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Fundamentally, the trouble with the Monty Hall problem isn't that analysis comes to the wrong answer, it's that people often come to the wrong model when reasoning about it informally.

It's not any harder to do the "correct" analysis than to write up a simulation. It's mostly just easier to convince yourself that the simulation matches the problem description when it reaches the unintuitive result.




Fundamentally, I think the real trouble with the Monty Hall problem is that the assumptions of the game are not clearly stated. Because of this, people come up with different models.


That's absolutely right; further, if you explicitly model the behavior of the game show host, you can exhibit models under which "it's better to switch" and models under which "it doesn't matter if you switch or not".


>models under which "it doesn't matter if you switch or not".

Could you provide an example? It seems obvious that a switcher wins exactly when a non switcher looses, which is 2 / 3 ?


Take a game show host who lets you choose a door, randomly reveals what is behind one other door, and then gives you an opportunity to change your choice. This game show host CAN (randomly) reveal the prize; he has equal probability of revealing ANY of the unchosen doors.

Say you are playing the Monty Hall game with this host. You choose your door, he opens another door, and it happens (purely by chance) that there is no prize there. Do you still believe that you have a 2/3 chance of winning if you switch to the other unopened door?


Isn't that a different problem entirely? The original is that the host reveals a door without the prize.

Aren't you modeling an entirely different problem as opposed to modeling the same problem with a different model, since the problem states the parameters and you are changing those?


> Aren't you modeling an entirely different problem...

Not really, but read on:

You correctly state that in the Monty Hall problem, the host reveals a door without the prize. That's the same situation which I described in my previous comment.

Try thinking about it this way: Say you are the contestant on that show. You have never played the game before, and you will never play it again. So you don't know how the host behaves. You pick your door, he reveals another door, there is no prize behind it. You would have to ask yourself: did he deliberately open that door because it had no prize? Or did he just happen to open a door that had no prize?

Your best estimation of your odds of winning changes completely depending on how you model the behavior of the host.

However, with any type of host, the situation whereby "contestant opens door with no prize, host reveals another door with no prize" can still occur, and regardless of whether you deem that the 'original' Monty Hall problem or not, it is the most interesting way to define the Monty Hall problem. Call it the extended Monty Hall problem if you want: the situation described above has occurred, and you have to both define a model for the behavior of the host (and game) and calculate your odds under that model.

Here's a challenge for you: Can you find a model under which the contestant has 100% chance of winning by not switching to the unopened door?


This modelling ambiguity is resolved by do calculus, which makes a clear distinction between intervention and observation: https://arxiv.org/pdf/1305.5506.pdf




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