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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.

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  • $\begingroup$ So he is always between 2 doors, can only open one of those 2 doors, you have to choose which, but he never open the one you choose, so he will open the other one, but if they are both goats, he has 50% chance of opening the door I chose after all? I'm not sure I understand the whole process. $\endgroup$ – stack reader Nov 16 '16 at 9:37
  • $\begingroup$ Sorry, I guess it is a little unclear. I will update it, you choose a door like in the classical Monty Hall problem. $\endgroup$ – Bougret Nov 16 '16 at 9:39
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    $\begingroup$ If host's between doors 1 and 2 (of say 10), I pick 1, and the car's behind 2, how will the host act? $\endgroup$ – Miff Nov 16 '16 at 9:59
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    $\begingroup$ How does the host move between rounds? $\endgroup$ – Especially Lime Nov 16 '16 at 10:03
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    $\begingroup$ To be as clear as possible about what isn't clear: 1. Does the host "go between two random doors" each round, or only at the start? 2. If only at the start, are the two "adjacent doors" on rounds other than the first the two doors adjacent to the door the host opened last time? 3. What happens if the two doors available to the host are (1) the one with the car and (2) the one you just picked? $\endgroup$ – Gareth McCaughan Nov 16 '16 at 16:17
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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.

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  • $\begingroup$ Correct, I added a bonus condition as it feels the problem conditions was too complex but not challenging enough. $\endgroup$ – Bougret Nov 17 '16 at 1:34
  • $\begingroup$ @Bougret how about my bonus solution? $\endgroup$ – stack reader Nov 17 '16 at 1:55
  • $\begingroup$ You can chose doors more efficiently. Think about the case where the host is directly next to the car. $\endgroup$ – Bougret Nov 17 '16 at 2:01
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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.

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