Darts

Props to Grimmet's "One thousand exercises in probability" for the approach.

Let the expected distance between darts on a disk of radius $r$ be $f(r)$ .

Consider a disk of radius $r+d$ , where $d$ is small. Note that $f(r+d)=\frac{r+d}{r}f(r)$ . We now calculate $f(r+d)$ in a different way.

The probability the two darts are within the old disk of radius $r$ is $\displaystyle(\frac{\pi\cdot r^2}{\pi\cdot(r+d)^2})^2=1-\frac{4d}{r}+O(d^2)$ .

The probability exactly one of them is within the old disk is $\displaystyle2\cdot\frac{\pi\cdot r^2}{\pi\cdot(r+d)^2}\cdot(1-\frac{\pi\cdot r^2}{\pi\cdot(r+d)^2})=\frac{4d}{r}+O(d^2)$ .

The probability both are on the new ring is therefore $O(d^2)$ .

Thus $f(r+d)=(1-\frac{4d}{r})\cdot f(r)+\frac{4d}{r}\cdot g(r)+O(d^2)$ , where $g(r)$ is the expected distance between a point on the boundary of a disk of radius $r$ and a uniformly picked inner point.

To find $g(r)$ , consider the above parametrization. The distance $s$ varies between $0$ and $2r\cdot cos{\theta}$ , the angle $\theta$ varies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ and the Jacobian determinant is $s$ , making $g(r)$ :

$\displaystyle g(r)=\frac{1}{\pi\cdot r^2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2r\cdot cos{\theta}}s\cdot s\cdot ds\cdot d\theta=\frac{32r}{9\pi}$

Therefore, $f(r+d)=\frac{r+d}{r}\cdot f(r)=(1-\frac{4d}{r})\cdot f(r)+\frac{128d}{9\pi}+O(d^2)\equiv f(r)=\frac{128r}{45\pi}+O(d)$ , making the answer $128$ .