Answer: The Thick Coin

Key Idea / Intuition

The beautiful insight is a classical theorem from solid geometry: the surface area of a spherical zone depends only on its height, not on where it sits on the sphere. So the probability of landing on edge equals the fraction of the sphere's surface covered by the "equatorial band" — which is simply proportional to the height of that band. Setting this fraction to $1/3$ is then a clean geometric calculation using the Pythagorean theorem.

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Formal Proof / Solution

Setup. Inscribe the coin (a cylinder of radius $r$, thickness $t$) in a sphere of radius $R$. The center of the sphere coincides with the center of the coin. A random point on the sphere is chosen uniformly; if the radius to that point hits the curved side of the cylinder, the coin lands on edge.

Archimedes' Hat-Box Theorem (Zone Area). For a sphere of radius $R$, the area of a spherical zone of height $h$ (the region between two parallel planes) is: $A_{\text{zone}} = 2\pi R h.$ This is independent of where the zone sits — only the height matters.

Geometry. The edge band corresponds to the zone of height $h = t$ (the thickness). The total surface area of the sphere is $4\pi R^2$. So: $P(\text{edge}) = \frac{2\pi R t}{4\pi R^2} = \frac{t}{2R}.$

Finding $R$ in terms of $r$ and $t$. The corners of the cylinder touch the inscribed sphere. By the Pythagorean theorem: $R^2 = r^2 + \left(\frac{t}{2}\right)^2.$

Setting $P(\text{edge}) = \frac{1}{3}$. We need: $\frac{t}{2R} = \frac{1}{3} \implies t = \frac{2R}{3}.$

Substituting into $R^2 = r^2 + (t/2)^2$: $R^2 = r^2 + \frac{R^2}{9} \implies \frac{8R^2}{9} = r^2 \implies R = \frac{3r}{2\sqrt{2}}.$

Then: $t = \frac{2R}{3} = \frac{2}{3} \cdot \frac{3r}{2\sqrt{2}} = \frac{r}{\sqrt{2}}.$

The ratio of thickness to diameter: $\frac{t}{2r} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \approx 0.354.$

Conclusion. The coin should be about 35.4% as thick as its diameter to have a $\frac{1}{3}$ chance of landing on edge. This is the answer von Neumann reportedly computed in his head in 20 seconds.

Sanity check: If $t = 0$ (flat disk), $P(\text{edge}) = 0$. If $t \to \infty$ (long cylinder), $P(\text{edge}) \to 1$. The formula $t/2R = 1/3$ interpolates cleanly between these extremes. ✓