Conference rusure::math

    
    Is it .........
    
    
    h=r/2
    		(r=radius of the coin , h=height of the coin)

    My money's on t(hickness) = d(iameter)/sqrt(3).
    
    Dick

     RE-1:
    If you do t=0, do you get a probability of 1/2 ? (works for me but I
    still may be wrong ..)

	If the coin is lying on one side, look at the potential energy
	needed to flip it over. With r=h/2 this energy is the same
	nomatter which side the coin is lying on.

	/�ke

    Only an engineer would build a "three sided coin." A mathematician
    would have used a die mod 3 (reduce to a previously solved problem).
    

look at .4 and .5 together.

If you take a cube and label one face heads and the opposite face tails, and the
other four faces edge, then the probabilities are

	heads = 1/6
	tails = 1/6
	edge = 4/6

although heads, tails and edge all have the same potential energy.  This
suggests that the area involved is significant.

This supports my intuition from statistical mechanics, where probabilities
generally depend on both potential energy and volume of phase space, equivalent
to energy here.

If you compute the areas for the solution of .1, you find that heads, tails and
edge all have the same area.

I'm still not entirely sure, based on the dynamics of the coin in bouncing and
coming to rest.  When it lands on heads or tails, all the rotational kinetic
energy has to be dissipated.  When it lands on an edge, some can be retained in
the rolling coin.  This may favor the edge.  But that is a matter of physics,
and very messy physics at that.

Inscribe the coin in a sphere.  I think you want the band in the middle
(projection of the "edge") to have the same area as the cap on an end
(projection of a "side").  That way, at any instant, a line straight down
from the coin's center [of gravity] has an equal probability of intersecting
each of the three regions.

This assumes that you randomly choose a moment for the coin's motion to be
stopped, and then whichever region is under the CG will come to rest on
the floor.  I'm not sure this is valid, since not all instants are equally
probable choices for the coin to first touch the floor -- which region is
more likely to be facing down when an edge hits?

    It is clearly a sphere with three concaved sphereical cuts, of the same
    size and equal distance from each other. It would come to rest in one
    of the three dished sections. 

    if you add two parabolic poits at the poles, it would almost bring to
    nill the possibility of landing spinnig on it pole.
    

It seems that some of you think there is too much physics involved
in this problem. Nevertheless, a few answers have been suggested
(.1, .2 and .4). As far as I can tell, they are all based on intuition
rather than analysis, and none of them is consistent with my own
answer (there was no answer in Poundstone's book). It may be that
different assumptions give different answers. Anyway, here is my
solution (rather lengthy, I'm afraid):

First, let us define the problem:

	Calculate the proportions of a homogenous coin that is
	thick enough to fall on its edge with probability 1/3.

	Hence, p(heads) = p(tails) = p(edge) = 1/3.

	It will be assumed that the surface on which the coin
	falls down is perfectly flat.

	The coin's proportions may be stated in terms of its
	radius, r, and its thickness, t.


Next, we define what is meant by randomly flipping the coin:

	The coin is flipped, which means that it is assigned a
	random orientation in space, at some point above the
	surface. It then falls to the surface, and comes to
	rest, possibly after bouncing, on either side or on
	its edge.

	The orientation of the coin in space is defined by
	the orientation of its cylindrical axis relative to
	a normal to the surface. In spherical polar
	coordinates, this orientation is specified by two
	angles, u and v. The angle u is the angle between
	the the axis and the normal, while v is the angle
	obtained by projecting the axis onto the surface,
	and comparing it to some reference axis in the plane.
	More simply stated, each possible orientation may be
	represented by a point on a spherical surface, the
	diameter of which is equal to the thickness of the coin.
	This is intuitively clear from rotating the coin freely
	in space. A simple figure will illustrate this.

	To summarise, the orientation of the coin is given
	by the two angles u and v, such that

		u : <0, pi>    (u=0: heads, u=pi/2: edge, u=pi: tails)      
		v : <0, 2 pi>  (not related to flip outcome)

	A random flip, then, is a flip such that all points on
	the sphere occur with the same probability at the
	moment of impact. Then we have not restrained the
	freedom of the coin in any way. Of course, the process
	is complicated by the fact that the coin may bounce.
	Although I cannot prove it (now), I think it seems
	reasonable to ignore bounces altogether without
	affecting the final result of the analysis. After all,
	some bounces merely bring the coin to rest in a state
	that is determined by the original impact angle (see below),
	and those that don't actually make the coin assume orientations
	that are distributed in the same way, or in exactly the
	opposite way, as the original orientation, and therefore
	should sum up to have the same effects as the distribution
	of orientations at the original moment of impact.

	It should also be noticed that the horizontal and
	vertical displacement of the coin relative to some
	reference point in the plane of impact does not
	affect the outcome of the flip. The same is true
	for the coin's revolution around its own axis.
	There are no other degrees of freedom.

	As for the kinetic characteristics of the coin, it
	can be assumed that these are symmetrically distributed
	and do not, on average, affect the outcome of the flip.
	This is also hard to prove, but seems very reasonable.


Determining the probability distribution:
	
	It is obvious that the only parameter affecting the
	outcome of a flip is the angle u. The angle v does
	not affect the outcome of the flip in any way.
	
	Now, by projecting the spherical surface representing
	all possible flips onto the plane of impact, we
	obtain the probability distribution in u:

		p(u) = (sin u)/2

	This step might not be totally obvious. However, it is
	clear that orientations for which u = pi/2 must be more
	common that orientations with u = pi/4, for instance. This
	is due to the fact that the coin is free to be in any of
	the orientations in v, which are projected onto a circle
	in the plane, the radius of which is proportional to sin u.
	A simple figure should clarify this. It is the same thing
	as saying that if a person is sent on a trip to a random
	point on the planet, he is much more likely to end up in
	the vicinity of the equator than near the north pole.

	Note that

		I p(u) du = 1

	where I is the integration symbol, and the probability
	distribution is integrated over the whole range of
	possible u, i.e. <0, pi>. Alternatively, integrate p(u, v)
	= (sin u)/(4 pi) over the spherical surface to get total
	probability 1. The factor 2 pi is a result of integrating
	a constant over all possible v.


Now, consider the outcome of a flip for different values of u:

	Let us define the angle w as that value of u, for which
	the coin is exactly balanced between heads and tipping
	over onto its edge. Then pi - w is the angle for which
	it is balanced between tails and edge. Hence:

		u : <0, w> = heads
		u : <w, pi - w> = edge
		u : <pi - w, pi> = tails

	We must find the angle w.

	In order to get a uniform mass distribution around the
	normal to the surface, so that the coin is balanced,
	it is necessary that

		tan w = 2r/t

	Again, a simple figure should illustrate this fact.


Finally, combine the findings above:

	I p(u) du = 1/3, when integrating over the "heads" interval.

	Using p(u) = (sin u)/2 gives w = arccos (1/3).

	Now, using tan w = 2r/t, we find that 2r = t tan w =
	= t tan (arccos (1/3)) = t sqrt(8), and hence:

		r = t sqrt(2)

	which defines the proportions of the coin. 

When I wrote .6, I confused .1 and .4.  .1 has h=r/2, which gives me the
equality in area, but .4 has h=2r which gives me the equality in energy.

In general, energy effects are more important than area effects, so I'd expect
.4 to be closer than .1.

Reply .7 introduces more physics, which is too complex for me in general.  Here
is a slight simplification.  If I give the coin a perfect traditional flip, it
will rotate with a uniform velocity around an axis perpendicular to the axis of
the cylinder.  Under these conditions .7 reduces to the lengths of arcs of a
circle and 

.7>probable choices for the coin to first touch the floor -- which region is
>more likely to be facing down when an edge hits?

can be ignored because of the uniform velocity.  

Within the limits of my algebra, I get 

	r = sqrt(3) * h / 2

which is the same as .2.

I think I have now agreed with everybody. :-)

    Let us put the following (from .-1) through a consistency test.
    (1) If t==0, i.e., an ideal two-sided coin
    (2) If t>>r, i.e., a verrry long cylinder relative to the area at the end
    
>		u : <0, w> = heads
>		u : <w, pi - w> = edge
>		u : <pi - w, pi> = tails
>
>	We must find the angle w.
>
>	In order to get a uniform mass distribution around the
>	normal to the surface, so that the coin is balanced,
>	it is necessary that
>
>		tan w = 2r/t
    
    (1) If t==0, tan w is +infinity, w = pi/2, and:		 P(�)
>		u : <0, w> = heads		<0,pi/2> heads    0.5
>		u : <w, pi - w> = edge		<pi/2,pi/2> edge    0
>		u : <pi - w, pi> = tails	<pi/2,pi> tails   0.5
    
    This looks valid to me.
    
    (2) If t>>r, tan w is near 0, and we have:
>		u : <0, w> = heads		<0,0> heads	    0
>		u : <w, pi - w> = edge		<0,pi> edge	  1.0
>		u : <pi - w, pi> = tails	<pi,pi> tails	    0
    
    This also looks valid to me. My money is on .10's �t = r / sqrt(2)�
    
      John

notes collision: it appears that .10 appeared while I was writing .11.

I don't have a problem with the math of .10, but I continue to think that the
physics is more important.

.10> It may be that different assumptions give different answers.

Yes, I think that is correct.

>	The coin is flipped, which means that it is assigned a
>	random orientation in space, at some point above the
>	surface.

When I flip a coin, I give it a linear momentum and an angular momentum.  If I
give it a perfect traditional flip, then in the terms of .10, v is a constant
(modulo pi) and u is uniformly increasing or decreasing.  This means that 

		p(u) = const
and 

>		p(u) = (sin u)/2

would not apply.  If my flip is imperfect, v will vary and u will deviate from
uniformity, but the formula above will still not apply, in general.

>	Let us define the angle w as that value of u, for which
>	the coin is exactly balanced between heads and tipping
>	over onto its edge.

This whole approach assumes that the coin has zero angular momentum.  It might
be possible to salvage it if we could assume that angular momentum were
uniformly distributed independently of u, but I could not rule out the
possibility that u and angular momentum are correlated, after the first bounce.

>	Although I cannot prove it (now), I think it seems
>	reasonable to ignore bounces altogether without

>	As for the kinetic characteristics of the coin, it
>	can be assumed that these are symmetrically distributed
>	and do not, on average, affect the outcome of the flip.
>	This is also hard to prove, but seems very reasonable.

Here is a hand-waving counter-argument.  A bounce will modify linear and angular
momentum on impact depending on their values and the value of u at impact (I
agree that, by symmetry, v and the third angular coordinate will have no
effect).  This change of linear and angular momentum will introduce correlations
between them and u at the point of the next impact.  Note that this might give
us yet another approach to a solution.  If we understood the bounce process, we
might be able to calculate the result of many bounces, ignoring the initial
conditions of the flip.

Here is another kind of counter-argument: If I give the coin a large angular
momentum about the horizontal spatial axis, then the bounces will tend to
convert this to angular momentum about the coin axis of minimal moment of
inertia.  If this is the axis of the cylinder, then the coin will tend to land
with the coin axis horizontal, u = pi/2, and roll away on its edge.

Computing the moments of inertia of a homogeneous cylinder is an easy problem,
but I don't feel up to it today.

I think that .13 has a number of valid arguments concerning the
physics of the flip. Being a physicist myself, I can appreciate
your concern with the influence of linear and angular momentum.
However, I maintain my opinion that, in a random flip, the
probability distributions in these parameters would be such
that they can be ignored from the analysis. A real in-depth
analysis would have to include friction coefficients etc,
which necessarily introduces further assumptions. Without
friction, the coin would retain its kinetic energy, and just
bounce back and forth without ever coming to rest.

Furthermore, I find it very hard to accept that, when you flip
a coin, the angle v can be assumed to be constant. A "perfect
traditonal flip", as you call it, has, to my knowledge, never
occurred. It would be particularly difficult to accomplish
such a flip using a thicker coin. Nevertheless, I think you
are right in assuming that the probability distribution in a
real-world implementation of a random flip would be affected
by the orientation of your thumb, so to speak. Therefore, the
probability distribution would not be entirely symmetric in v.
Would you agree that the "correct" answer must be somewhere in
between these two extremes, and, for a thicker coin it should
be closer to the case where v is random?

Hi,

I believe that .2 has the "correct" answer.  At least that's what I came
up with on my analysis.

.10 has a pretty fancy solution. But the error in thinking in .10 is
that for some reason u is more likely to end up at one angle that at some
other.  I don't understand this reasoning.  I agree that angle v is
irrelevant.  Once v is eliminated then why isn't u just another random
angle?  The analogy is of wandering the earth randomly and ending closer to 
the equator than to a pole requires that angle v be taken into account.
If you eliminate v then you are really wandering a great circle. As such
if you take a random walk you are no more likely to wind up at the equator
than at a pole - both are just pairs of opposing points on a circle. 
Besides, the answer for .10 just feels wrong: if you have a one inch 
diameter coin then it would be only 0.35 inches thick.  This doesn't fit
the bill.

Rather than sit back and argue these points and refer to simple diagrams
that aren't presented (! :-) I went ahead and manufactured coins with
some suggested thicknesses.  Then I tossed them and *measured* the results.
Here's what I got.

I made three coin proportions out of two types of material.  For thicknesses
I chose:

        d/sqrt(2)       too thick
        d/sqrt(3)       just right
        d/2*sqrt(2)     too thin
        
For coin stock I used:

        wood            which in some previous life might have been a broom 
                        handle but now is just wood 0.9 inches in diameter
                        
        plastic pipe    2.0 inch diameter lightweight PVC tubing.  I used
                        this because it is easy to get the right dimensions
                        and none of my wood stock is this big around.  Also,
                        by using a "hollow" coin I allow plenty of room for
                        us folks to argue about the results!

The six coins' thicknesses (in inches) are:

        proportion              wood            plastic
        --------------          ----            -------
         d/sqrt(2)              0.64            1.41
         d/sqrt(3)              0.51            1.15
         d/2*sqrt(2)            0.32            0.71
         
Then I proceeded to toss each coin 100 times and log the results:

   wood            heads           tails           edges
   -----------     -----           -----           -----
   d/sqrt(2)       25              21              54
   d/sqrt(3)       30              38              32
   d/2*sqrt(2)     42              51               7

   plastic         heads           tails           edges
   -----------     -----           -----           -----
   d/sqrt(2)       28              26              46
   d/sqrt(3)       35              31              34
   d/2*sqrt(2)     51              37              12

Clearly "d/sqrt(3)" is the winner!

If you want to toss these coins yourself I have them in LKG1-3.  It isn't
like tossing a regular coin.  When you toss these babies they roll away.
I started tossing them on my kitchen table but they rolled off onto the 
floor.  So I tossed them on the floor.  There they became fair game for
the cats who sent my wonderful coins under the refrigerator!  I'm down
there with a stick and a flashlight digging around in the lint all for the
sake of science trying to retrieve my disks.

Also I got to observe first hand how the effects of horizontal motion
change the fall of the coins.  For the flips listed above I tried to give
the coins a good flip with the center of mass going straight up and down.
If you flip them horizontally they tend to stand up and roll.  I can take
the whole handful of six coins and roll them in a bowling ball motion and
have all six finish on edge!  Consider this bait for arguing what 
constitutes a fair flip. ;-)  Interestingly, the plastic (d/2*sqrt(2)) got
way more edges because on a lot of bounces the plastic ring shot away from
the landing spot and picked up a roll.  The wood in that proportion didn't
do that at all.

FWIW,
Chuck

Thanks for doing the experiment. I was just waiting for someone
crazy enough to do it...

I haven't had time to study all your results yet, but I'd like
to clarify the point I was making about the probability
distribution:

  Imagine that the centre of the coin is fixed. If you allow
  the coin to rotate freely around this fixed point, you will
  sketch the contours of a sphere. In my opinion, a random
  flip should be defined as a flip in which every point on
  the surface of the sphere occurs with the same probability.
  Otherwise, you would have to introduce some assumption
  that restricts the flip in some way. It could be argued
  that "a perfect traditional flip" restricts the sphere to
  a very narrow part of it, i.e. basically a circle.

  Now, if you designate one point on the sphere as the north
  pole, you could identify that point with angle u = 0. That
  would be heads. Unquestionably, there is only one such
  point on the sphere. This is different to u = pi/2, which
  is a circle on the spherical surface, i.e. "the equator".
  The probability of the coin being aligned along this
  circle is proportional to the length of the circle, i.e.
  p(u, v) integrated over all v. At the north pole, the same
  argument applies, but the circle is reduced to a point,
  which means virtually zero radius, and hence zero
  probability.

  Thus, the angle v has no impact on the probability
  distribution in v, but it does affect the distribution in u.

>  that restricts the flip in some way. It could be argued
>  that "a perfect traditional flip" restricts the sphere to
>  a very narrow part of it, i.e. basically a circle.

Indeed, this explains the difference.  Your "complete flip" also
accounts for flips wherein the coin's cylinder axis is horizontal and
the rotation of the flip is also about the cylinder axis.  With a flip, 
or a roll, like that even a normal coin should land on its edge!

Regards,
Chuck

  Well, using a "perfectly traditional coin" having d = 25 t,
  the coin should land on its edge approx 4% of the time.

  However, this requires the surface to be perfectly flat.
  Using a somewhat thicker coin, it is easily verified
  empirically that a rough surface will push the odds in
  favour of heads/tails.

  Again, different assumptions will give different answers.

  In the real world, my result would probably not be the
  best one, but then again, this is the math conference,
  and not the real world... 

    Re .15
    
    Chuck, well done in persevering with the experiment (especially the bit
    where you verified my answer :-) I was guilty of not showing my
    working, something I'm always criticising others for, but all I used
    was geometry. I assumed that just before coming to rest it was poised
    on an edge, any rotation about the tangnt of contact being ignored -
    assuming that to be equally likely in either direction. I'm also
    assuming that tangent to be stationary - no spin. So the dimensions are
    given by t/d = tan-1(30) = 1/sqrt(3). I think there is a similar
    derivation buried in the physics in a previous reply.
    
    I cut up some 22mm copper tube to perform the experiment, but got
    bored. I couldn't interest any of the children in a science project
    either.
    
    Does the Scientific American still have that Amateur Scientist column?
    Perhaps you should enter your experiment.
    
    Re .18 (1/25 coin)
    
    On the same principle I make it 2.5% ( 2/(25*pi) ). But still very
    high, I agree. I suspect that ignoring spin in my derivation is more
    critical than requiring a flat surface.
    
    Dick

I commend you on actually *trying* it.  Good spirit !

Here's another idea on how to try it.  Use real coins, and either glue several
together, or just put a rubberband around them (and we can argue about how
much the band destroys the results).

/Eric