A coin of diameter d and thickness t is flipped so that it rotates uniformly around a horizontal axis parallel to a face. A coin of diameter d and thickness t is flipped so that it rotates uniformly around a horizontal axis parallel to a face. 
a) Find the ratio d/t so that the probability it lands on its edge is 1/3. Ignore aeordymics and bouncing. [Answer: 1.73]
b) Redo under the assumption that the coin is tossed so that any orientation is equally likely. [Answer: 2.83]
I must admit that I am having trouble understanding what is even being asked here.
Like for a) I assumed that since it is rotating around the horizontal axis that we are only interested in the length of thickness / diameter and not worried about areas of the faces on the coin since it is rotating around the axis. Is this correct?
So I thought the answer would just be:
$$P[\text{lands on edge}] = \frac{2t}{2d+2t}$$
since there are 2 diameters and 2 thicknesses on opposite sides of the coin. We could just work with 1 I guess since it's proportional.
$$\frac{1}{3} = \frac{t}{d+t}$$
$$\frac{d+t}{t} = 3$$
$$\frac{d}{t} + \frac{t}{t} = 3$$
$$\frac{d}{t} + 1 = 3$$
$$\frac{d}{t} = 2$$
which doesn't look right and, as I said I don't really understand this question but I tried to answer it?
 A: The line of C.G. should cut the two circular arcs containing the edge as their chords.
\begin{align*}
  P(\text{lands on its edge}) &= \frac{\alpha}{\pi} \\
  &= \frac{2}{\pi} \tan^{-1} \frac{t}{d}
\end{align*}
(a)$\quad \displaystyle \frac{1}{3} = \frac{2}{\pi} \tan^{-1} \frac{t}{d}$
(b)$\quad \displaystyle \frac{1}{2} = \frac{2}{\pi} \tan^{-1} \frac{t}{d}$

A: (a) Consider the way the coin lands.

As you rotate the coin in the diagram clockwise through angle $x$ until the other diagonal is vertical, any such position means the coin lands on its edge. The same applies, of course, for the opposite edge of the rectangle so the probability of landing on the edge is $\dfrac{x}{\pi}$. So we want $\dfrac{1}{3} = \dfrac{x}{\pi}$ and thus, $x=\pi/3$. From the geometry of the rectangle, we have $$\dfrac{d}{t} = \tan{(\pi/2-x)}=\tan{\pi/6} = \sqrt{3}.$$
$$\\$$
(b) Imagine that the coin is fixed inside a minimal sphere, as in the diagram, so that the coin touches the sphere on its two circular edges.

Having the coin land at any orientation with equal probability means having the sphere land so that all points of its surface are equally likely to touch the floor first. If the point is in the shaded area then the coin lands on its face. So the required probability is 
$$P(\text{Coin lands on face}) = \dfrac{\text{Area (shaded) of two spherical caps}}{\text{Surface area of sphere}}.$$
The formula for the area of a spherical cap is $2\pi r h$ where $r = \dfrac{1}{2}\sqrt{d^2+t^2}$ and $h = r-\dfrac{t}{2}$. Thus we need
$$\dfrac{2}{3} = \dfrac{2\pi r h}{2\pi r^2} = \dfrac{h}{r} = \dfrac{\frac{1}{2}\sqrt{d^2+t^2} - \frac{1}{2}t}{\frac{1}{2}\sqrt{d^2+t^2}}.$$
Simplifiying gives $d^2 = 8t^2$ hence $\dfrac{d}{t} = 2\sqrt{2}$.
