# What is the distribution for various polyhedral dice all rolled at once?

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.]

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.