Sum of roll of N dice

[Allen]

Supposing you know N in advance, it's easy enough to precalculate the probability distribution (as started above).. Then add up the individual probabilities to create an empirical cumulative distribution. Then you can generate 1 random number (from 0 to 1) and get the exact sum of dice just by a lookup table.

For example, if your random number is .855666 go down the cumulative pdr until you find a number greater than the rolled number.. that's your sum.

r-nomial coefficients here:https://oeis.org/A063260

sum prob cumulative probability
4 : 1/1296 = 0.000771604938272 0.000771604938272
5 : 4/1296 = 0.00308641975309 0.00385802469136
6 : 10/1296 = 0.00771604938272 0.0115740740741
7 : 20/1296 = 0.0154320987654 0.0270061728395
8 : 35/1296 = 0.0270061728395 0.054012345679
9 : 56/1296 = 0.0432098765432 0.0972222222222
10 : 80/1296 = 0.0617283950617 0.158950617284
11 : 104/1296 = 0.0802469135802 0.239197530864
12 : 125/1296 = 0.096450617284 0.335648148148
13 : 140/1296 = 0.108024691358 0.443672839506
14 : 146/1296 = 0.112654320988 0.556327160494
15 : 140/1296 = 0.108024691358 0.664351851852
16 : 125/1296 = 0.096450617284 0.760802469136
17 : 104/1296 = 0.0802469135802 0.841049382716
18 <---------------------- 0.855666
18 : 80/1296 = 0.0617283950617 0.902777777778
19 : 56/1296 = 0.0432098765432 0.945987654321
20 : 35/1296 = 0.0270061728395 0.97299382716
21 : 20/1296 = 0.0154320987654 0.988425925926
22 : 10/1296 = 0.00771604938272 0.996141975309
23 : 4/1296 = 0.00308641975309 0.999228395062
24 : 1/1296 = 0.000771604938272 1.0