Been thinking about custom dice this morning!
The probability of Fog of War (3d6 >=16) is 4.6296(repeating)%.
I was wondering if you made custom dice with the following specs:
- 6 sides
- only 2 faces (FoW, continue)
would there be an arrangement of face counts such that rolling exactly three
FoW would have the same probability as
3d6 >= 16...
The following arrangement works:
Die 1: 5 FoW, 1 Continue
Die 2: 2 FoW, 4 Continues
Die 3: 1 FoW, 5 Continues
Probablistically, your odds of rolling FoW are:
Die 1: 83.3%
Die 2: 33.3%
Die 3: 16.6%
and your odds of
all three coming up FoW at the same time, and therefore ending your turn, are:
4.6296296%
EDIT - turns out the site I consulted was rounding their results without saying so. That arrangement works perfectly.
Anyone else have an opinion?
(of course some scenarios modify the FoW roll, you're shit out of luck for those!
)