I was idly musing over the unpublished nameless system from Paul
MacLean and the Yog-Sothoth.com crew, and a trick occurred to me…
That system's primary mechanic involves coin flips: the easiest
grade of task needs one flip to come up heads (50%), then two in a row
(25%), then three in a row (12.5%) for the hardest.
(There was also an alternative system based on 3d6, but it didn't have
the same odds, so I Diskard it.)
But the players, playing only once a month and not keeping the system
reliably in mind, found themselves flipping all three coins and
saying, well, they got at least one head, so that's a success, right?
But actually we can build any multiple of 12.5% with three coin flips,
and we don't even have to keep them in order. Flipping three coins
and counting heads, the possible outcomes are:
- No heads (TTT), ⅛
- One head (HTT, THT, TTH), ⅜
- Two heads (HHT, HTH, THH), ⅜
- Three heads (HHH), ⅛
And so we can combine them!
If you want a probability of ⅛: you need to get three heads.
If you want ¼: you need to get three, or zero, heads.
If you want ⅜: you need to get exactly two heads.
If you want ½: you need to get two or three heads.
If you want ⅝: you need to get zero, two or three heads.
If you want ¾: you need to get one or two heads.
If you want ⅞: you need to get one, two or three heads.
Yes, of course anyone reading this blog probably has a d8 within reach
(I certainly do). But it's still fun. (And if you want an actual value
from 1 to 8, flip the coins in order and read them as binary.)
Sadly you can't do this as readily for a d6 or d12, since you can't
generate the divisor 3 from powers of two. Or rather you can, but you
have a potentially infinite series of extra flips: in effect, roll a
d8, but reroll 7s and 8s.
And you can't do the same trick with four coins: the outcomes have the
probability 1-4-6-4-1, because of course it's Pascal's
Triangle,, and you
can't select some of those to add up to 3. (Though again you can read
them in order for a number from 0 to 15.)