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-4

According to Dave Arneson, D&D uses a d20 because he liked the die, rather than for any mathematical reason. I think this means that trying to faithfully duplicate d20 probability with d6s is over-thinking the basic idea of "choose numbers by throwing dice," especially if doing so involves complicated math. Here is a simpler way to get results from 1 to ...


1

An exhaustive generation of the possibilities can be done. It is not 100% clear what you mean by "dice totalling at least 5", I've chosen to interpret this as "sum up to 5 dice" and that means one would have at least one success with a pool of 5 or more dice (and at least 2 successes with a pool of 10 dice). In principle, the algorithm is simple: # ...


6

While @AgentPaper has provided a very nice formula, reality is slightly different. I used a small python script to calculate the average number of results over 1000000 tries. import random, math rep = 100000 data =[] for n in range(1,11): result = [] for _ in range(rep): R = sorted([random.randint(1,6) for __ in range(n)]) S = ...


6

This is actually pretty simple to calculate. First, there is a 1/3 chance of each dice being a 5 or 6. So you will get (dice pool size) / 3 successes from that to start. After that, you need to calculate how many successes you get from adding other dice together. Since every result that isn't a 5 or 6 is going to be a 1, 2, 3, or 4, that means that your ...


-2

I was taught two methods (way back in the 80's when I started out). One is the number on the bottom of the face (which will the same on all three showing faces) or you can pick it up and look at the bottom face and the result is the number NOT listed. In all cases the number would be the same. :)



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