Coin Problem.

[mfgoode]

Here is a problem on probability that I have often thought about.
A coin has the probability to fall 1/3 on its head, 1/3 on it tail and 1/3 on its edge (if it is thick enough)
What wil be its dimensions given that the radius is R ?
Mally

[xilman]

Originally Posted by mfgoode

Here is a problem on probability that I have often thought about.
A coin has the probability to fall 1/3 on its head, 1/3 on it tail and 1/3 on its edge (if it is thick enough)
What wil be its dimensions given that the radius is R ?
Mally

Just for clarification, can we tighten up the specification?

Is the coin of uniform density throughout? If not, how is the density distributed?

Is the coin a perfect right-cylinder? If not, what is it's geometry?

Are the coin and surface on which it lands of perfect rigidity? If not, describe how each may deform.

Is the surface horizontal? If not, what is its orientation?

Does the coin land with zero angular momentum? If not, what is the distribution of this vector quantity?

Is the orientation of the coin and the surface at first contact uniformly distributed over the complete sphere? If not, what is the distribution of their mutual orientation?

All these questions are designed to point out that the original problem is non-trivial to solve. Indeed, the problem may be intrinsically unsolvable.


Paul

[mfgoode]

[QUOTE=xilman][Just for clarification, can we tighten up the specification?

[Is the coin of uniform density throughout? If not, how is the density distributed? ]

>It is of uniform density.

[Is the coin a perfect right-cylinder? If not, what is it's geometry?]

>Yes it is

[Are the coin and surface on which it lands of perfect rigidity? If not, describe how each may deform.]

>Normal conditions

[Is the surface horizontal? If not, what is its orientation?]

> The surface is horizontal.

[Does the coin land with zero angular momentum? If not, what is the distribution of this vector quantity?]

>On edge at least it has zero angular momentum.

[Is the orientation of the coin and the surface at first contact uniformly distributed over the complete sphere? If not, what is the distribution of their mutual orientation?]

>I don't quite understand this question but I would say uniform distribution.

[All these questions are designed to point out that the original problem is non-trivial to solve. Indeed, the problem may be intrinsically unsolvable.]

>Not exactly Paul. As we say in Chemistry all 'conditins under normal temp and pressure'

The problem is a simple one and theoretically possible to solve by simple trig.
Mally

[alpertron]

The problem is not that simple. We have to know how the force that maintains the coin bouncing disappears. So we have to consider that we have air, which makes the problem too complex. The "normal conditions" you say above does not describe the impulse absorbed by the floor.

[mfgoode]

ALPERTRON:
You must accept what is given as possible. You don't examine a centipede by counting if the number of its legs is exactly 100 or more or less. By actually counting the number will not enable you to get a better idea of what a centipede actually is.

Similarly with this coin problem. I have made a Google search and admit it's difficult to find a solution. Personally I failed. But that does not mean it is not reasonable. This problem had been posed to millions of readers ot the TOI and no one has given any hypothetical objections as to the conditions the coin may be subjected to.

We are not scientists or engineers to lay down the pros and cons of the
problem and its various parameters. We are mathematicians who dissect a problem whether its hypothetically possible ot not.

Therefore we must accept it as it is given. its is difficult as its given and there is no sense making it more complicated.
So given that a coin on being tossed 'Can' come to rest in three basic positions viz: heads, tails, or on edge what are its dimensions in terms of its radius. In other words what is the dimension of its edge(thickness) if the possibility is 1/3 for each position.
Its a simple Trigonometry problem!
Hint at what angle will it be inclioned to topple over?
Mally

[alpertron]

Your problem is not about physics. You can put that thick coin in your hand a meter over the floor with the head side pointing towards the floor. Assuming that R is at most 5 cm, after bouncing several times, it will rest in any of the the three positions you cited above.

Then you start approximating your hand to the floor. After the distance is less than R, chances are more than 1/3 that the head side will continue pointing to the floor, and this chance increments when you continue approximating your hand to the floor.

The edge position has more probability than you think at a first moment. How do you think that bicycles work? If after bouncing part of the impulse makes the coin start spinning, the gyroscopic inertia will try to put the edge against the floor, thus increasing the probability that after things calm down the winning position is "edge".

Sorry for my bad English. I could make this post ten times as long if I were able to write in Spanish. But from the above you can see that the problem is not trivial at all.

[alpertron]

Originally Posted by mfgoode

We are not scientists or engineers to lay down the pros and cons of the
problem and its various parameters. We are mathematicians who dissect a problem whether its hypothetically possible ot not.

My degree is "Electronic Engineer", and I had to study a lot of physics so I only wanted to warn you about reality.

Maybe you want to talk about some ideal coin in ideal conditions that you have to specify.

[Richard Cameron]

Originally Posted by mfgoode

Here is a problem on probability that I have often thought about.
A coin has the probability to fall 1/3 on its head, 1/3 on it tail and 1/3 on its edge (if it is thick enough)
What wil be its dimensions given that the radius is R ?
Mally

and

Originally Posted by alpertron

Maybe you want to talk about some ideal coin in ideal conditions that you have to specify

What is your aim here Mally? My first thought on reading your post was

get some tins of different shapes;
drop them on the kitchen floor a few hundred times each;
examine the distributions of end and side orientations.

ie I would start from empirical evidence.


My approach might produce an answer to your original query, but it would of course give no mathematical insight into the idealised problem. It would probably also mark the lino.


Richard

[mfgoode]

Originally Posted by alpertron

My degree is "Electronic Engineer", and I had to study a lot of physics so I only wanted to warn you about reality.

Maybe you want to talk about some ideal coin in ideal conditions that you have to specify.

Thank you Alpertron for your comments. The ideal conditions are that whatever affects the coin causing it to be face down applies to the edge on position making the probability the same viz: 1/3.
You must take it for given that this is possible without introducing any parameters, elasticity, torque, momentum, etc.

Quote:
[Originally posted by Richard Cameron;
ie I would start from empirical evidence.
My approach might produce an answer to your original query, but it would of course give no mathematical insight into the idealised problem. It would probably also mark the lino.
Richard]
No Richard save your lino, but sharpen your pencil instead !
This is a trigonometric problem and may I add that it is not a concoction from my fevered brain. It appeared in MIND SPORT column by Mukul Sharma in the Times of India. Perfectly plausible answers which were abs. correct were sent in.
I have presented many such problems myself which were published in this weekly column.
To get an idea of it, take one of your beer cans standing on its base and tilt it and find at what angle it is on the point of turning on its side. The centre of gravity will shift from its centre in an arc and come in a straight line (always vertical) with the edge. The theory is that when the projected CG. gets out of the base area it will tilt. This applies to any shape including the bicycle. If the vertical line of the CG is moved out of base area the object will tilt.
That should enable you to derive the width of the edge from its radius.
Mally

[alpertron]

Originally Posted by mfgoode

Thank you Alpertron for your comments. The ideal conditions are that whatever affects the coin causing it to be face down applies to the edge on position making the probability the same viz: 1/3.
You must take it for given that this is possible without introducing any parameters, elasticity, torque, momentum, etc.

Suppose you drop the coin vertically but not with the faces or edge making an non-null angle with the floor (it is practically impossible to have non-null angles).

When the coin touches the floor it will bounce, but since the other side of the coin didn't touch the floor yet, that point continues falling, generating a torque (one part of the coin is falling while the diametrally opposite side is going up), so when the coin bounces you will see that it spins. In presence of air we would have to consider Magnus effect but even with no air this problem can be very tricky, since it is possible that after multiple bounces the thick coin lands and starts running over its edge, increasing the probability that the edge touches the floor when the coin finally rests as I said in a previous post.

So I think that this problem in particular is extremely compilcated even in ideal conditions.

[mfgoode]

Originally Posted by alpertron

Suppose you drop the coin vertically but not with the faces or edge making an non-null angle with the floor (it is practically impossible to have non-null angles).
snip....

So I think that this problem in particular is extremely compilcated even in ideal conditions.

Thank you alpertron for your observations.
Since this leads us to nowhere in particular I reproduce two answers which were accepted.

1) For the probability to be the same, since the coin is round, a vertical line through the centre of gravity should pass through a corner when the coin is tilted by 30* angle from the 'standing on edge' position..
This way, out of possible 360*, one third (120*) will correspond to the coin eventually resting on the edge.
There fore the width of the coin should be 2R/sq.rt.3 where R is the radius of the coin.

2) In order to have a probability of 1/3 in a cylindrical coin, the centre of the coin must subtend an angle of 60* with the thickness of the rim.
Rest is trigonometry. Thus thickness must be 2R/sq rt 3 or 1.547 R nearly.
N.B. * stands for degrees.
Mally .