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So I got interested in the Monty Hall problem - I understand what it's about, but somehow I can't wrap my head around the idea of the final choice not being 50/50. More precisely: we all know (or maybe most of us do) the roulette progression systems which were popular a while ago - they told you that you should wait for a streak of some three-or-more consecutive reds or blacks and then bet on the contrary, always staying with this colour and eventually it has to be drawn as with every other spin, the chances become higher for the streak to be broken.

Of course the system is dumb and was shortly afterwards popularly neglected as a fallacy (though surprisingly many people still believe it works) as every spin is independent and even though you had a million reds in a row, in the next spin the odds for red, black and green are still the same as they had been these million spins ago.

And here I see an analogy to Monty Hall - sure, if we were proposed "Do you choose doors k and l (both at the same time!) or m?", it would be a straightforward 2/3 to 1/3. But why don't we cast aside the fact of some door having a goat and treat the two remaining doors as a simple 50/50 just like we do in roulette? Why aren't we interested in the past in roulette while it does matter in Monty Hall? Isn't the two remaining doors - one with a car - just "another spin"?

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Think of it like this.

Imagine the croupier has 2 roulette wheels; one with 12 red and 24 black slots and one with 24 red and 12 black (you can have 1 or 2 green if you like).

BEFORE you place your bet, you do not know which wheel the croupier will use (neither does he).

You place your bet and if you choose red, the croupier will tell you he is going to use wheel 1; vice-versa if you choose black.

THEN you are offered the opportunity to change your bet.

This is EXACTLY the same situation as the Monty Haul problem.

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The money hall problem is much more straightforward with 1000 doors.

Now, you choose 1, the presenter opens 998 doors which don't contain a prize, leaving 1 remaining unopened door which you didn't choose.

You can now choose to either open your door (which you picked at the start, so a 1/1000 chance its the correct one) or you can open the other door the presenter specially chose to remain un-open (a 999/1000 chance of being the prize).

If you imagine the 1000 doors stretching in front of you, and you pick the first one, the presenter then runs around opening doors, leaving say no.345 closed. It should be quite clear that is much more likely that this 345 door is the winning one rather than the no.1 you happened to choose first.

The same applies with 3 doors, but its less obvious. Lets say you can only choose door 1 to start with. Divide the doors into two groups: (1) and (2,3). If you stick you are betting door 1 has the prize, while if you switch you are betting that (2,3) has the prize, since the presenter will open one non-winning door from (2,3) leaving you with the winning one (unless 1 was the winning door).

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Because Monty Hall isn't a ball.

The roulette ball knows nothing. It rolls around the wheel without knowing which numbers came up before, or which numbers you would like to come up.

Monty knows where the prize is and where the goats are. He will always open a door showing a goat, because he always can.

I think we can all agree that in the standard version, you have a $\frac{1}{3}$ chance of picking the prize on your first guess.
Now in the $\frac{2}{3}$ chance where you didn't pick the prize, he tells you something. He tells you where the prize isn't and thus where it is. You forced his hand. There was one and only one door he could open. And there was a $\frac{2}{3}$ chance of you forcing his hand.
Only when you did pick the prize, he doesn't impart information on you. But there's only a $\frac{1}{3}$chance of that happening.

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The answer is simply that the events in this case aren't independent. We wouldn't be ignoring some independent iteration of the same experiment in the past, but an integral part of the present one.

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Thanks. But why should we even take the empty door into account? I mean - why does a door which is surely empty raise the chance of the other one being with a prize? It makes sense in the very beginning when all the doors are unknown to us but when we do know one of them is empty, why don't we cancel its odds from the odds of the not-chosen one just as well as we cancel the door itself from our choice spectrum? – Straightfw Dec 2 '12 at 17:10
@Straightfw: There are very many explanations of the Monty Hall problem, on this site, at Wikipedia and elsewhere. I'm not going to write yet another one here. You had asked specifically about the difference between this and roulette, and I answered that. To answer this new question you pose in the comment, I suggest that you read through some of the existing explanations, and if you still have a question then, formulate it more specifically, as you did the roulette question, and post it as a question of its own. – joriki Dec 2 '12 at 17:17
@Straightfw Because the empty door gives us information about the state of the current system, and that's what changes the probabilities. The events 'door 2 is empty' and 'door 3 is empty' aren't independent, so when we find out that door 2 is empty we've gained knowledge. On the other hand, the events 'flip 2 of this coin comes up heads' and 'flip 3 of this coin comes up heads' are independent, so knowing the first says nothing about the second. – Steven Stadnicki Dec 2 '12 at 17:18
@Straightfw: To explain what I mean by "more specifically": In the present question, you'd given a reason why one might give another answer, namely the analogy with roulette, and asked about the difference that invalidates that analogy. That's more specific than just saying "why don't we do $X$" without explaining why one might think that one should do $X$, as you did in your comment. – joriki Dec 2 '12 at 17:22

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