Suppose I have a die in the shape of a rectangular prism that has eight edges with a length of $1$ unit and four edges with a length of $2$ units so that there are two faces that are $1$ by $1$ and four faces that are $2$ by $1$. If I place this die on the table at a random angle, what is the probability that it will land on a small face?
This problem is very hard to start on; however, I have managed to formulate a strategy. I am going to need to use integrals to calculate this probability. I can represent all possible landing angles of the die by letting $\theta$ represent its "horizontal tilt" and $\phi$ represent its "vertical tilt":
I'm going to need to integrate over $\phi$ and $\theta$ to account for all possible angles for each from $0$ to $2\pi$, but I have no idea how to find a "probability function" that will do this. My best guess was to find a function $f(\theta,\phi)$ whose output is a value from $0$ to $1$ that expresses the probability of the event $S$ that the die lands on a small face given the vertical angle $\phi$, which is written $P(S|\phi)$. Of course, $P(S|\phi)$ will be in terms of $\theta$ and $\phi$. Then my answer will be $$P(S)=\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}P(S|\phi)d\theta d\phi$$
But I have no idea how to find $P(S|\phi)$. This problem is a little bit of physics and a little bit of probability, and I'm not quite sure how to go at it from here.
Any help or hints are appreciated!
Assuming the shape has uniform density (material composition), if the prism is rotated to a random orientation (uniform distributed) then lowered to a surface to rest on a face determined by the centre of gravity, the probability that it is so lowered onto a particular face will be proportional to the size of the solid angle for the apex of the rectangular pyramid whose base is the face and apex is the center of the prism.
When $a,b$ are lengths for the sides of the rectangular face, and $c$ the third dimention of the prism, then that solid angle is (in steradians) :
$$4\arcsin \left(\dfrac{ab}{\sqrt{(a^2+c^2)(b^2+c^2)}}\right)$$
And since the solid angle of a sphere is $4\pi$ steradians, you now have the probability.
$$\pi^{-1}\arcsin \left(\dfrac{ab}{\sqrt{(a^2+c^2)(b^2+c^2)}}\right)$$
As a reality check, for a cube ($a{=}b{=}c$) this would be
$$\pi^{-1}\arcsin \left(\dfrac{1}{2}\right) \qquad = \dfrac 16$$