3 Sided Coin

P(Head)=P(Tail)=P(Side) = 1/3

So that:

P(Head)+P(Tail)+P(Side) = 1

I wrote an essay on it, and from experimental results, it seems that the
only way to do this is to construct a coin so that the height of the
cylinder equals the radius of it.

Any help is appreciated.

-JZS

--
dfg...@sdergdfg.com

In practice, a cylindrical ruler of triangular
cross-sections would do the trick.
Then one "rolls" it. Of course, better is
an ordinary die in the shape of a cube,
labeled H, H, T, T, S, S. And one could
do that with a ruler; give it hexagonal
cross-sections, that is, making it easier
to roll.

Is there a mathematical way to construct a right circular cylinder to function
as a fair coin with P(h)=P(t)=P(s)=1/3?

I know of know precise way of doing this other than by experimental trials
(empirically).

Len

Of course this isn't what you had in mind,
but virtually any 3-sided object can function
that way.

Label its sides H,T,S, and toss it repeatedly
(independently) in triplets of tosses, until
getting a triplet in one of the following sets:
h={HTS, HST}, t={THS,TSH}, s={SHT,STH}.

This guarantees that p(h)=p(t)=p(s), for *any*
positive p(H),p(T),p(S), because each of the
triplet possibilities has the same unknown
probability p(H)*p(T)*p(S).

You may need to toss for quite a while though,
since the mean number of tosses until getting
(h or s or t) is E[N]=1/[6*p(H)*p(T)*p(S)].

(E[N]>=27/6=4.5, attaining this minimum only when
p(H)=p(T)=p(S), and can obviously be arbitrarily
large unless there is a reasonable degree of
symmetry.)

I think it's worth noting how we can start with
probabilities that are "unknown" due to a lack of
symmetry of the *object*, and by *our* symmetric
behavior end up with exactly known probabilities.

--
r.e.s. (Spam-block=XX)

JZS wrote in message <7l2go9$jaq$1...@news-02.meganews.com>...

Of course. :)

I was wondering what would be the best way with still preserving the *coin*
shape.

Comments are welcome!

__

Three Sided Coins

We are all very familiar with two-sided coins. We carry them around
everyday in our pockets, feed them to videogames, and some of us even use
them (by flipping) to make important decisions. Additionally, we know that a
fair two-sided coin has equal probability (50%) of landing on either its
side designated Heads, or its other side, Tails. Would this same idea be
possible with a three-sided coin?
What would a three-sided coin look like? It would be similar to a two-sided
coin, but would have a greater height, thus creating a third side (Edge).
When trying to determine how to construct a fair three-sided coin, it is
important to consider the ratio of height to diameter (H:D). It is very rare
for a two-sided coin to land on its edge; conversely, if you were to flip a
poster tube, it would almost always land on its edge, rather than either
side. Therefore, one must determine the ratio of height to diameter that
makes the probability of landing on any of the three sides of the coin
uniform. If the ratio H:D is very small, the edge landing probability
approaches 0, as in the example of the two-sided coin. If the ratio H:D is
very large, then the probability of landing on the edge approaches 1, as in
the case of the poster tube.
I , and a friend set out to determine what H:D ratio would be required to
create a fair three-sided coin. David made three coins in his machine shop.
The coins were made smooth on all sides, to reduce uneven distributions of
mass, which could slightly bias the outcomes. The ratios of H:D were as
follows: .375, .5, and .667. Using our experimental data, we attempted to
determine the edge landing probability as a function of the ratio H:D.
Our statistical observations were used to establish the associated
probabilities. Generally, statistical estimates for probabilities are valid
if the experiment can be repeated a number of times under similar
circumstances. In this case, repetition was possible.
How does one flip a three-sided coin? A two-sided coin only has one main
axis of rotation. However, a three-sided coin has two axes: the axis of the
cylinder, and the axis orthogonal to that. Until the correct ratio H:D is
determined (to create a fair coin), we certainly can’t flip the three-sided
coin the same as the two-sided coin. Therefore, I flipped the three-sided
coin on a line between the two axes, reasoning this would be a fair way to
continue.
A fair three-sided coin would have P(Head)+P(Tail)+P(Edge) = 1 -- the
landing probabilities would be uniformly distributed. On July 30th, 1998, I
flipped each of the three coins 3,000 times, which yielded percentages for
each coin landing on Edge. When H:D = .375, Edge was obtained 4.6%. When H:D
= .5, Edge was obtained 32.6%, and when H:D = .667, Edge was obtained 43.3%.
The coin that fit the theoretical model the best was H:D = .5, or to write
it differently when H=Radius. In general, symmetry often suggests a uniform
distribution function, as was the case with this experiment.
Coins have a wide variety of uses, and three-sided coins could have
ramifications in several areas. Coins are often used to simulate random
number generators, and a three-sided coin could add something extra to
sports and gaming. Sports that begin with the flip of a coin to determine
possession (i.e. football) would be impacted because the opportunity exists
for the teams to not select the side that comes up (two teams, three sides).
Does the referee then determine who gets the ball? Is the decision based on
fan input? With a three-sided coin, perhaps Edge would mean that someone
other than the two teams makes the decision. A three-sided gaming coin would
certainly alter the winning percentages in Las Vegas! And three-sided coins
would very dramatically impact the designs of coin operated machinery.
I am pleased with the results of this experiment, and plan to repeat it in
the future with some modifications. Variables to consider might include rate
of spin, density and composition of coin, flipping surface, and coin design
(i.e., smooth versus serrations). While three-sided coins may never be used
on a daily basis, they create some intriguing possibilities with statistical
probabilities.

If you look at the cylinder from the side, so that its profile is
a rectangle, then I'd say the height-to-diameter ratio must be such
that a diagonal of the rectangle is vertical when the end faces are
at 60 degrees from the horizontal (thus the other diagonal becomes
vertical when the cylinder is tipped further over to 120 degrees).

This allows a complete revolution of the cylinder, seen sideways,
to be partitioned into six equal angles within each of which the
centre of mass lies between the vertical planes passing through
the profile's two corners that are respectively the lowest points
(the point of contact) as the each radius bounding the angle is
passed.

By solving a trivial equilateral triangle with diagonal length and
cylinder height equal to 2 and 1 respectively the diameter works
out to be sqrt(3).

In practice though you may also have to consider gyroscopic effects.
For example if the component of spin is mostly about its central axis
then this may impart an extra tendency to flip its centre of mass one
way preferentially.

You'd probably be better off using a standard six-sided die with each
opposite pair of faces identified with one of the three alternatives.
You could also use a three sided pretzel (a stack of similar aligned
triangles increasing in size from a point to a maximum then back to
a point).

Cheers

John R Ramsden (j...@redmink.demon.co.uk)

Hmmmm... Wouldn't that produce a cylinder that looks kind of squattish?
What will be the surface areas for each side, if you adopt that particular
scheme?

| Any help is appreciated.

Assuming the coin smacks into a flat surface (e.g., the floor, or a
tabletop) over and over again, before coming finally to rest, shouldn't
the real question be whether the coin has equal surface area available
to each of its three sides?

But regardless of the *real* answer, if you are planning on mass-producing
a 3-sided die (such as you seem to have proposed, as for a boardgame
intended for mass-consumption), what you REALLY want is something that
'looks' fair, even if it isn't. If that is the case, then the height of
the cylinder ought to be approximately equal to its diameter!
_____

BTW, would the probability change if --- instead of throwing the die upon
a flat surface --- one were to throw it upon a very wide but gently curved
surface?
--

<snip>

Interesting!

Nevertheless, as true as that may be, the human eye takes comfort in
seeing what 'appears' to be an equal amount of surface area. Humans
rarely think of the invisible sides that cannot be inferred from the
apparent sides, hence the perplexing eight-sided (octohedral) die, but the
pleasing appearance of the dodecahedron. Thus, my prediction that there
would be more market acceptance for 3-sided dice that have heights that
are 'equal' in proportion to their diameters, or maybe something like 1.5
x radius. :) As you pointed out in your highly enjoyable essay, the
proof is in the pudding: radius ought to be equal to height, for a
supposedly 'fair' 3-sided die.

A wonderful essay, if I may add.

--

In that case, the coin's proportions must be such that an enscribing
sphere is cut into three equal areas by the intersection with the
edges of the coin.

John

It seems like an intuitively reasonable claim, but it falls apart when you
consider the behavior of real coins. For this claim to be true, real coins
have to have an appreciable probability of landing and staying on their
edges. The same is true for rods.

For instance, a nickel is approximately 1/16" think and 13/16" in diameter.
According to this claim, it should have better than a 7.6% chance of coming
up edges. If it was even a fraction of one percent, it would happen
frequently enough that whenever someone said "Heads or tails?" people would
respond, "What if it comes up edges?" You can bet (no pun intended) that Las
Vegas would take advantage of this possibility (just like the green spot in
roulette).

If real behavior departs so drastically from theory, the theory is either
wrong or incomplete. If it is this incomplete, then it must be enhanced to
make it useable. If you wish to claim that it is sufficiently accurate
except for extreme radius to thickness to ratios (and a nickel only has a
ratio of 6.5) then you really need to be able to describe the factors which
become so dominant for those ratios that deviate from theory AND show that
they are either driven to negligible levels or that they cancel out for the
ratios you are working with.

John Bailey wrote in message <37760680....@news.frontiernet.net>...

http://revolution.martini.nu/

He addresses the 3-sided coin along with many other interesting topics here.

---K. Roberts (DrKRo...@aol.com)

It's humorous and interesting when this happens. Unfortunately, it has
happened numerous times when researchers (with or without knowing)
cross-cite each others work as though the others work was the result of
independent results.

DrKRoberts wrote in message
<19990627191849...@ng-fq1.aol.com>...

>---K. Roberts (DrKRo...@aol.com)

It wasn't bad. The thing that I liked about it was that we had a mathematics
major willing to attempt to determine something through emperical
observation and then seek a mathematical explanation for the observed
phenomenon. So frequently, what you get instead is someone that comes up
with a paper result and then is unwilling to even consider asking if the
results agree with direct observation.

>Well, it's not the first time anybody goes *whooops*!

Don't worry, it won't be the last, either.

>One question that remains is the 'time' factor: even if a 3-sided coin of
>arbitrary thickness lands on its arbitrarily thick edge, how long will it
>remain standing? There are residual shockwaves on the tabletop, vortexes
>(ahem, vortices) of wind from the path the coin took, and ongoing,
>lasting vagaries in the wind flow, and the wind pressure that must,
>eventually, serve to overcome the stability of the sorry coin, and take
>it down --- if the coin is too thin or too lightweight to resist!
>

I think the much bigger factor is the degree of stability and the range of
landings and momentum combinations that can take it from the side it SHOULD
have come up (based on the claim of equal spherical area segments) to the
more stable side. It is easy to visualize that for a nickel that SHOULD have
landed on its edge, virtually any momentum in the coin will carry it past
the point of no return and into the region of one of the faces. Likewise,
most of the time that the coin lands on one of the faces, there is
insufficient momentum to change that result and, if there is, it is
sufficient to carry it past the "edge" region and on to the other face.

<snip>

>I think the question is ultimately one of exposed areas, and if
>residual shockwaves and unpredictable movements of the media *do*
>have an effect, they probably cancel each other out in the wrong
>run.

If this is true, then you must conclude that a nickel should end up on it's
edge about one throw for every twelve that it lands on one of the faces.
Otherwise, these other effects most certainly do NOT cancel out in the long
run and the question is NOT ultimately one of exposed areas.

>
>For instance, we could rephrase the question by visualizing a very large
>hollow sphere in zero-G, and then ask ourselves, "what are the odds that
>a very small 3-sided die moving in any given direction, spinning or
>tumbling or otherwise, will eventually come to rest, on one side (as
>opposed to either of the other two sides)?"
>--

Is the die inside this sphere? Is the sphere evacuated? By saying "very
large" and "very small" you seem to be imposing the condition that the die
never touches the sphere. Is this the case? What constitutes "coming to
rest" and what constitutes "on one side"?

| A flurry of discussion takes place and then someone says, "Here is

| It's humorous and interesting when this happens. Unfortunately, it

So, naturally, we have to figure in the thickness of the medium through
which the coin is thrown, and the thickness of the medium wherein the coin
finds itself standing, and the currents and waves and eddies and vortexes
(vortices) that oppose it. :)

Now, for any *regular* nickel, how often do we a see a coin at rest
suddenly change its state by bouncing back up again and flipping over?! ;)

For instance, we could rephrase the question by visualizing a very large

Thus, the possibilities are strictly limited to the actual available
surface areas of the sides exposed.

--

<snip>

Sure, for a split second. :) Does that count? What if it remains up for
a full second? Or spins about for a while before falling? :)

In the real world, all those things cancel out in about 5 or 10 seconds.
I think that a flipped nickel (or a thrown die) achieves its rest state
within 5 or 10 seconds.

In a Euclidean universe of extremely large dimensions, where there is only
a massless die, and a massless sphere, it might not make that much of a
difference if it is on the outside or the inside. :) hehe -But as I
was visualizing it in my head, I was thinking of the die starting out in
the center of the sphere, with an arbitrary spin and arbitrary velocity.

Oh, that's a good point!

By 'very large' and 'very small' I was also thinking of a die moving for
a very long time, in a straight line, and if this was so, with the
sphere hardly moving in relation to the die, would the die eventually come
in contact with the surface of the sphere, and come to rest, in contact
therewith? The odds ought to be equal to the surface areas of each of the
sides of the die.
--

No. Because the original question and all subsequent discussions deal with
constructing a "fair 3-sided die" where the probability of it "ending" up
heads, tails or edges are all equal.

Matthew Montchalin joked:

Okay.... Well, in that case, hmmmm....
--

You may be correct. I mused on the problem overnight based on my own
misgivings and discovered your post this morning. The other factor
which must be considered in the potential energy of the coin. A coin
will come to rest in a position which is a local minima for its
potential energy. If a thin coin lands on its edge, it is almost
certain to fall over because the energy well for the edge of a thin
coin is so shallow. I am exploring how to express this in a
quantitative model.
I had thought the three sided coin problem finessed the energy
well-barrier problem through the symmetry of equal subtended areas,
but upon reflection, that may not be the case.

John

No, actually its my site.

Why so harsh?

Don't be one of those people who take the fun and exploring side out of
mathematics and make it sterile.

Thanks.

-Dej

How is a nickel's thickness equal to its radius?

It could have to do with the fact taht the coins we made were out of a heavy
plastic, as opposed to iron. (more bounce Im thinking)

-Dej

Well, you are making some considerable assumptions about how the coin is
tossed when you make that assumption -- specifically, you're assuming
that the kinetic energy of the coin is negligible. Reasonable, perhaps,
for a rough estimation, but not reasonable for a "correct" answer in the
mathematical sense.

I think it's legitimate to say that it's possible to define a
non-symmetric die (i.e. a "coin") that will give "fair" results for any
precisely _specified_ tossing method (mind you, this is quite a
non-trivial problem!). On the other hand, if the tossing method is not
specified precisely (as is _always_ the case in real life), then the
only way to guarantee fairness is to have a symmetric die.

As a note, it's possible to make a symmetric, volume-containing,
three-sided die. As an example, take a six-sided die, divide the sides
up into three pairs of adjacent sides, and round each pair off on a
belt-sander to make a smooth rounded surface. Then number the sides.

- Brooks

Oh, what the heck, a moderately-sized hollow sphere that is *bigger*
than the *somewhat* smaller 3-sided die, starting out right smack in
the center. So long as there is enough room for it to tumble about
a bit.

or moderately sized

William L. Bahn wrote:
| Is the die inside this sphere? Is the sphere evacuated?

For the sake of simplicity, let's assume there's a vacuum instead of
molasses.

I was assuming that the die will move in a straight line that eventually
intersects with the surface of the sphere.

Let us further assume that *any* contact between three points or more
(this part is hazy?!), however brief, will be sufficient to drain the
die of all energy and motion. In short, the die will stick to the inner
surface of the sphere, once it strikes it. If this is the case, then
the odds are 100% that *some* part of the die will eventually strike
the surface of the sphere.
--

Who said anything about a nickel's thickness being equal to its radius? Go
back and reread the post a little more carefully.

Keep in mind that this sub-thread is dealing with the claim that the
probabilities are equal to the fractional area of the enclosing sphere that
is associated with each "side" of the coin. So far, that has only been
offered as a self-evident claim. It is merely my contention that everyday
physical observations are adequately inconsistent with this claim that
cannot be accepted on face value.

You're not out of the woods yet.

HOW will it stick? If it touches at an angle, does it freeze at that angle?
I don't understand the "three points" constraint. Will it bounced
elastically if fewer than three points contact? In that case, it would
probably never come to rest since that requires an EXACT alignment of the
coin at the moment of contact.

Yes, I think that is what the claim probably boils down to.

If a 3-sided die were to be contained within a very closely fitting
sphere, any given motion would likely result in contact with one of the
three Faces, or with one of the two Edges. If a face of the die is in
contact with the surface of the sphere, I will assume that it is at rest.
If a face is in any other way in contact with the sphere, I will assume
that it is teetering, and therefore in a precarious or unstable position.

(Perhaps I can invoke some special law of nature to explain that any die
found to be standing on an Edge is or ought to be unstable, and will or
should eventually teeter over, and come to rest upon one of its faces.
Thus, for a 3-sided die, we will rule out all instances of landing upon
an edge, and once it lands upon its edge, it will stay that way.)

Okay, I have never seen any six-sided die thrown upon a flat surface come
to rest so that it stands upon one of its edges, let alone one of its
corners. Only upon a face. :)
--

If a 1-sided die (ahem, a small sphere) were inside a great big hollow
sphere, then it should, under the same proposed circumstances, either
ping around for eternity, or come in contact with the surface of the
sphere and meet at exactly 1 point, right? This may be a trivial case,
but it can be summed up by a special rule applying to it all by itself:
if a 1-sided die *ever* comes into contact with the inner surface of a
hollow sphere, we may as well consider it to have completed its journey,
and declare it at rest. :)

If a 2-sided die (er, ah, a Nickel) were inside a great big hollow sphere
in zero gravity, then it should eventually come to rest in such a way that
its circumference is 1) tangentially in contact with only 1 point on the
surface of the sphere, or else 2) with the center of one of its two faces
coming as close as possible to the surface, and this would make for a
plenitude of points in contact. I prefer to invoke a law to avoid
thinking about a nickel that is tangentially in contact with the sphere:
"If a multi-sided die ever contacts the inner surface at only 1 point,
it is declared unstable, and will move off in some other direction."

| HOW will it stick?

How about something magical like monopolar magnetism? :)

Um, do you think it is possible to touch at an angle? For instance, a
small ball inside a big ball cannot; nor can a nickel inside that same
hollow sphere. Now, as for a 3-sided die being able to touch at an
angle... Hm

It is just a rule to invoke so as to avoid billions and billions of
calculations to figure out where all that energy goes as it bleeds off
from every itty bitty bump or bounce: "If the die makes contact with the
surface at only 1 point, it will bounce off in some other direction."

Are you asking if there is conservation of spin, momentum, and that sort
of thing? Ouch, that kind of stuff hurts my head...

I was just hoping that the coin would tumble around a bit, bouncing
around, until such a time that, presto, 3 or more points were in contact.
Then I'd invoke this rule by declaring the coin has finally come
to rest. There has got to be an easier way to do all this than keep track
of vectors and momenta and centers of gravity and stuff. Hence, the rule
about 3 points. Since it seems to work, and it saves a lot of calculation,
it is an eminently practical sort of rule.
--

If you don't think that a nickel can touch the inside of a hollow sphere at
an angle, then you need to seriously revisit you understanding of hte
geometries involved.

How do you conclude that it "seems to work"? You can "hope" all you want.
You can make up special rules all you want. With each rule you impose
conditions that make the model less usable. I think you have completely
forgotten what the goal of this model was all about anyway, and now you are
just trying to throw patches at it to overcome the multitude of gross flaws
in it.

Go back to a nickel thrown on a flat table. Will you agree that the odds of
the nickel making the initial contact with the table exactly on its face are
very low? The coin will almost always have some tilt to it. If one edge
touches the table while the other edge is even a micron away, the nickel has
touched at one point (since we are talking about mathematical constructs, we
are assuming perfectly rigid bodies). According to your rules, a nickel
tossed onto a table would, in practice, bounce forever with only a
vanishingly small chance of ever coming to rest.

>--

>

If a nickle touches at 1 point, and we invoke the rule, it has not yet come
to a state of rest.

Or else you should revisit the rule that anything touching with only
1 point is not yet at a state of rest: "If a multi-sided die *ever*
contacts the inner surface at one [and only one] point, it is declared

>>| I don't understand the "three points" constraint.

It precipitates the fall of the multi-sided die. Without having to figure
in gravity, elasticity, momentum, energy, &c.

For instance, let us throw a nickle onto an ordinary tabletop, and track
it a rate of a thousand times a second. We note that what you object to,
actually does happen: the nickel lands upon its edge, mirabile dictu!
How long will it take for the nickle to fall if it is spinning on this
point of contact? Since the nickel eventually falls, the rule is not
disproved, but vindicated.
--

<snip>

The only thing I will assent to, is that the nickel will bounce off on its
first contact. In short, the initial contact is unstable, and the nickel
is not yet at a state of rest. Unless we use a high speed camera, it is
going to pretty hard gathering empirical evidence to support your claim
that there was no recoiling from a flat-impact.

Why do you think that? How fast is the camera you are using?

No, because a nickel typically assumes one of any number of random
orientations, several times, before finally falling flat on its face.

Actually, the 'small' chance of coming to rest, increases with time.
--

But your statement is that it cannot touch at an angle.

>contacts the inner surface at one [and only one] point, it is declared

Which means that it will never come to rest - or more coorectly, that the
chances of it ever coming to rest are vanishingly small.

What do you mean by "precipitates the fall"? According to your description,
it is just a means to immediately bring the coin to rest in the highly
unlikely event that it should ever happen to be perfectly aligned at the
moment of contact.

What rule is vindicated? What "actually happens" is occurring under the
rules of reality - not YOUR rules. YOUR rule is in direct contradiction with
the behavior of real coins thrown onto a table top. You have specifically
created a set of rules so as to avoid "having to figure in gravity,
elasticity, momentum, energy" and everything else. According to your rules,
if the nickel lands on its edge - "IT IS DECLARED UNSTABLE AND WILL MOVE OFF
IN SOME OTHER DIRECTION." But, guess what? A real coin doesn't do this
(unless it has so much excess kinetic energy that it bounces, but gravity
will quickly bring it back down and eventually it will touch at one point
and remain in contact (the point of contact will probably move) and it will
NOT move off in some other direction and then it will eventually lose energy
until it comes to rest. But YOUR rules would prevent this entire mechanism
from occurring and you do not even permit these mechanisms to be employed
(since you want an evacuated sphere in free space).

Bottom line: You cannot create a set of arbitrary rules and then use
observations that occur under a completely different set of rules (namely,
reality) and then claim this as vindication that your rules work.

Get real!!! Use your head!!! Think!!! For it NOT to have any tilt, the axis
through the face of the coin must be perfectly vertical (for the face to
touch at more than one point) or perfectly horizontal (for the edge to touch
at more than one point) at the moment of contact.

Not according to YOUR rules. You have NO mechanism that makes the 2nd bounce
and different from the 1st bounce and different from the 1000000000000000th
bounce beyond the vanishingly small probability of it striking in perfect
alignment.

Not according to YOUR rules.

Matthew Montchalin wrote in message <7lbu6g$903$1...@news2.OregonVOS.net>...

>Are we going to skip the first 30 seconds, or not? If not, then we should
>be looking at this at every THOUSANDTH of a second, and evaluating the
>likelihood that, subject to ANY kind of initial throw, it somehow finds
>itself lyind down or standing up, bouncing, spinning, or flying. There
>will be all kinds of axes under those circumstances.
>
>--
>

>Keep in mind that this sub-thread is dealing with the claim that the

>is associated with each "side" of the coin. So far, that has only been

Since my earlier post, I have calculated the relative magnitudes of
the two effects which will determine the behavior of a three sided
coin. One effect is scaled to the fractional area of the enclosing
sphere and the other is scaled to the depth of potential energy wells
associated with the coin landing on a face or edge. Net net, my
conclusion is that there is a significant potential energy well
difference for a coin which is dimensioned for equal probability of
landing on a side or edge on the basis of equal fractional area of the
enclosed sphere. My interpretation of the calculation is that the
three way coin cannot be fair, its probability will be dependent on
the excitation provided during the coin toss. If the coin is tossed
so that there is minimum energy the probability will be close to
0.3333. To the extent that it is tossed in a way that adds more
energy, enabling the state to escape from the initial landing
configuration, it will transition to a lower energy state.

Coin with a mass M effectively located at its cg.
Coin thickness t and diameter d.
Maximum potential energy is M*g*sqrt(t^2+d^2)/2 (barrier between
energy wells)
Minimum potential energies are M*g*t/2 and M*g*d/2
Note that the ratio of potential energies is exactly the aspect ratio
of the coin's cross section.

Recapping the calculation of the fractional area of the enclosed
sphere.
Surface of sphere as a function of angle A:
S = 2*pi()*r^2*sin(a)da|0 to A = 2*pi()*r^2*cos(a)|0 to A =
2*pi()*r^2[1 - cos(A)]
p(h) = p(t) = (1-cos(A))/2 = (1 - cos(arctan(d/2)))/2= (1 - t/sqrt(t^2
+ d^2))/2
p(e) = 1-2*p(h) = t/sqrt(t^2 + d^2)

For p(e)= 1/3, d/t = sqrt(8) = 2.828427 There is almost a 3:1 ratio
of energy associated with a face vs an edge. If there is kinetic
energy available as the coin comes to rest, the coin will move to the
lower energy state. Since this effect depends on whether energy is
available, the outcome of the coin toss become technique dependent and
cannot be considered fair.

John
http://www.frontiernet.net/~jmb184

But it almost always bounces away on the first contact. That's why it
is so practical to wait 30 seconds before evaluating. That way it will
have bled off its energy by bouncing, spinning, and flipping about. Again,
if we are to be practical, we ought to skip the first 30 seconds since its
state will be unpredictable during those 30 seconds.

Since it rapidly changes its state, we'll have to be honest enough to
inject some kind of 'uncertainty principle' into the equation.

Yes: the coin will eventually achieve an end state that is vertical. Are
you worried about how long it takes to do so? I thought we were not going
to figure in the variable T ('time') into the equation. If your real
complaint is that it is going to take a short time, or long time, then we
will have to insert T into the equation.

Yes.

Yes, re-orienting itself from one unknown or unpredictable state (the
'uncertainty' principle) into another unknown or unpredictable state.

That is true. But, otoh, I haven't defined its original motion, or mass,
or center of gravity, either.

Because T is not available for use in the equation.

It would be more 'realistic' if we were to look at the thing from its
initial state, which is not available to us. Remember, we start with the
assumption that its original motion was truly Random. For a 3-sided Die
encompassed by a loosely fitting sphere, the odds are 100 percent that it
will --- after an unknown period of time T --- come to rest 'on' the surface
of the sphere.

Oh, in case we were not using the terms in the same way, I view a 3-sided
die as being cylindrical, with a Top, Bottom, and Side: there are only
two edges, however -- the interfaces separating the Side from the Top or
the Bottom.
--

>
>By solving a trivial equilateral triangle with diagonal length and
>cylinder height equal to 2 and 1 respectively the diameter works
>out to be sqrt(3).
>

I came to this conclusion a few years ago and produced a such a coin by
sticking together a number of UK two pence pieces together. I recorded a few
hundred tosses and did a chi-squared test which seemed to support the idea.

Martin

Under your rules, it is unpredictable forever. Under the rules of the real
world, it bleeds off energy and moves rapidly to a rest state. Your rules do
not have a reliable energy bleed mechanism. You have stated that you didn't
want to deal with energy and so left it out of your model in favor of a
specially fabricated rule to make up for it. In your model, under your
rules, the coin bleeds of NO energy by bouncing, spinning and flipping
about. None. You have specifically excluded that possibility from your
model. The fact that this is what happens in the real world only goes to
show how useless your model is. Under the rules of the real world, we know
that each bounce removes energy and that there is a limit to how much energy
can be removed before the coin must come to rest. Under your rules, any
bounce that fails to grab the coin and bring it immediately to rest bleeds
off no energy - so the coin as just as much energy as it did before it
bounced. If the fifty millionth bounce failed to grab it, then the coin has
just as much energy as it did prior to the first bounce and the odds of it
being grabbed on the next bounce are identical to it being grabbed on the
first bounce.

Consider the case of a bouncing ball. If the contact between the ball and
the ground is perfectly elastic, then the ball will bounce forever. If the
model I am using says that we will ignore energy loss mechanisms such as
friction and non-perfectly elestic collisions, then I cannot look at the
real world case, note that balls DO come to rest after a reasonable amount
of time and them claim that they will do so under my model. They won't. The
rules of my model are not in agreement with the real world in this regard.

The one "energy-loss" mechanism you have put into you model is very
analogous to my model above if I say that I will pick one point on the ball
and IF that point happens to be the the point that touches the ground, then
the ball will stick to the ground. Otherwise it will bounce away again.

The probability of that one point being the exact point that makes contact
is vanishingly small - about as small as the probability that all of the air
in this room will just happen to randomly move to the corner of the room.

How the hell do you know that it will eventually achieve an end state that
is vertical? You have NOTHING in your model that makes it tend toward that
state.

WHY? You have NO mechanism to ensure that that happens! It's like saying
that the odds are 100 percent that a given comet out in the Oort cloud will
eventually impact the Earth.

^^^^^^^
Since we refuse to employ Time as a variable, it is 'unpredictable' *period*
--- whether for a split second, or an eternity; we will not figure in Time.

Fair enough, there ought to be a description of its energy state, and what
happens when it makes contact through just 1 point, as opposed to more
than 1.

Objection well taken. There *ought* to be some way of figuring out how
much energy bleeds off. (Now, if by colliding with the surface, however,
it takes on MORE energy, what should we make of that?)

I didn't want to have to spend weeks and weeks calculating the changes in
velocity from acceleration, deceleration, elasticity, spins, &c.

And similarly I left out time. Think about it. All eternity can pass
in the mere blink of an eye. If it is theoretically *possible* that it
would change from one state to any other state, merely by bouncing off
the surface of the sphere, then --- eventually --- or in the blink of
an eye --- it will have achieved the state whereby the condition is met.

To summarize the most important 'assumptions' of this model:

I. The initial state (location, mass, velocity, spin) of the 3-sided die
is unknown. But we do know the shape, or at least we have an idea,
of how the 3-sided die ought to look. Think of it as a massless
particle having the properties of movement and indivisibility and
individuality. Treating a 3-sided die as a subatomic particle is
obviously unsatisfying to anybody in the real world, but at least
it is a first step (of *some* kind).

II. And we know that the surrounding sphere is somewhat bigger than
the 3-sided die. Enough to allow 'free' movement.

III. We assume that the 3-sided die is moving, and will move, to such
a degree, and place, that it will, eventually, come to 'rest'
(whatever that means) on the surface of the sphere.

IV. We are interested in probabilities of the various end-states of where
this particular 3-sided die might end up.

V. I suggested that it bounces off in some *other* unpredictable (and
therefore purely random) direction. You correctly pointed out that
something ought to happen with every contact, and it should be a
little bit *more* impressive than simply changing its X, Y, Z
components.

Naturally, the model *would* work better if we were supplied with more
information, like mass, momentum, spin, velocity, and all that kind of
stuff. Um, elasticity, too. But if we are deprived of this information,
and must make up a simple rule to account for 90% of it, then it isn't
that bad of a rule.
--

Then the answer came. Construct a wankel engine shape in the middle as your
starting crossection. Then extrude the faces to the left and to the right to
form a point at each end.

Read the experimental trial as the "face down"

You will obtain your 3-sided coin.

Rich V.