Probability of a self-intersecting brokenline.

On the plane starting at some point `O`, pick a random angle uniformly and draw a unit segment `OA`. Again pick a random angle uniformly and draw a unit segment `AB`. Repeat two more times to get `BC` and `CD`. What is the probability of this broken line `OABC` self-intersects?

Answer.

The angle of the segment at `O` is doesn't matter. So without loss we can assume it is just horizontal. Now denote `alpha = angle OAB` and `beta = angle ABC`. By symmetry we only need to consider `alpha in (0^o, 180^o)` and `beta in (0^o, 360^o)`.

When `alpha in (90^o, 180^o)`, there is no chance of `OABC` to self-intersect.

When `alpha in (60^o, 90^o)`, self-intersection happens if and only if `beta in (0^o, 180^o - 2 alpha)`.

When `alpha in (0^o, 60^o)`, self-intersection happens if and only if `beta in (0^o, 90^o - alpha/2)`.

Calculating the probability of intersection by considering relative areas of the region described above, we get 

The probability of self-intersection is `1/12`.