Take the 5 Platonic solids from a set of Dungeons&Dragons dice. These consist of a 4-sided, 6-sided (conventional), 8-sided, 12-sided, and 20-sided dice. All start at the number 1 and count upwards by 1 to their total.

Roll them all, and take their sum (minimum sum is 5, max is 50). Do so multiple times. What is the distribution?

Obviously they will tend towards the low end, since there are more lower numbers than higher. But will there be notable inflection points at each boundary of the individual die?

[Edit: Apparently, what seemed obvious isn’t. According to one of the commentators, the average is (5+50)/2=27.5. I wasn’t expecting this. I’d still like to see a graph.]

**Answer**

I wouldn’t want to do it algebraically, but you can calculate the pmf simply enough (it’s just convolution, which is *really* easy in a spreadsheet).

I calculated these in a spreadsheet*:

```
i n(i) 100 p(i)
5 1 0.0022
6 5 0.0109
7 15 0.0326
8 35 0.0760
9 69 0.1497
10 121 0.2626
11 194 0.4210
12 290 0.6293
13 409 0.8876
14 549 1.1914
15 707 1.5343
16 879 1.9076
17 1060 2.3003
18 1244 2.6997
19 1425 3.0924
20 1597 3.4657
21 1755 3.8086
22 1895 4.1124
23 2014 4.3707
24 2110 4.5790
25 2182 4.7352
26 2230 4.8394
27 2254 4.8915
28 2254 4.8915
29 2230 4.8394
30 2182 4.7352
31 2110 4.5790
32 2014 4.3707
33 1895 4.1124
34 1755 3.8086
35 1597 3.4657
36 1425 3.0924
37 1244 2.6997
38 1060 2.3003
39 879 1.9076
40 707 1.5343
41 549 1.1914
42 409 0.8876
43 290 0.6293
44 194 0.4210
45 121 0.2626
46 69 0.1497
47 35 0.0760
48 15 0.0326
49 5 0.0109
50 1 0.0022
```

Here $n(i)$ is the number of ways of getting each total $i$; $p(i)$ is the probability, where $p(i) = n(i)/46080$. The most likely outcomes happen less than 5% of the time.

The y-axis is probability expressed as a percentage.

* The method I used is similar to the procedure outlined here, though the exact mechanics involved in setting it up change as user interface details change (that post is about 5 years old now though I updated it about a year ago). And I used a different package this time

(I did it in LibreOffice’s Calc this time). Still, that’s the gist of it.

**Attribution***Source : Link , Question Author : Marcos , Answer Author : Community*