Why is my solution to the 'thick coin' problem incorrect? I've been working through Fifty Challenging Problems in Probability by Frederick Mosteller.
For problem 38 my answer differs from the solution. Although I understand the solution, I don't know where I went wrong in my answer. So, where did I go wrong?
Problem

How thick should a coin be to have a $1/3$ chance of landing on [its] edge?

My solution
Consider a cylinder with radius $r$ and height $h$. Then the following 2 equations hold:


*

*$2\pi r^2 = 2/3$, i.e. the surface area of the heads and tails equals
the probability of not landing on the edge, $2/3$

*$2\pi rh = 1/3$, i.e. the surafce area of the edge equals the probability of landing on the edge, $1/3$


Solving these yields


*

*$r = 1/\sqrt{3\pi} = 0.326$

*$h = \sqrt{3\pi}/6\pi = 0.163$


Therefore, the coin with $1/3$ chance of landing on it's edge has a height $\approx 25\%$ of its diameter.
Given solution

The simplifying conditions that spring to mind are those that correspond to inscribing the coin in a sphere, where the centre of the coin is in the centre of the sphere. The coin itself is regarded as a right circular cylnder. Then a random point on the surface of the sphere is chosen. If the radius from that point to the centre strikes the edge, the coin is said to have fallen on edge.
To simulate this in reality, the coin might be tossed in such a way that it fell on a thick sticky substance that would grip the coin when it touched, and then the coin would slowly settle to its edge or its face.
A key theorem in solid geometry simplifies this problem. When parallel planes cut a sphere, the orange-peel-like band produced between them is called a zone. The surface area of a zone is proportional to the distance between the planes, and so our coin should be $1/3$ as thick as the sphere. How should the thickness compare with the diameter of the coin?
Let $R$ be the radius of the sphere and $r$ that of the coin.
The Pythagorean theorem gives
$R^2=r^2+R^2/9$
  ...
  [So] $R/3 = \sqrt2r/4 \approx 0.354r$
And so the coin should be about 35% as thick as the diameter of the coin.

 A: It seems the discrepancy here, as in some other geometric probability "paradoxes", comes from using implicitly different probability distributions.
For simplicity, let's work in two dimensions. Using the picture below, let the $h \times 2r$ rectangle be your coin's cross-section, $h$ - coin thickness, and the $R$-radius circle be the "shell" in the authors definition.

The event "coin lands on edge" is defined by the authors as:

A random point [uniformly distributed] on the surface of the circle is chosen. If the radius from that point to the centre strikes the edge, the coin is said to have fallen on edge.

Your definition is:

A random point [uniformly distributed] on the surface of the rectangle is chosen. If the point falls on the edge, the coin is said to have fallen on edge.

Since cord length and arc length are not proportional, the two definitions result in two different distributions. Now, don't trust my geometry, but for example with $h=3, R=3$ I am getting:
ratio of sides:bottoms of rectangle $= 3:5.2$
ratio of arcs corresponding to sides:arcs corresponding to bottoms $= \arcsin 1/2 : (\pi/2 - \arcsin1/2) = 1:2$
For $X \sim Unif$, $P(X \in A) \propto P(|A|)$, so the same ratios will apply to probabilities, leading to discrepant results.
