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After seeing the popularity of the standard $3$ door problem, Monty thought to put a twist in the story.

There are $N$ doors, $1$ car, $N-1$ goats.

We need to choose any one of the doors. After we have chosen the door, Monty deliberately reveals one of the doors that has a goat and asks us if we wish to change our choice.

After we decide our choice, Monty then again reveals one more door that has a goat and asks us if we wish to change our choice (both 1st and 2nd).

This goes on. What strategy should we follow? Keep switching?

And if we keep switching, is it okay to switch to some of the previous choices (provided they are still not revealed!!)

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  • $\begingroup$ And Monty is a mercurial person!! He can decide to end the game at any moment. You have to be satisfied with whatever in hand at that moment. So please suggest me some trick so that at any moment I have the maximum possible chances of getting a car. $\endgroup$ – maulik May 13 '16 at 13:34
  • $\begingroup$ x @maulik: In that case we have to assume that Monty would end the game immediately if your first choice was a goat. So if he does not do that, stick to that door forever! $\endgroup$ – Henning Makholm Jun 8 '16 at 22:28
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You should stay with your initial choice until $N-2$ doors have been opened. Then switch to the single door you can switch to.

With this strategy you are sure to win except when your initial choice happened to be the prize door. In other words, your chance of winning is $\frac{N-1}{N} = 1-\frac1N$.

If you switch any earlier, your chance of having the winning door picked immediately before your last chance to switch will increase -- which will decrease the advantage of switching in the end, and will therefore reduce your overall winning chances.

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The rules of your extension are a little unclear: You say that Monty allows us to change all our previous choices? So if you've chosen a door once, he can never open it? When does the game end?

If the game continues until all doors but the one you have chosen and one more are closed, the best strategy is to stick with your choice until 2 doors are left and then change, then you win if you didn't choose the car in your first choice.

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  • $\begingroup$ The parenthetical in OP's last sentence makes it clear that Monty might open a door you had previously chosen. Nonetheless, your point stands. $\endgroup$ – Ceph May 12 '16 at 21:46
  • $\begingroup$ I don't agree that "it doesn't matter what you do until 2 doors are left". Suppose there are 4 doors in total. Then the probability of pointing at the prize is $1/4$ at first; after switching once it is $3/8$, and after switching for a second time it is only $5/8$ -- whereas if you stay once and then switch once the chance of winning is $3/4$. $\endgroup$ – Henning Makholm May 12 '16 at 21:49
  • $\begingroup$ I initially though so to, but then I started thinking about what it could mean that Monty allows you to change previous choices and then I became uncertain, and thought I would document that. $\endgroup$ – Henrik May 12 '16 at 21:51
  • $\begingroup$ @HenningMakholm Which is why I've aleady deleted that paragraph. $\endgroup$ – Henrik May 12 '16 at 21:51

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