Neat! All in successful rolls?
The math for the number of "slots" I'm using is basically dividing the number of participants by 3; the integer result is the number of 'A' slots; then if the remainder (i.e. 'modulus') is 1 you add a C slot, and if it's 2 you add a B and a C slots.Kitsu Moshumaru wrote: ↑Tue Aug 17, 2021 1:11 amBy the way, can someone here explain the math?
I wish to be able to help with future events.
... and reading that it sounds more complicated than it should.
Basically, for 23 participants for instance, you'd have 21 / 3 = 7 A slots. Two participants would remain, so 8 B slots and 8 C slots.
Does that sound clearer?