# How thick should a cylindrical coin be for it to act as a fair three-sided die?

When flipping a coin of radius $r>0$ and thickness $t>0$ in the real world, there is some non-zero probability of getting neither heads nor tails, but instead landing on the thin lateral side. My question is, how thick does this lateral face need to be relative to $r$ so that the coin has equal chance to land on each of its 3 faces? That is, how do we choose $t$ so that the chance of landing on the lateral is $\frac{1}{3}$?

Clearly this problem requires some simplification, so we may assume the coin does not bounce and lands on a flat surface, and, well, any other assumptions implicit in the next paragraph:

I have heard from a friend that there is a method of solving this which entails centering the coin in a sphere, and adjusting $t$ so that a randomly chosen line segment between the center and a point on the sphere has an equal probability of intersecting each face of the coin. Is this a common modeling technique? How should we choose the radius of this sphere? Does it not matter given that the coin is fully contained?

And secondly, how can we proceed to find $t$ as a function of $r$ once this model has been established?

• That's really interesting - I don't have much background in applied maths but it intuitively makes sense as a modelling technique: rotate the sphere so that radius point to the earth's centre of gravity, assume the surface is perpendicular to this, then the coin will rest on whichever face the radius pierces. Given that certainly I would say the radius is immaterial as long as the coin's edges are touching the sphere. I guess you would just need to equalise the 3 regions the coin's edges form on the sphere? Nov 24, 2016 at 23:36
• The radius of the sphere won't matter. Nov 25, 2016 at 0:14
• "Assume a spherical cow." This problem depends so strongly on the method of throwing, and the properties of the surface, that any theoretical answer will be worthless as a practical guide. But I suppose you know this. Nov 25, 2016 at 1:58
• Numberphiles video on this is very related. This problem in particular is discussed at 4:30. heres the video Nov 26, 2016 at 1:13

As correctly stated in the comments, this problem could be solved (only in the very simplified scenario described in the OP) by considering that, once we have centered the coin into a sphere, the contact point of the coin with the ground identifies a specific radius of the sphere, oriented vertically and normal to the landing surface. Since any radius passes through the center of mass of the coin, that radius necessarily points towards the face on which the coin will finally land. This can be better visualized by considering the case in which the coin touches the surface in a manner that the radius exactly points towards one of its edges. Here is a figure of the section of the cylindrical coin (yellow), represented inscribed in the sphere (green). By simplicity, I used a sphere whose radius is equal to the diagonal of the rectangular section of the cylinder (however, as already noted in the comments, the radius of the sphere does not matter). In the section, the round faces of the coin correspond to $a$ and $a'$, and its lateral surface corresponds to $b$.

If the coin touches the landing surface with its edge, its center of mass is aligned with the vertical radius. This case, in the simplified scenario (no bouncing, no inertia, and so on), corresponds to the only theoretical possibility in which the coin would remain in equilibrium on its edge. On the other hand, if the radius points towards the circular face corresponding to $a$ (i.e., if the center of mass is above this face), the coin will finally land here. Similarly, if the radius points towards the lateral surface corresponding to $b$ (i.e., if the center of mass is above this surface), the coin will finally land here. The same considerations can be made in the symmetric case that the coin touches the landing surface in the other edge, between $a'$ and $b$.

Therefore, in order to have probabilty equal to $1/3$ that the coin lands on the lateral surface $b$, we have to calculate $b$ so that the surface of each of the two spherical caps corresponding to $a$ and $a'$ is equal to $1/3$ of the total surface of the sphere. The surface of a spherical cap is $2 \pi R h$, where $R$ is the radius of the sphere and $h$ is the height of the cap. In our case, we have $h=(2R-b)/2$, so that the surface of each of the two spherical caps is $\pi R (2R-b)$. Setting this equal to $\displaystyle \frac{4}{3} \, \pi R^2$ we obtain

$$\pi R (2R-b)=\frac{4}{3} \, \pi R^2$$

from which we get

$$b=\frac{2}{3} R$$

To calculate $a$, we can simply note that $a^2 + b^2=(2R)^2$, so that

$$a^2=4R^2- \frac{4}{9} \, R^2=\frac{32}{9} \, R^2$$

from which

$$a=\frac{4}{3} \sqrt2 R$$

Lastly, because the radius of the coin is $\displaystyle a/2=\frac{2}{3} \sqrt2 \, R$, we can conclude that, to obtain an equal probability of landing among the three faces, the ratio between the radius of the coin and its thickness must be $\sqrt2$.

Again, it must be pointed out that these calculations cannot be considered valid in a realistic scenario, where a number of confounding factors contribute to determine the final way of landing.

• love this question! sadly was very late when got to it so didn't get around to doing the calculation myself, but whoever came up with putting the coin in a sphere to get the answer was having a good day I think, a very nice method! Nov 25, 2016 at 11:57
• Thank you to MattG88 and MathVandal for your comments. I would also add that this question is not only nice, but is also intriguing because it yields a very simple final result in terms of the ratio between coin radius and thickness. Nov 25, 2016 at 12:36
• Also note that such problem could be reformulated with a number of interesting variations and generalizations. Nov 25, 2016 at 17:35
• With a ratio of $\sqrt{2}$, if the coin is dropped onto a hard surface and bounces repeatedly, $a$ and $a^\prime$ will each come up more than $b$. This is because the energy to get from $a$ to $b$ is greater than the energy required to do the reverse. Thus, the coin will become "trapped" in $a$ more often than in $b$. Dec 28, 2019 at 5:35
• Using a 3D printer, we made "coins" with this ratio and dropped them onto a hard surface 3179 times. The edge came up only 10% of the time. Which means heads and tails got about 45% each. Mar 6, 2020 at 3:52

Unless we admit mass and gravity, it’s hard to argue that we’re talking about a “coin toss”. Once we admit mass and gravity, then we must admit friction (else our coin will never settle). If we seek a geometry that will result in three equally probable rest states after an injection, then a dissipation of energy, then the corresponding potential wells must have equal depth. This results in a cylinder with r = t/2 - the diameter and length are equal. But we’re not done yet. The wells may have equal minimum energy, but the terrain between them is different. If we admit any momentum normal to gravity, then the “rolling” state will have heavy preference over “heads” or “tails”.

This proposed 3-die is fundamentally different from the 2-state coin or the next one up, a 4-die (a tetrahedron). With both, each state is separated from each of the other possible states by identical terrain, so each offers an equally probable end state after a sufficiently energetic upset.

My solution is similar, and I found many other sources of the same solution later in the literature. I started with a rectangle, as you'd see the coin from the edge side. It's circular, and therefore symmetrical by any line that goes through the centre, and symmetrial around the point in the centre. Therefore I figure a rectangular depiction of the coin is sufficient to create a model.

To make it equally equal that the coin fall on its edge, or on either side, it is sufficient that a line through the edge of the rectangle and the centre of the rectangle, on two edges of the shorter side, describe a 60 degree angle at the centre. This gives 120 degrees of probability of either sides of the rectangle to fall on, and 60 degrees probability on the edge, but also on the other side of the rectangle, so the two edges add up to 120 degrees as well.

Geometry gives you r^2 = D^2 + s^2, where r is half the diameter of the rectangle, D is half the diameter of the coin, and t is half the height of the cylinder. If you take r as given, for instance as 1, then s=1/2, since sin (30) = 1/2, and and D is sqrt(3)/2.

There is an other solution: the equal area solution. This presupposes that equal area surfaces will attract, so to speak, equal area of landings.

The area of one circular side is PiDD=2PiDt, which is the area of the planer edge of a cylinder. Divide both sides by PiD, and you get D = 2t, which means the diameter of the coin is twice the distance between the two faces of the coin, that is, it is the height of the cylinder.

Experimental results do not justify any of the three solutions. This is so because of the third law of thermodynamics: entropy, according to which every object will seek, so to speak, its lowest possible energy state. In translation this says, that if the coin falls on its edge, or on its side, in an elastic collision of the surface it lands on, and therefore it bounces, then chances are bigger that it will come to rest on a circular side than on its edge surface.

I worked out the mathematical models myself; my friend Geza Gyuru has come up with why the math models do not yield an experimental result due to the third law of thermodynamics.

This is an old and charming question. The thick coin is a cylinder. Cut it vertically by its diameter. You get a rectangle by section with sides $$e\wedge D\$$. The thickness is $$e$$ and $$D$$ its diameter.

When you flip the coin the decision variable are the angles that form the rectangle diagonals. These angles determine if the rest position will be on some head-tail or on the edge. Name $$\vartheta_e$$ the angle between diagonals having the edge as the opposite side.

The fact is that the angle is a random variable having uniform distribution U[0,2$$\pi$$]. $$\$$Thus $$\frac{2\vartheta_e}{2\pi}=\frac{1}{3}$$ and $$\vartheta_e=\frac{\pi}{3}$$. But $$\tan(\vartheta_e/2)=e/D$$ and $$e=D\cdot\tan(\pi/6)=D/\sqrt{3}$$.

I've seen some wrong answers having high score for this question. I do not know, but I hope this can be ammended.