# How to determine the probabilities for a cuboid die?

Imagine we take a cuboid with sides $a, b$ and $c$ and throw it like a usual die. Is there a way to determine the probabilities of the different outcomes $P_{ab}, P_{bc}$ and $P_{ac}$? With $ab$, $bc$, $ac$ I label the three distinct faces of the die (let us not distinguish between the two opposide sides).

I would guess that the probabilities can not be calculated solely by the weights of the different areas (something like $P^\text{try}_{ab}=ab/(ab+bc+ac)$).

I believe this is a complicated physical problem which can in principle depend on a lot of factors like friction of air and the table, the material of the die, etc. However, I would like to simplify it as much as possible and assume we can calculate the probabilities just by knowing the lengths of the different sides.

For a start, let us assume that two sides $b=c$ are equal (that is, we only have two events $ab$ and $bb$). Now with dimensional analysis we know that the probabilities $P_{ab}$ and $P_{bb}$ can only be functions of the ratio $\rho=a/b$. We also have $P_{ab}(\rho)+P_{bb}(\rho)=1$ and we know, for example, that (i) $P_{ab}(0)=0$, (ii) $P_{ab}(1)=2/3$ and (iii) $P_{ab}(\rho\rightarrow\infty)=1$.

My question is: is there a way to determine $P_{ab}(\rho)$?

Bonus: Since I am too lazy to perform the experiment, would there be a way to run this through some 3D rigid-body physics simulation and determine the probabilities by using a huge number of throws (of course this is doomed to fail for extreme values of $\rho$)?

Remark: Actually, the function $P^\text{try}$ given above fulfills all three properties (i)-(iii). For $b=c$ we have

$P^\text{try}_{ab}=2\frac{ab}{2ab+b^2}=\frac{2\rho}{1+2\rho}$

(the additional factor of 2 comes from the fact, that we have four sides $ab$ instead of two like in the asymmetric die above)

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I suspect that this is a hard problem. – dmckee Oct 20 '12 at 19:38
I love when scientists use that sentence. – Colin K Oct 20 '12 at 22:31
Could the answer depend on what you assume for the distributions of initial state parameters? For example, do you weight all potential axes of rotation equally? Or are you more likely to impart a spin around, say, the principal axis with the smallest moment of inertia? But since all that seems hard, I'd be tempted to assign a potential energy to each final configuration, assume some equal-likelihood microstate stuff, define a temperature, and say the answer follows a Boltzmann distribution. – Chris White Oct 21 '12 at 8:33
@Chris: If I had to give initial conditions, I would choose the rotation axis randomly (uniform). Your approach of using a Boltzmann distribution sounds interesting, although I have no idea how you would justify it and how you would define your temperature. But let's try to work it out as far as possible: the potential energy of the configuration $ab$ is just $Mg b/2$ while the energy of configuration $bb$ is $Mg a/2$. Instead of defining an artificial temperature, we can also define a length $\lambda$ related to the temperature... (continuing in next post) – shark.dp Oct 21 '12 at 9:09
I think general solution to your conditions would be $P_{ab}=f(ab,bc,ca),P_{bc}=f(bc,ca,ab),P_{ca}=f(ca,ab,bc)$, where $f$ is a function which is symmetric in its second and third arguments, and goes to zero when its first argument goes to zero. For example $f(x,y,z)=x^n/(x^n+y^n+z^n)$ for any $n$ is also a solution. – user10001 Oct 21 '12 at 14:07

I think a reasonable first approximation can be made like this: choose an arbitrary orientation for the die, and figure out, if the die were released in that orientation with its lowermost point resting on a surface, which side would it fall on? That can be easily calculated; you just draw a line going straight down from the center of mass, and whichever face it passes through, that's the one that will land down. We can then assume (this is the unjustified approximation) that the orientation of the die before it begins its final tip is uniformly random, so the problem is reduced to finding the proportion of all possible directions which pass through each face. Hopefully it's clear that that is just the solid angle subtended by the face divided by the total solid angle, $4\pi$.

For a cuboid, we can figure out the solid angle of a face by doing an integral,

$$\Omega = \iint_S\frac{\hat r\cdot\mathrm{d}A}{r^2}$$

as given on MathWorld. Take the face to lie in the plane $z = Z$ (capital letters will be constants) and to have dimensions $2X\times 2Y$. The integral then becomes

\begin{align}\Omega &= \int_{-Y}^{Y}\int_{-X}^{X} \frac{(x\hat{x} + y\hat{y} + Z\hat{z})\cdot\hat{z}}{(x^2 + y^2 + Z^2)^{3/2}}\mathrm{d}x\mathrm{d}y \\ &= Z\int_{-Y}^{Y}\int_{-X}^{X} \frac{1}{(x^2 + y^2 + Z^2)^{3/2}}\mathrm{d}x\mathrm{d}y\\ &= 4\tan^{-1}\biggl(\frac{XY}{Z\sqrt{X^2 + Y^2 + Z^2}}\biggr) \end{align}

This gets divided by $4\pi$ to produce the probability. Translating into your notation, and multiplying by 2 to take into account the two opposite sides under one probability, this becomes

$$P_{ab} = \frac{2}{\pi}\tan^{-1}\biggl(\frac{ab}{c\sqrt{a^2 + b^2 + c^2}}\biggr)$$

$P_{bc}$ and $P_{ca}$ are the same under the appropriate permutation of the labels.

Some sanity checks show that this is at least a reasonable candidate solution:

• $P_{ab}$ is symmetric in $a$ and $b$
• $P_{ab}$ is directly related to $a$ and $b$ and inversely related to $c$
• $P_{ab} \to 1$ as $a,b\to\infty$ or $c\to 0$
• $P_{ab} \to 0$ as $a,b\to 0$ or $c\to\infty$
• For three equal sides,

$$P_{aa} = \frac{2}{\pi}\tan^{-1}\biggl(\frac{a}{\sqrt{3a^2}}\biggr) = \frac{2}{\pi}\tan^{-1}\frac{1}{\sqrt{3}} = \frac{1}{3}$$

• For two equal sides, $b = c$, as in your simplified example:

\begin{align}P_{ab} &= \frac{2}{\pi}\tan^{-1}\biggl(\frac{ab}{b\sqrt{a^2 + 2b^2}}\biggr) = \frac{2}{\pi}\tan^{-1}\biggl(\frac{\rho}{\sqrt{\rho^2 + 2}}\biggr) \\ P_{bb} &= \frac{2}{\pi}\tan^{-1}\biggl(\frac{b^2}{a\sqrt{a^2 + 2b^2}}\biggr) = \frac{2}{\pi}\tan^{-1}\biggl(\frac{1}{\rho\sqrt{\rho^2 + 2}}\biggr)\end{align}

1. $P_{ab}(0) = 0$ as expected
2. $P_{ab}(1) = \frac{2}{\pi}\tan^{-1}\frac{1}{\sqrt{3}} = \frac{2}{\pi}\frac{\pi}{6} = \frac{1}{3}$ for one particular pair of opposite sides, as expected
3. $P_{ab}(\infty) = \frac{2}{\pi}\tan^{-1}0 = \frac{2}{\pi}\frac{\pi}{2} = 1$, as expected
• $P_{ab} + P_{bc} + P_{ca} = 1$ can be shown using the identity

$$\tan^{-1} x + \tan^{-1} y = \tan^{-1}\frac{x + y}{1 - xy}$$

from which follows

\begin{align}\tan^{-1} x + \tan^{-1} y + \tan^{-1} z &= \tan^{-1}\frac{(x + y)/(1 - xy) + z}{1 - z(x + y)/(1 - xy)} \\ &= \tan^{-1}\frac{(x + y + z - xyz)/(1 - xy)}{(1 - xy - zx - zy)/(1 - xy)} \\ &= \tan^{-1}\frac{x + y + z - xyz}{1 - xy - zx - zy} \end{align}

The relevant products are

\begin{align} x + y + z &= \frac{ab}{c\sqrt{a^2 + b^2 + c^2}} + \frac{bc}{a\sqrt{a^2 + b^2 + c^2}} + \frac{ca}{b\sqrt{a^2 + b^2 + c^2}} \\ &= \frac{(ab)^2 + (bc)^2 + (ca)^2}{abc\sqrt{a^2 + b^2 + c^2}} \\ xy &= \frac{ab}{c\sqrt{a^2 + b^2 + c^2}}\frac{bc}{a\sqrt{a^2 + b^2 + c^2}} \\ &= \frac{b^2}{a^2 + b^2 + c^2} \text{and cyclic permutations, so}\\ xy + yz + zx &= 1 \\ xyz &= \frac{abc}{(a^2 + b^2 + c^2)^{3/2}} \\ x + y + z - xyz &= \frac{[(ab)^2 + (bc)^2 + (ca)^2](a^2 + b^2 + c^2) - (abc)^2}{abc(a^2 + b^2 + c^2)^{3/2}} \neq 0 \end{align}

so you wind up with $\tan^{-1}\frac{x + y + z - xyz}{0} = \frac{\pi}{2}$, giving $\sum P = 1$.

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As you stated, the assumption about the uniform distribution is questionable, but I think this should give a more realistic answer than the fraction-of-area approach. Thank you for the great and detailed answer! For convenience, I have provided a plot of both functions (simple area approach vs. solid angle approach). – shark.dp Oct 22 '12 at 14:48
I like the approach. But I just can't believe that with $a = b = 2$, $c = 1$, the result will be the unstable configuration a full 40 percent of the time. Now to find some rectangular prisms to toss around... – Chris White Oct 23 '12 at 4:24
Yeah, I share the same objection. Perhaps it could be improved by postulating some distribution over $x-v$ phase space and then accounting for the possibility that the die gets trapped in the potential well of one of its stable configurations, but that's more than I have time to do at the moment. – David Z Oct 23 '12 at 5:39