[extropy-chat] alt dot fair dice

[Hal Finney]

Spike writes:
> Can other shapes be made such that there is
> equal probability of any face downward?  I can
> think of one: a five sided pyramid shaped
> solid (four triangular faces and one square
> face).
> ...
> A two sided fair die is a coin, but that suggests
> another three sided die: a cylinder, like a
> really fat coin, equally likely to land on edge
> as on either side.  For that matter one could 
> grind arbitrarily many flat sides on a cylinder,
> so that the cylinder gets longer and thinner
> as the number of flat sides gets larger.

That's very interesting.  I remember watching people play Dungeons
and Dragons when I was a kid, using many different kinds of fancy
polyhedral dice.  I think they were all regular polyhedra, though.
I haven't seen suggestions for fair, non-polyhedral dice.

I can suggest a path to analyzing the second case, a polygonal,
thick coin, which is simpler because it is more symmetrical.  Imagine
circumscribing a sphere around the coin, with the center of the sphere
at the geometrical center of the coin (which will be the center of
mass of the coin), and all of the vertices of the coin on the sphere.
Then I would suggest that the probability of landing on a face is the
spherical area of that face.  I.e. if you think of the face as a polygon
drawn on a sphere, the area of that polygon corresponds to the probability
of that face.

This is because if you imagine the coin randomly oriented in space,
and draw a line straight down through the center, the probability of
that line falling through a given face is proportional to the spherical
area of that face.

A coin based on an n sided polygon corresponds to an n+2 sided object.
The n edge faces are square, and the top and bottom faces are n-gons.
So you need to take the formula for the area of a polygon on the sphere,
and adjust the thickness of the coin until the spherical area of each
of the two n-gons equals the spherical area of each of the squares.

Once upon a time I was pretty familiar with such formula, back in my
math contest days, but it would take me a while to do it now, so if
anyone else is motivated they might be able to come up with a solution.

Hal