[RPG] Dealing with large amount of dice rolls for a single damage roll without making it drag on

There's a built-in explode function, but you have to use it before doing your other processing. Specifically, use FOOdBAR to have BAR evaluated as a sequence and used to generate FOO "dice" with arbitrary side counts and values. Here, DICEd([explode d6]) does the job, turning an exploded d6 into the base for the actual roll. (explode doesn't separate dice that you give it — you have to do that, or you'll get a 5d6 roll that only explodes, once, on 30.)

You can't compare sequences to a threshold to get a list or sum of values against that threshold — you'll just get the sequence summed and checked against the threshold. Instead, use [count VALUES in SEQUENCE] for each of the valid values (3-6). And as it turns out, since every 6 triggers a new die and all those dice count, you can just double-count 6s once exploded. So really that should be [count {3..6, 6} in ROLL].

The results, as best I can manage, aren't pretty. Each explosion will either fumble harder (1), make no progress against fumbling (2-5), or break even and get a new chance to do the same thing (6), and the more dice you start with, the more chances for explosions. There is no provision for fumbles to peter out in explosions, as one would expect from a great success; the only question is whether the chain ends with a 1 or not.

As other answers have stated, this method is equivalent for the higher/highest die but is ambiguous as to the value of the lower die.

That said, here is an intuitive way to think about this and convince yourself that the methods are equivalent for the higher/highest die: Instead of "rolling a die" by immediately tossing it onto a surface and waiting for it to settle, replace that with the two-step process of (1) rattling it under an opaque cup, and (2)removing the cup to reveal the die. The key here is that any amount of time can go between steps (1) and (2).

Now think about the roll two and reroll method, specifically as follows:

• Rattle three dice under three cups
• Reveal two dice (you have now "rolled" two dice)
• Discard the lower of the two revealed
• Reveal the third die (you have now "re-rolled" the second die-- nothing says it has to be the same physical die, and we assume all dice are fair)
• Choose the highest of the two non-discarded dice

Vs the roll three method:

• Rattle three dice under three cups
• Reveal three dice (you have now "rolled" three dice)
• Choose the highest of the three dice

Is anything fundamentally different happening here? No. Nothing is really changing except the order of the reveals. You will always end up selecting the same highest die. But again, there is ambiguity for the identity of the second die.

I find thinking about reroll situations in this way (i.e., separating the generation of the number from my knowledge of it) helpful to sharpen my intuition about how randomness actually works.