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### Coin Landing on Edge

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Date: 11/20/2001 at 22:45:54
From: Robert
Subject: Coin Probability

I am trying to assess how to determine if there is a way to calculate
the odds of a coin dropped from a known height landing upright, on its
edge, rather than heads or tails.

Due to the extreme random nature of all the parameters involved we may
apply certain constraints to the problem:

1) A fixed height, let's say 3 ft.
2) A quick release mechanism that drops the coin vertically each time
in the same way
3) Known surface - i.e. flat granite table
4) Known properties of the table and coin, to estimate energy
absorption
5) Quiescent Room conditions - only air resistance a factor
6) The different possible angles of approach and its effect on the
elastic bounce
7) Let's assume we use a nickel. Flat edge (no ridges), and we know
the weight, Center of mass, elasticity, and surface area of the
coin's edge relative to the coin's flat sides.

Is this result possible? Is there a closed form solution to this
problem? I assume that their may be a way to calculate this, but do
not know how. I also assume that the odds are astronomically high and
that this result is not impossible.

Thanks.
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Date: 11/20/2001 at 23:27:57
From: Doctor Jubal
Subject: Re: Coin Probability

Hi Robert,

Thanks for writing to Dr. Math.

You're right that this is a very random problem, and we will have to
impose some constraints on it to make it even close to mathematically
tractable.

Your constraints are a good start, but let me impose one more that
will make no sense at first and then I'll justify why it's reasonable.

Let's say that the coin is "sticky" - once it strikes the table, it
doesn't bounce back. It just settles onto one side or the other, or
occasionally the edge.

Of course this is a nonsensical assumption. Flip a coin and watch it;
the coin bounces back. But if the coin is dropped from a very small
height, it will bounce back so little that it makes no difference.

You will now protest that you're dropping the coin from a much larger
height, and it does bounce back, and so this assumption makes no sense
at all. I agree with you, but think about it this way: when the coin
bounces back, it does so in a way that's terribly chaotic and random
and generally impossible to predict the exact details of. So, however
high it bounces back, it's as if we're dropping the coin again from
some new height. But the coin loses energy on each bounce, so the new
height is less than the previous one, and sooner or later, it will
bounce back to a height that's so small that the next rebound will be
negligible. Then we're back to our approximation that the coin is
"sticky" and doesn't bounce at all.

The real reason for making this assumption, though, is that it makes
the problem mathematically tractable. A sticky coin will settle onto
its edge if, when it strikes the stable, its center of gravity is
located directly above the edge. If instead the center of gravity in
located above one of the faces, it will settle onto that face.

Picture the coin as being inscribed inside a sphere such that both
rims of the coin touch the sphere at all points. There's a narrow
"belt" around the "equator" of the sphere between the rims. We drop
the sphere at some random orientation onto the surface, and so any
part of the sphere is equally likely to tounch the surface first. If
the part of sphere touching the table is part of the "belt" between
the rims, then the coin's center of gravity lies directly above the
edge, and the coin will settle onto its edge. On the other hand, if
the sphere lands outside the belt, it will settle onto a face.

So to solve this problem, you need to

(a) Figure out the area of the "belt." You'll need some calculus to
do this.
(b) Divide by the total surface area of the sphere to figure out the
probability that the sphere lands on the belt.

more, or if you have any other questions.

- Doctor Jubal, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Probability
High School Probability

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