Coin Landing on Edge
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. Any insight would be helpful. Thanks.
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. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/
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