Back in H.S. I wondered about this for any polyhedron, and guessed a face's probability of landing down is proportional to the solid angle it subtends from the center of gravity. That might be almost true if it landed on a soft surface, but for a hard surface there might be other things to consider. --Dan On 2012-09-20, at 5:48 PM, Veit Elser wrote:

On Sep 20, 2012, at 11:20 AM, Warren Smith <warren.wds@gmail.com> wrote:

...

--Here's a different problem: Devise a convex polyhedron with 5 faces, such that the probability it lands on each face, is 1/5. I have a solution in mind which in fact works for any number N>=4 of faces (here N=5), although I'm not sure whether we should accept my solution. The question inside the question is: "what is the right probabilistic model?" And the answer inside the answer is not so obvious to me.

--Warren D. Smith

I made a 2-sided dice (a right circular cylinder) that is fair according to a simple physical model and proved to be fair in actual experiments.

On Sep 21, 2012, at 12:32 AM, meekerdb <meekerdb@verizon.net> wrote:

On 9/20/2012 9:02 PM, Dan Asimov wrote:

...

Back in H.S. I wondered about this for any polyhedron, and guessed a face's probability of landing down is proportional to the solid angle it subtends from the center of gravity.

And whether the projection of the CG onto the plane of the face falls within the face. :-)

Brent That's close to my physical model. It came up in connection with estimating the probability of a coin landing on its side. I had my students design a cylindrical dice that has equal odds for heads, tails and "side". But because this was a physics course we did not make the ad hoc assumption that the rotation group is sampled uniformly in a toss. Instead we used the coin tossing model where the cylinder axis rotates rapidly about a horizontal axis. Non-uniform sampling is especially valid for symmetric objects, where the three moments of inertia are equal and there is no free precession. So I don't find Michael Kleber's "fairness by symmetry" very compelling. I bet if one "rolled" a cubic dice more like the standard coin toss one could strongly bias the outcome (against the faces intersected by the rotation axis).

-Veit

If you make a longish pentagonal prism, and cap the ends with pentagonal pyramids, the resulting 15-faced die has 0 probability of landing on one of the triangular faces, and 1/5 of landing on each of the rectangular faces. All this is by symmetry, and doesn't depend on the physical model except that we expect the die to topple off an unstable face. This of course is not a solution to the D5 problem Warren proposed. I wonder if there is a pentahedron that is fair regardless of physical model. My intuition is "no", but I don't see any way to prove that. Andy, I haven't frequented casinos. Don't they require you to throw dice from a cup? On Fri, Sep 21, 2012 at 9:59 AM, Andy Latto <andy.latto@pobox.com> wrote:

On Fri, Sep 21, 2012 at 9:13 AM, Veit Elser <ve10@cornell.edu> wrote:

I bet if one "rolled" a cubic dice more like the standard coin toss one could strongly bias the outcome (against the faces intersected by the rotation axis).

Provided you don't think this bias is almost completely undone by requiring the die to bounce against an irregularly-shaped surface before it lands, there are any number of casinos willing to take your bet. You don't even need a strong bias; a bias of 5% or so on the individual dice should be big enough to get the 2% bias you need on the total to make money at the craps table.

Andy andy.latto@pobox.com

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I wonder what is known about the fairness of dice -- and even more, of flipping coins -- in practice. Supposing good practices are used (no spinning, lots of flips, etc.), how fair is an American penny? A 1€ coin? Etc. I've "noticed" some biases in my experience but I don't know that those would stand up to statistical scrutiny. Charles Greathouse Analyst/Programmer Case Western Reserve University On Fri, Sep 21, 2012 at 12:57 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:

There was a discussion of questions related to polyhedral stability by Conway and others on math-fun in June 1997, under subject headings including ' Thoughts about "fair dice" ', ' oddohedrons ', ' Odd dice ' .

I have a file of it available if anyone wants to take a look; though it should be easy to locate in the archive.

Fred Lunnon

On 9/21/12, Allan Wechsler <acwacw@gmail.com> wrote:

...

If you make a longish pentagonal prism, and cap the ends with pentagonal pyramids, the resulting 15-faced die has 0 probability of landing on one of the triangular faces, and 1/5 of landing on each of the rectangular faces. All this is by symmetry, and doesn't depend on the physical model except that we expect the die to topple off an unstable face. This of course is not a solution to the D5 problem Warren proposed.

I wonder if there is a pentahedron that is fair regardless of physical model. My intuition is "no", but I don't see any way to prove that.

Andy, I haven't frequented casinos. Don't they require you to throw dice from a cup?

On Fri, Sep 21, 2012 at 9:59 AM, Andy Latto <andy.latto@pobox.com> wrote:

...

On Fri, Sep 21, 2012 at 9:13 AM, Veit Elser <ve10@cornell.edu> wrote:

I bet if one "rolled" a cubic dice more like the standard coin toss one could strongly bias the outcome (against the faces intersected by the rotation axis).

Provided you don't think this bias is almost completely undone by requiring the die to bounce against an irregularly-shaped surface before it lands, there are any number of casinos willing to take your bet. You don't even need a strong bias; a bias of 5% or so on the individual dice should be big enough to get the 2% bias you need on the total to make money at the craps table.

Andy andy.latto@pobox.com

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On 9/21/2012 9:34 AM, Allan Wechsler wrote:

If you make a longish pentagonal prism, and cap the ends with pentagonal pyramids, the resulting 15-faced die has 0 probability of landing on one of the triangular faces, and 1/5 of landing on each of the rectangular faces. All this is by symmetry, and doesn't depend on the physical model except that we expect the die to topple off an unstable face. This of course is not a solution to the D5 problem Warren proposed.

I wonder if there is a pentahedron that is fair regardless of physical model. My intuition is "no", but I don't see any way to prove that.

I think the answer is yes based on a kind of continuity argument. Consider a three sided prism with equilateral triangular endcaps. If the prism in long compared to the triangular ends, then each rectangular face will come up with probability 1/3. If you make the prism short enough the probability of the rectangular sides will go to zero and the ends, by symmetry, will each be 1/2. So there must be a length at which the rectangular sides each have probability 1/5 and that will imply the ends also have probability 1/5. Note that this kind of reasoning generalizes to higher N. Brent Meeker

Andy, I haven't frequented casinos. Don't they require you to throw dice from a cup?

On Fri, Sep 21, 2012 at 9:59 AM, Andy Latto<andy.latto@pobox.com> wrote:

...

On Fri, Sep 21, 2012 at 9:13 AM, Veit Elser<ve10@cornell.edu> wrote:

I bet if one "rolled" a cubic dice more like the standard coin toss one could strongly bias the outcome (against the faces intersected by the rotation axis).

Provided you don't think this bias is almost completely undone by requiring the die to bounce against an irregularly-shaped surface before it lands, there are any number of casinos willing to take your bet. You don't even need a strong bias; a bias of 5% or so on the individual dice should be big enough to get the 2% bias you need on the total to make money at the craps table.

Andy andy.latto@pobox.com

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@Brent: I think your reasoning is flawed. It is true that for any given, fixed physical model, the triangular prism can be "tuned" to make all the faces equally likely. But under different physical assumptions, the aspect ratio required might be different. Forgive me if I am going over ground that you already understand, but I want to make this really clear, so I'll give two different, fairly plausible physical assumptions, and hope it'll be evident that the "fair triangular prism" under one assumption might be different from the one required by the other. In both models we will assume that the die tumbles enthusiastically enough that its orientation when it first strikes the table is randomly distributed. In "the inelastic model", at the moment that the die first contacts the table, all kinetic energy is dissipated: a perfectly inelastic collision. The die stops dead in space, and then topples onto the face under its center of gravity. Under this assumption, it's fairly clear that the probability associated with each face is proportional to the angle subtended at the center of gravity by that face. To this model we oppose "the almost elastic model". Here, at the moment of impact, almost all kinetic (including rotational) energy is conserved. The die experiences an impulse at the impacting corner, normal to the surface of the table and almost exactly enough to leave its energy unchanged after the collision. If the collision were _perfectly_ elastic, the die would continue to bounce forever, so we imagine that an infinitesimal energy tax is paid at each bounce. This assumption is much harder to analyse, but a secondary assumption helps: imagine that at the next impact, the die's orientation and angular momentum are distributed randomly around the space of accessible configurations given the total amount of kinetic energy available. This is equivalent (I think) to imagining a sphere, whose center is the die's center of gravity, and whose radius is the maximum height the center could reach with the amount of energy the currently has. At each collision, the sphere shrinks a little, until finally it begins to intersect the body of the die itself. At any moment, the die is destined to come to rest only on a face that has some points inside the critical sphere. The region of the die's surface that is within the critical sphere we may call the accessible region. As the critical sphere shrinks, the accessible region does also, until some collision separates the accessible region into two or more subregions. At that moment, the die's destined resting-face is restricted to just one of those subregions, and our secondary assumption is that each subregion has an associated probability proportional to its subtended solid angle. As the die rattles to a rest, the accessible subregion may undergo further subdivisions, until finally the die becomes fated to land on a single face. That was a very long explanation. The point I'm trying to make is that we have no reason to expect the two models to yield the same probabilities. In the inelastic model, the probability associated with each face depends only on the geometry of the exterior; but the almost-elastic model depends crucially, as well, on the distribution of mass inside the die. Therefore we should not be surprised to find that even for dice made of a homogeneous material, the choice of model has a real effect on the probabilities when the die cannot be proven fair by symmetry arguments. This means that in real-world conditions, a die tuned for fairness if it's made of cellulose acetate bouncing on hard felt might turn out not to be fair when made of bone and thrown on an oak tabletop. There is, in fact, no single _mathematical_ criterion of fairness that can be applied outside of the realm of perfect symmetry. On Fri, Sep 21, 2012 at 3:19 PM, Michael Kleber <michael.kleber@gmail.com>wrote:

On Fri, Sep 21, 2012 at 3:06 PM, meekerdb <meekerdb@verizon.net> wrote:

...

On 9/21/2012 9:34 AM, Allan Wechsler wrote:

...

If you make a longish pentagonal prism, and cap the ends with pentagonal pyramids, the resulting 15-faced die has 0 probability of landing on one of the triangular faces, and 1/5 of landing on each of the rectangular faces. All this is by symmetry, and doesn't depend on the physical model except that we expect the die to topple off an unstable face. This of course is not a solution to the D5 problem Warren proposed.

I wonder if there is a pentahedron that is fair regardless of physical model. My intuition is "no", but I don't see any way to prove that.

I think the answer is yes based on a kind of continuity argument. Consider a three sided prism with equilateral triangular endcaps. If the prism in long compared to the triangular ends, then each rectangular face will come up with probability 1/3. If you make the prism short enough the probability of the rectangular sides will go to zero and the ends, by symmetry, will each be 1/2. So there must be a length at which the rectangular sides each have probability 1/5 and that will imply the ends also have probability 1/5. Note that this kind of reasoning generalizes to higher N.

Sure, but it doesn't seem obvious that the fair height for a die rolled, say, underwater would be the same as for a die rolled in air.

--Michael

...

Brent Meeker

...

Andy, I haven't frequented casinos. Don't they require you to throw

dice

...

...

from a cup?

On Fri, Sep 21, 2012 at 9:59 AM, Andy Latto<andy.latto@pobox.com> wrote:

On Fri, Sep 21, 2012 at 9:13 AM, Veit Elser<ve10@cornell.edu> wrote:

...

I bet if one "rolled" a cubic dice more like the standard coin toss one could strongly bias the outcome (against the faces intersected by the rotation axis).

Provided you don't think this bias is almost completely undone by requiring the die to bounce against an irregularly-shaped surface before it lands, there are any number of casinos willing to take your bet. You don't even need a strong bias; a bias of 5% or so on the individual dice should be big enough to get the 2% bias you need on the total to make money at the craps table.

Andy andy.latto@pobox.com

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...

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On 9/21/2012 4:53 PM, Allan Wechsler wrote:

@Brent: I think your reasoning is flawed. It is true that for any given, fixed physical model, the triangular prism can be "tuned" to make all the faces equally likely. But under different physical assumptions, the aspect ratio required might be different. Forgive me if I am going over ground that you already understand, but I want to make this really clear, so I'll give two different, fairly plausible physical assumptions, and hope it'll be evident that the "fair triangular prism" under one assumption might be different from the one required by the other.

In both models we will assume that the die tumbles enthusiastically enough that its orientation when it first strikes the table is randomly distributed.

In "the inelastic model", at the moment that the die first contacts the table, all kinetic energy is dissipated: a perfectly inelastic collision. The die stops dead in space, and then topples onto the face under its center of gravity. Under this assumption, it's fairly clear that the probability associated with each face is proportional to the angle subtended at the center of gravity by that face.

It's not that clear to me, but in any case this hypothetical is so nomologically impossible as not be of interest. There's no way short of magic to remove all the angular momentum of a body by contact forces at a point.

To this model we oppose "the almost elastic model". Here, at the moment of impact, almost all kinetic (including rotational) energy is conserved. The die experiences an impulse at the impacting corner, normal to the surface of the table and almost exactly enough to leave its energy unchanged after the collision. If the collision were _perfectly_ elastic, the die would continue to bounce forever, so we imagine that an infinitesimal energy tax is paid at each bounce. This assumption is much harder to analyse, but a secondary assumption helps: imagine that at the next impact, the die's orientation and angular momentum are distributed randomly around the space of accessible configurations given the total amount of kinetic energy available. This is equivalent (I think) to imagining a sphere, whose center is the die's center of gravity, and whose radius is the maximum height the center could reach with the amount of energy the currently has. At each collision, the sphere shrinks a little, until finally it begins to intersect the body of the die itself. At any moment, the die is destined to come to rest only on a face that has some points inside the critical sphere. The region of the die's surface that is within the critical sphere we may call the accessible region. As the critical sphere shrinks, the accessible region does also, until some collision separates the accessible region into two or more subregions. At that moment, the die's destined resting-face is restricted to just one of those subregions, and our secondary assumption is that each subregion has an associated probability proportional to its subtended solid angle. As the die rattles to a rest, the accessible subregion may undergo further subdivisions, until finally the die becomes fated to land on a single face.

That was a very long explanation. The point I'm trying to make is that we have no reason to expect the two models to yield the same probabilities. In the inelastic model, the probability associated with each face depends only on the geometry of the exterior; but the almost-elastic model depends crucially, as well, on the distribution of mass inside the die. Therefore we should not be surprised to find that even for dice made of a homogeneous material, the choice of model has a real effect on the probabilities when the die cannot be proven fair by symmetry arguments. This means that in real-world conditions, a die tuned for fairness if it's made of cellulose acetate bouncing on hard felt might turn out not to be fair when made of bone and thrown on an oak tabletop.

But that doesn't follow from your argument above. Those two cases are both instances of your "almost elastic" model and so should produce the same probabilities. Brent

There is, in fact, no single _mathematical_ criterion of fairness that can be applied outside of the realm of perfect symmetry.

On Fri, Sep 21, 2012 at 3:19 PM, Michael Kleber<michael.kleber@gmail.com>wrote:

...

On Fri, Sep 21, 2012 at 3:06 PM, meekerdb<meekerdb@verizon.net> wrote:

...

On 9/21/2012 9:34 AM, Allan Wechsler wrote:

...

If you make a longish pentagonal prism, and cap the ends with pentagonal pyramids, the resulting 15-faced die has 0 probability of landing on one of the triangular faces, and 1/5 of landing on each of the rectangular faces. All this is by symmetry, and doesn't depend on the physical model except that we expect the die to topple off an unstable face. This of course is not a solution to the D5 problem Warren proposed.

I wonder if there is a pentahedron that is fair regardless of physical model. My intuition is "no", but I don't see any way to prove that.

I think the answer is yes based on a kind of continuity argument. Consider a three sided prism with equilateral triangular endcaps. If the prism in long compared to the triangular ends, then each rectangular face will come up with probability 1/3. If you make the prism short enough the probability of the rectangular sides will go to zero and the ends, by symmetry, will each be 1/2. So there must be a length at which the rectangular sides each have probability 1/5 and that will imply the ends also have probability 1/5. Note that this kind of reasoning generalizes to higher N.

Sure, but it doesn't seem obvious that the fair height for a die rolled, say, underwater would be the same as for a die rolled in air.

--Michael

...

Brent Meeker

...

Andy, I haven't frequented casinos. Don't they require you to throw dice from a cup?

On Fri, Sep 21, 2012 at 9:59 AM, Andy Latto<andy.latto@pobox.com> wrote: On Fri, Sep 21, 2012 at 9:13 AM, Veit Elser<ve10@cornell.edu> wrote:

...

I bet if one "rolled" a cubic dice more like the standard coin toss one could strongly bias the outcome (against the faces intersected by the rotation axis).

Provided you don't think this bias is almost completely undone by requiring the die to bounce against an irregularly-shaped surface before it lands, there are any number of casinos willing to take your bet. You don't even need a strong bias; a bias of 5% or so on the individual dice should be big enough to get the 2% bias you need on the total to make money at the craps table.

Andy andy.latto@pobox.com

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On Fri, Sep 21, 2012 at 1:09 PM, Dan Asimov <dasimov@earthlink.net> wrote:

The only trouble is, depending on the kind of craps game it is, after enough wins like that they'll either escort you out of the casino telling you to never return, or they'll beat you up in a back alley.

Sure eventually you'll get barred. But my friends on the MIT blackjack team took 7 figures from the casinos before that happened. And since it was known that blackjack was beatable at the time, but the casino thinks that craps is not beatable, it would take them even longer to bar you. And to answer Alan, in casino craps dice are thrown from your hand; dice cups are not used. But the dice are required to bounce off of the irregularly shaped back wall of the table. Andy

--Dan

On 2012-09-21, at 6:59 AM, Andy Latto wrote:

...

I bet if one "rolled" a cubic dice more like the standard coin toss one could strongly bias the outcome (against the faces intersected by the rotation axis).

Provided you don't think this bias is almost completely undone by requiring the die to bounce against an irregularly-shaped surface before it lands, there are any number of casinos willing to take your bet. You don't even need a strong bias; a bias of 5% or so on the individual dice should be big enough to get the 2% bias you need on the total to make money at the craps table.

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-- Andy.Latto@pobox.com

Which brings up the old problem: Does there exist a polyhedron with no stable face on a tabletop? (I.e., for which the center-of-gravity's projection to the plane of any face lies outside that face.) The standard argument for why no such polyhedron exists is that it would keep rolling forever, so be a perpetual motion machine. Some time ago there was no known proof purely by geometry. Does anyone know if that's still the case? --Dan On 2012-09-20, at 9:32 PM, Brent wrote:

And whether the projection of the CG onto the plane of the face falls within the face. :-)

I think you need to distinguish between no stable position vs. no stable face. The perpetual motion argument says there must be a stable position, but it does not say that the position is one with a face flush against the surface the polyhedron is resting on. Tom Dan Asimov writes:

Which brings up the old problem: Does there exist a polyhedron with no stable face on a tabletop? (I.e., for which the center-of-gravity's projection to the plane of any face lies outside that face.)

The standard argument for why no such polyhedron exists is that it would keep rolling forever, so be a perpetual motion machine.

Some time ago there was no known proof purely by geometry. Does anyone know if that's still the case?

--Dan

On 2012-09-20, at 9:32 PM, Brent wrote:

...

And whether the projection of the CG onto the plane of the face falls within the face. :-)

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