What Andy says is true provided the thrower can control both the stable spin mode and the orientation of the spin axis. If the thrower can't control the orientation of the axis, the orientation of the die at first strike will still be random. On Wed, Jan 30, 2019 at 5:38 PM Andy Latto <andy.latto@pobox.com> wrote:
Assuming the die has 3 distinct moments, there are two stable ways to have the die spinning as it is tossed. I don't see any reason to expect the probabilities will be the same for these two different ways to throw the die. Or the unstable ways, but those are harder for the thrower to reproduce accurately.
Andy
On Wed, Jan 30, 2019 at 5:32 PM Allan Wechsler <acwacw@gmail.com> wrote:
Even more than that: two dice with exactly the same geometry and density can have different probability distributions depending on the coefficient of restitution. If all the kinetic energy of a die is absorbed the
instant
it hits the table, then stable positions with a smaller basin of attraction will be favored, compared to the same geometry with a very bouncy die. In very bouncy dice, positions with larger basins get more chances to "steal" from less favorable positions. For example: the probability of a nickel landing on its edge will always be small, but it will be even smaller if the nickel is bouncier.
On Wed, Jan 30, 2019 at 5:02 PM Tomas Rokicki <rokicki@gmail.com> wrote:
I believe at least most of the dice that are fair but oddly shaped derive fairness through symmetry; each face is isomorphic to some other face through some spatial rotation or mirroring.
I think there's much more to fairness for non-symmetric dice than solid angle from center of gravity. For instance, it's easy to make a polyhedra where a face has a positive solid angle from the center of gravity, yet the center of gravity does not lie over the face when the face is flat (and thus the dice will never be stable on that face).
-tom
On Wed, Jan 30, 2019 at 1:39 PM Dan Asimov <dasimov@earthlink.net> wrote:
Let F be a face of a bounded convex polyhedron K in R^3. Assume K is solid with constant density. Then:
----- What is the probability
p = p(F; K)
that if K is randomly thrown onto a horizontal surface, it will land on the face F ? -----
My guess back in high school was that
p = (1/4π) * (solid angle(F), subtended from the center of gravity of K).
I recently saw for sale some novelty dice that are irregular hexahedra with planar faces, advertised as having equal probabilities for all faces. So I'm wondering if this is just an issue of that solid angle fraction, or whether there is more to it. Maybe this depends on how "randomly thrown" is defined? A priori it seems that this might be possible, say, if some faces would be unlikely to be landed on if our polyhedral die K were rolling on the horizontal surfaces with enough velocity (or end-over-end angular momentum).
So the random throws might have different statistical outcomes if there are many with high energy.
Or not?
—Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- -- http://cube20.org/ -- http://golly.sf.net/ -- _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun