + d6 + 42
18d6: d6 × 3 + d6 + 49
19d6: d6 × 4 + d6 + 49
20d6: d6 × 5 + d6 + 49
21d6: d6 × 4 + d6 + 56
22d6: d6 × 5 + d6 + 56
23d6: d6 × 4 + d6 + 63
24d6: d6 × 5 + d6 + 63
```

Here is how well the approximation works for the arch-mage; the 18d6 distribution is in red and the d6×3 + d6 + 49 in blue: