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)
Answer
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:
$$\begin{align}P_{ab} &= \frac{2}{\pi}\tan^{-1}\biggl(\frac{ab}{b\sqrt{a^2 + 2b^2}}\biggr) = \frac{2}{\pi}\tan^{-1}\biggl(\frac{\rho}{\sqrt{\rho^2 + 2}}\biggr) \\ P_{bb} &= \frac{2}{\pi}\tan^{-1}\biggl(\frac{b^2}{a\sqrt{a^2 + 2b^2}}\biggr) = \frac{2}{\pi}\tan^{-1}\biggl(\frac{1}{\rho\sqrt{\rho^2 + 2}}\biggr)\end{align}$$
- $P_{ab}(0) = 0$ as expected
- $P_{ab}(1) = \frac{2}{\pi}\tan^{-1}\frac{1}{\sqrt{3}} = \frac{2}{\pi}\frac{\pi}{6} = \frac{1}{3}$ for one particular pair of opposite sides, as expected
- $P_{ab}(\infty) = \frac{2}{\pi}\tan^{-1}0 = \frac{2}{\pi}\frac{\pi}{2} = 1$, 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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