How EXACTLY do you calculate an Empowered spell result with an odd number of dice?

[TuggyNE]

I'm usually very interested in dice, math and probability but this issue never came to my mind.

It was just something sparked by another thread that was discussing the best spells to Empower, and why; it seemed that spells that give low numbers of d4s are best, which led to the realization that, at those ranges, it's non-trivial to calculate exactly what the result is. (Take time stop, or enervation, for example.)

Also I am, in general, highly dedicated to best practices.

I understand your point, but there are some problems that work against a smooth distribution, like the rules on rounding and the fact that dice give inevitably discrete results.

Yeah, D&D's rounding rules are messed up, and die quantization is unavoidable, but I believe it's possible to get a really good match, even if it's not perfect.

I just noticed that you already did: 3d4+(3d4/2) named "partial multiplication". This should give the right distribution (except the fact that 3d4/2 is rounded) and avoid 'holes' in distribution.
Didn't it work?

It sort of does. It avoids gaps, certainly, and it closely matches what I suspect the ideal distribution to be, but it's not quite right, probably due to rounding. It's also not unambiguously correct: a strict reading of the rules may indicate it's disallowed, and it's not the only distribution that is similarly close.

OTOH, there's only really one other serious candidate: minimal partial multiplication, which is a shortcut to reduce the amount of division you have to do at the table. But here's a shortlist graph anyway.