The Monty Hall Arena

Question

Monty Hall had enough of all these pesky mathematicians winning cars.

Quite demotivated, he modified his show:

• There is a finite number $n$ of doors
• There is one car behind a door, otherwise there is a goat behind each
• You want the car
• All the doors are arranged in a circle
• Monty Hall doesn't want to move too much, so he will always go to an adjacent door
• Door opened are removed
• If possible he will open a door with a goat
• He will never open the door you picked
• If there are goat in both adjacent doors he will open one with a 50% probability

The show will proceed as follows:

• The host of the show goes between two random doors
• You pick a door among all the doors
• The host opens one of his adjacent door and asks you if you want to change your door, he then repeats the process until there is only two doors left

For instance there are 5 doors 1 2 3 4 5, 1 and 5 are adjacents:

• If the host is between the doors 2 and 3 and open 3, the adjacent doors become 2 and 4
• If the host is between the car and your door, he will open the door with the car and you will lose
• The host will always, for every round, open an adjacent door
• He will always, for every round, have a 50% probability to open an adjacent door if there is a goat in each and neither are your door
• If the host is between door 2 and 3 and there is a car in 2 and a goat in 3 he will always open 3 if it is not your door

What is the winning strategy?

Bonus challenge: you can only change three times.

Answer 1

Let me give it a try.

Every turn, you pick a door that is NOT adjacent to him, which ensures that that you will not lose the car until the end.
If I understand correctly, every time he opens a door, the door next to it will be come the new adjacent door, and the one he didn't open will remain the other adjacent door.
Once only 2 doors are left, choose the one that has been adjacent to him for the longest time without being opened.

Attempt at the bonus

First time, you choose the door in front of him, dividing the doors in half between each others(or 1 side with 1 more door if N is not even).
Let him work his way through the doors until he arrives next to you, then choose the door in front of him again and repeat the process again.
Worst case scenario : He opens n/2 doors, then n/4 doors before you have to make your final choice. But that would actually be the best case scenario since it would mean that he never opened the door on one side and its almost sure to be the car. Else, there will be less than 1/4 of the doors left and at that point you should choose the one that has been adjacent to him for the longest time without being opened.
This should provide a success rate higher than 75% of the original scenario.
In the case that being allowed to change 3 times actually meant we could pick 4 doors, then we could further remove 1/8 of the doors and get an even higher success rate.

Answer 2

Assuming this is supposed to work like the classical Monty Hall problem but with more than one step, we can get a higher chance to win the car than $\frac{2}{3}$.

Choose the first door randomly. The chance that it contains the car is $\frac{1}{n}$. The chance that any other door contains the car is $\frac{n-1}{n}$. Now stick to the door you have chosen in each step until the very last one, where only 2 doors are left. The chance for the car behind the door you have chosen at the beginning is still $\frac{1}{n}$. But after eliminating $n-2$ wrong doors the chance that the other door contains the car is $\frac{n-1}{n}$. Therefore you should now switch to that door.