Find the probability

[Guest]

A parking lot consists of 2020 spaces, numbered from 1 to 2020 and lined up in a single line. Сars arriving at the parking lot are parking in the following way: The first one chooses the place randomly. Every other car chooses the space from which the distance to the nearest car is the greatest (one of these spaces with equal probability).Find the probability that the last car(2020th) will take the space with number 1?

[HallsofIvy]

I would start by looking at simple cases.

If there were one parking space, the probability is obviously 1.0. The first car must
take that space and is the last car.

If there were two spaces, the probability the first car takes space 2 is 1/2 and then the second (last) car must take space 1 so the probability is 1/2.

It's when there are three spaces that it starts getting complicated! Obviously if the first car parks in space 1 then it impossible for the last car to park there. If the first car parks in space 3 the second car must park in space 1 so it is impossible for the last car to park there. So we only need to consider the case where the first car parks in space 2, which has probability 1/3. Then if the second car parks in the first space the last car can't park there. In order that the last car park in the first place the second car must park in space 3 which has probability 1/2. So the probability is (1/2)(1/3)= 1/6= 1/3!.

Hmmm! I think that, at this point I would conjecture that with n cars parking this way, the probability space 1 is open for the last car is 1/n!. I think I would try to prove that by induction on n.