Here’s how you can play D&D with just that little six-sided dice that came in your monopoly box.

(I think I”ve got the approach with the fewest expected rerolls, there are still going to be some but it will do in a pinch (or you could just buy my dice app I guess…)) #Dice #DnD #ttrpg

To be clear, I do not recommend doing this. It will slow the game down a bunch.

But, the math works, the results will be fair.

Okay so I thought of a way to reduce the expected roll to 1.333etc when trying to get a d4 result, but it’s not as straightforward to describe and complicates the process. OTOH it does also cap the rolls for that part to 2.

Basically:
If you roll 1-4, that’s the result, if not remember the result and roll again.
If the second roll isn’t 1-4, then compare first and second rolls:
5,5 = 1
5,6 = 2
6,5 = 3
6,6 = 4

@Sophie You can do this on the d8 to guarantee three rolls or less too
@dx yup, and the d20
@Sophie I’m not sure you can guarantee the d20 in any number of rolls. Or d10. Because 6^N is never divisible by 10.
@Sophie (That is of course theoretically speaking. I would stake any amount of money on a fair d20 eventually resolving under this scheme, and quickly too)
@dx I think you’re right, I’m very close to getting nerd-sniped by trying to think of a guaranteed limit to rerolls but I’m going to just do myself a favour and… not. lol
@Sophie Having just gone down that garden path myself, I’ll save you the trouble: I’m pretty sure to guarantee number of rolls you need 6^N to be divisible by the number of faces on the simulated die, where N is any integer. If you consider each additional roll as adding x6 leaf nodes per existing leaf node to a graph of possibilities, that makes it easier to visualize. (1/2)

Since 6^N always ends in 6, it can never be cleanly divisible by 5, so there is no way to guarantee a map.

I wondered about d7 whether that’s possible with d6 and guaranteeing number of rolls, because 7xN sometimes ends in 6, but I tested up to 5000d6 without luck. (2/2)