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I'm having trouble understanding something about the monty hall problem. If monty opened one door before you arrives, then you would have a 50/50 chance, whichever door you picked, because there are only 2 doors to pick from - right?

So, what about if you arrive, and monty said "I assumed you would pick door 1, so I opened door 3"? In that case, picking door 2 would be the same as if you had picked door 1, and then switched, so you should have 2/3 odds on it. But if monty hadn't said anything and you just picked one of the two doors, then your chances would be 50/50... or would they? Does monty assuming that you would want to pick door 1, change the probabilities even if you don't know about his decision?

What about if someone else picks a door and tells monty, who opens another one, and you then have the option to choose - without knowing which door the original person picked? What are the probabilities then? Does one of the doors secretly have a 2/3 probability? That doesn't make any sense to me.

Can someone explain this? Because I'm really not getting it.

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6 Answers 6

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I'm assuming that you understand the argument in the original problem (if not, read the other answers!).

This question is actually deeper than it may seem. The big issue here is that we commonly mean two different things by probability: the likelihood, in some abstract sense, of the world being in different ways, or our state of uncertainty about the world.

The changing probability in the Monty Hall problem reflects not changes in the way the world is, but changes in the state of our knowledge about the world. This is the source of most confusion about the problem.

You ask if, in the case that a separate person had played the game with Monty earlier (but left two doors closed) whether somehow one of the doors would secretly have a $\frac{2}{3}$ probability of having the prize, and express that this seems like it would be really weird. It is really weird, and the weirdness comes about by a conflation of these two ideas of probability.

Let's make this more extreme to make it clearer. At the end of the last game, Monty revealed which door had the prize, but for some reason, the other guy left without taking it. He then closes all the doors again, and now you (who haven't seen anything) come on stage. Shouldn't one of the doors "secretly" have a probability $1$ of having the prize? The door behind which the prize is located is determined, there is nothing random about the state of the world. But of course you can't find it. In fact this is true in the original problem as well! The probabilities reflect the state of your ignorance; you know no information to make you favor one door over another. That's why being told information "changes" the probabilities: because the probabilities were expressions of the state of your knowledge in the first place, not facts about the world.

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Thank you! An immediate +1 for having actually read my question. –  Benubird Dec 17 '13 at 17:24

The idea here is not to look at what happens at the moment of the desition. Since monty never reveals the car. If you had it wrong in the first time but later decide to switch then you win.

So if you allready know that you are going to switch before the game starts when do you lose? You only lose when you had it right the first time. However if you had it wrong the first time then when you switch you will have to change to the winning door.

Therefore the probability of winning with the switch strategy is the same as the probability of not getting it right the first time which is $\frac{1}{3}$

While the probability of losing is the same as the probability of getting it right the first time which is $\frac{1}{3}$

On the other hand the no switching strategy requires you to actually get it right the first time which has a probability of $\frac{1}{3}$

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One way I like to make the Monty Hall Problem clearer is imagine if there were $99$ goats and $1$ car. Pick a door. Now Monty reveals $98$ doors with goats in them, and gives you the option to switch. Do you think that the door you picked is one of the $99$ goats or the car? Do you still think that switching gives you a $50/50$ chance of getting the car? Of course not! $99$ times out of $100$ you picked a goat, and you'll want to switch.

What does that tell you about $2$ goats and $1$ car?

EDIT: Sorry, I suppose I didn't read the question very thoroughly. In the first scenario, you don't have enough information about Monty's method of choosing. If Monty chose as randomly as you would, the chances are unchanged. If Monty sees it as "Car or not car," then you have $50/50$ odds.

If you suggest a door, there's a $2/3$ chance it was a goat, in which case Monty has $50/50$ goat/car. In the other case, it's $1/3$ you chose a car, with $100\%$ Monty chooses a goat, so you still get $2/3$ a goat is chosen and $1/3$ a car is chosen.

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That does not help at all. What if the 98 doors were open when you arrived? Does one of the doors still have a higher probability? –  Benubird Dec 17 '13 at 16:55
    
If Monty chose one totally randomly as you would, then you still have favorable odds switching. If we don't know Monty's methods, in mathematical probability it is most likely he chose totally randomly, and you may as well treat it that way, so switching still gives you a $99\%$ chance of car (in my more extreme scenario) –  Tim Ratigan Dec 17 '13 at 17:03
    
And I apologize in advance... most people see Monty Hall, get really excited, and don't bother to directly address what you're asking (myself included). –  Tim Ratigan Dec 17 '13 at 17:04

The easiest way to explain the Monty Hall problem is this:

You start by picking a door. There are two possibilities here:

Case 1: You choose the right door. (Probability: $\frac{1}{3}$) In this case, if you switch, then you lose; if you stay, then you win.

Case 2: You choose the wrong door. (Probability: $\frac{2}{3}$) Monty then eliminates a bad door, leaving only two: your door (the wrong one), and the right one. So, you will win if you switch, and lose if you stay.

So, if you choose to switch doors after Monty eliminates one, then you win if you chose incorrectly the first time and lose if you chose correctly; hence there is a $\frac{2}{3}$ probability that you win. Does that make sense?

By the same logic, if you choose to keep your door then you will win if you chose right, and lose if you chose wrong; thus there is a $\frac{1}{3}$ probability that you will win if you choose to keep your door.

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The problem here is that the two doors do not have the same probability. Let me give you an analogy: suppose you want to pick the best student for a scholarship and want to select it out of two universities(with the probability the best student is in either university being equally likely). University A picks its representing student randomly while university B picks its representing student by doing a bunch of exams and a rigorous elimination period. Clearly you would be better of picking the student that has been through this elimination period since he is the best student of unversity B while the other must not necessarily be the best of college A.

The same happens with our problem. Suppose you chose door A and monty reveils door B. Then door C is not the same as door A. since we know that out of doors B and C, door B is a losing door. Therefore if the winner was in either B or C then C would be the winning door. So we have more information on door B. That is why you cant say both have odds $\frac{1}{2}$

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Re paragraph 2: I don't think it would be the same. The 2/3 probability of winning depends on the sequence of events: first you choose a door, THEN Monty reveals a goat, THEN you switch. If Monty reveals a goat first, without giving you a chance to switch, then of course your probability of winning is 1/2.

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