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A classical coin has almost no chances of ending its course on the side when tossed. A round pencil with both ends flat has no chance of ending its course on the tip, when tossed.

What would be the proportions of a cylinder that has 1/3 chances to land on one end, 1/3 chances to land on the other end and 1/3 chances to land on the side?

I think my basic approach by integration is useless, I'm not even sure how to start.

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  • $\begingroup$ Wouldn't you just want the surface area of the side to be the same as the area on the ends? If an end has area $\pi r^2$ , then construct the side so that it has surface area=$\pi r^2$ If your cylinder had radius r, then the area of each of the sides would be $\pi r^2$ . If the height were =$r/2$ , then its surface area would be $2\pi r r/2$, so each would have the same area. $\endgroup$ – DBFdalwayse Sep 6 '13 at 6:47
  • $\begingroup$ I guess $2r = h$. $\quad r\,$: radius. $\quad h\,$: height.$\quad$ It's just a conjecture. $\endgroup$ – Felix Marin Sep 6 '13 at 6:50
  • $\begingroup$ @DBF No, it would work if it was a sphere, but the distance between each point of the coin and the center of the object is not constant, this approach is useless, I tried. $\endgroup$ – SteeveDroz Sep 6 '13 at 6:54
  • $\begingroup$ I don't think so, I think it depends of the angle covered by the surface from the center of gravity. That's all I could figure out. Imagine a real coin with the dimensions you said. It won't have the properties I want. $\endgroup$ – SteeveDroz Sep 6 '13 at 7:01
  • $\begingroup$ With a cylinder, if it lands on the curved face it can roll without changing the side in contact with the surface on which it lands. If it lands on an end there is a symmetric instability. That is hard to analyse. $\endgroup$ – Mark Bennet Sep 6 '13 at 7:17
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This is a physics problem, so the answer depends on the simplifying assumptions that we make in order to reduce it to geometry. It turns out that if the coin lands without bouncing, and it is dropped from a random orientation, then the fair aspect ratio (height/diameter) is $1/2\sqrt2$. On the other hand, if the coin is flipped vigorously, then the fair aspect ratio is $1/\sqrt3$. And if the coin bounces, the answer is more complicated.

Source: Ee Hou Yong and L. Mahadevan (2011), "Probability, geometry, and dynamics in the toss of a thick coin", American Journal of Physics 79 (12) http://www.seas.harvard.edu/softmat/downloads/2011-13.pdf

See also the previous discussion at Calculating the probability of a coin falling on its side and Find thickness of a coin

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The center of mass model ($\frac{h}{R}=\frac{1}{\sqrt{3}}$) gives the best statistical results, but not great. Here is an excellent paper that discusses the physics of bouncing: https://statweb.stanford.edu/~owen/courses/306a/threesidedcoin_amstat.pdf

The authors, who developed a model that included the effects of $2$ bounces, rejected tosses that resulted in an edge landing that caused a roll over one foot.

I suspect that if a soft surface were used (like a pool or craps table), the center of mass model would be more accurate.

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Let your cylinder have radius r and height =$r/2$ . Then the area on each of the ends is $\pi r^2$, and the surface area of the side is $2\pi rh=2\pi rr/2=\pi r^2$, i.e., each would have the same area as the others.

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    $\begingroup$ I'm pretty sure it's not only a question of surface but that it depends of the distance of that surface with the center of gravity. A surface close to the center is more stable than the same surface a bit further. $\endgroup$ – SteeveDroz Sep 6 '13 at 6:57
  • $\begingroup$ So maybe it is more of a physics problem? Then I'm out of my comfort zone. $\endgroup$ – DBFdalwayse Sep 6 '13 at 6:59
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I think the answer depends on the way you toss the „coin“.

Let’s regard the coin as a body of rotation around the $y$-axis generated by a rectangle with the vertices $(\pm r, \pm h/2)$ and say we toss it so it rotates randomly around the $z$-axis. When the coin hits the floor, physics tells us, that to see on which side it falls, we have to drop the perpendicular from the mass centre. If it points into the direction of one of the faces, the coin will fall on that face and if it points in the direction of the side, the coin will fall on the side.

Now we assume that the angle of rotation and so also the direction of the dropped perpendicular, at the point of impact, is uniformly distributed over $[0, 2π]$. In the $x$-$y$-projection denote the angle $(b/2, -a/2), (0, 0), (r, h/2)$ as $α$ and the angle $(-r, h/2), (0, 0), (r, h/2)$ as $β$ (so that $α+β=π$). Now if top, bottom and the side of the coin should have the same probability, under the above assumption $2α$ (the angle range for outcome ‚side‘) must be equal to $β$ (the angle range for outcome ‚top‘ as well as for ‚bottom‘). Thus $$α+β = 3α = π ⇒ α = \frac13π, β = \frac23π\\\frac{2r}{h} = \frac{\sin (β/2)}{\sin (α/2)} = \sqrt{3}$$

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Ok, Here is the problem that I see with the above answers. If you take the cylinder as described by bodo, for example, and balance it on the rim so that the center of mass has the maximum amount of potential energy without rotation then you have already have chosen 2 of the 3 "sides" of the coin. If "heads is +h and tails is -h, r = edge, and S is the surface that this cylinder is balanced on. Therefore, if the contact point on S is any point on tails, then the coin becomes a one of heads or edge. Equally, if the contact point on S is any point on heads, then the coin becomes a one of tails or edge. I am sure with extra motion more edge problems will come up.

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