this post was submitted on 15 Dec 2023
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[–] candybrie@lemmy.world 1 points 1 year ago* (last edited 1 year ago) (1 children)

To keep the same probabilities, you can only reduce and only to one that is a factor. E.g. d20 can be equivalent to d10, d5, d4 and d2.

Multiplying the rolls messes things up. As an example, for d12 as a d6xd2 you have double the chance to roll 2, 4, and 6 and no chance to roll 7, 9, and 11.

You could make the equation a little more complicated (6×(d2-1))+d6 to make it work.

[–] fushuan@lemm.ee 1 points 1 year ago* (last edited 1 year ago)

You are absolutely right, I was thinking d6d2 as: the D2 rolls 1, it's d6. The D2 rolls 2,its 6+d6. That's not what my math said so my bad!

Edit: your equation is what I had in my mind, which is sorta what we do to roll d100.