After the first car is parked, the circular parking lot is no longer circular since it has a head and a tail.

The probability of parking in any spot of the parking lot is 1/(n-1) where n is the length of the parking lot segment e.g. after the first car parks you have a unbroken segment of 98 spots. Since the segment is symmetric, you only need to calculate half the positions and multiply by 2 at the end.

This challenge is based on a puzzle we heard from Yaniv Shmueli, who heard it himself a few years ago.

Let's assume that cars have a length of two units and that they are parked along the circumference of a circle whose length is 100 units, which is marked as 100 segments, each one exactly one unit long.

A car can park on any two adjacent free segments (i.e., it does not need any extra maneuvering space).