This is a popular gag in movies, but I wonder how likely it really is.

What is the probability that a uniform cylindrical coin (with radius $1$ and height $h$) lands on its side? If the ground were some kind of superglue so that it stopped moving as soon as it touched the ground, and the final result is declared to be "heads" or "tails" or "side" depending on which part of the coin is on the upward ray originating at the center, then I would expect that the probability would be $\dfrac h{\sqrt{4+h^2}}$, or with $h=0.183$ for the relative thickness of a nickel (the thickest of US coinage), $p=9.16\%$. (This is the relative solid angle of the side as projected out onto the circumsphere of the cylinder.) However, I think this is a significant overestimation, because the stable equilibrium at the side is much more energetic than the one on heads or tails (when $h<2$), so subsequent bounces would push it toward the faces more than the side.

What model could make a decent estimate of what the actual probability is here? If the bounce coefficient is zero, the above analysis should still be correct, since it will land and then fall toward the side underneath the center of gravity. But real surfaces have a higher bounce coefficient, perhaps returning 10-20% of the initial kinetic energy after each bounce, so there is a possibility of other funny stuff happening after the first bounce which I don't have a good handle on. Since it has never happened in reality in my experience, I would have to guess at most $p<10^{-5}$, but since one can balance a coin without too much difficulty, it is certainly nonzero.

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    $\begingroup$ According to adsabs.harvard.edu/abs/1993PhRvE..48.2547M, about 1/6000. $\endgroup$ – Rations Dec 17 '15 at 5:07
  • $\begingroup$ Do you assume the coin is dropped with random angular velocity about all three axes? $\endgroup$ – Carl Witthoft Dec 17 '15 at 14:49
  • $\begingroup$ @Carl My guess is that the answer is not too senitive to magnitude of dimensional parameters, so either a 3D Gaussian distribution around zero or a uniform distribution inside a sphere with spread $\omega$ should do for the distribution of the initial angular velocity. Then again, the moments of a cylinder are not all equal so I don't know if it is more natural to distribute the angular momentum inside a sphere instead. Since the angular position is also randomized it probably all balances out, though. $\endgroup$ – Mario Carneiro Dec 17 '15 at 23:52

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