The forest 2

ALL CREDIT FOR THIS SOLUTION GOES TO THE MAN WHO CALLS HIMSELF " @Brian Charlesworth " (I don't think it's his real name, it's just an alias. His real name is probably Einstein ;) )

This solution is copy pasted from The Forest

Without loss of generality, fix a starting point on the circumference. Draw a diameter through this point, and let the central angle $\theta$ measured relative to the starting point vary uniformly from $0$ to $\pi$ . (By symmetry we only have to consider "exit" points on one half of the circle). This will cover all the "exit" points on one half of the circle. Now, using Law of Cosine, the length $L(\theta)$ of a chord for a given value of $\theta$ is given by

$L^{2} = r^{2}+r^2- 2 r^2\cos(\theta)) = 2r^{2}(1 - (2\cos^{2}(\frac{\theta}{2}) - 1)) =$

$4r^{2}(1 - \cos^{2}(\frac{\theta}{2})) = 4r^{2}\sin^{2}(\frac{\theta}{2})$

$\Longrightarrow L = 2r\sin(\frac{\theta}{2})$ for $0 \le \theta \le \pi.$

Thus, using the MVT, the expected length of a path out of the forest will be

$\displaystyle\dfrac{1}{\pi}\int_{0}^{\pi} 2r\sin(\frac{\theta}{2}) d\theta = \dfrac{4r}{\pi}\int_{0}^{\frac{\pi}{2}} \sin(u) du = \dfrac{4r}{\pi}(-\cos(\frac{\pi}{2}) - (-\cos(0))) = \dfrac{4r}{\pi}.$

Note that the substitution $u = \frac{\theta}{2}$ was made, giving us $2 du = d\theta$ with $u$ going from $0$ to $\frac{\pi}{2}$ as $\theta$ went from $0$ to $\pi$ .

I know the ambiguity: The randomizing method is either choosing a point at random on the circle's circumference or choosing it as the intersection between the circle and a randomly chosen angle from A. See Martin Gardner for more info.