Far away from home

There has been something bugging me about this question, is it right to assume a $2D$ circle for this problem?

Here is an illustration of the cases considered when assuming a 2D circle, as used in the other solutions:

As you can see, the cases are not evenly spaced. By right, there should be a higher possibility of teleporting a distance of $\pi$ as compared to a distance of $2\pi$ .

However, there is no doubt the answer is $\pi$ due to symmetry, I even verified it with a little calculus to make sure:

$\frac { \int _{ 0 }^{ 2\pi }{ 8\pi x\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } } }{ \int _{ 0 }^{ 2\pi }{ 8\pi \sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } } } =\pi$

EDIT: This solution is dumb and bad, please ignore it.

Let $d$ be the expected distance someone will have to walk home.

Since our creatures will be walking straight home, this can be reduced to a two-dimensional problem.

The shortest possible distance is if the creature teleports right back into its house ( $0\ m$ ). The longest possible distance is if the creature teleports to the other side of the planet ( $2\pi\ m$ ). Since the possible outcomes are equally spaced apart (by $\epsilon$ ) and the chances are uniform, we can find the answer with the average of the maximum distance and the minimum distance:

$d = \frac{max + min}{2} = \frac{2\pi+ 0}{2} = \boxed{\pi}$