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.