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Monty Hall Problem

Michael C said:
Yes, you can consider it like that if you wish. It doesn't change the result: of the 200 times you actually play the game, in 100 of them the car is behind the door you chose and in 100 of them the car is behind the other closed door. The probability of winning is 1/2.
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Why? Each time you play, you pick the right door with 1/3 probability. Hence, if you play 200 times you'll probably hit the car around 66 times.
 
Herzblut said:
Why? Each time you play, you pick the right door with 1/3 probability. Hence, if you play 200 times you'll probably hit the car around 66 times.
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We're talking about the case where Monty doesn't know where the prize is, but always opens a door you didn't pick, and in this situation, you just know that he revealed a goat.

If you say you're playing 200 times, then that means that you started choosing 300 times, and in 100 of them, Monty revealed the prize by his random guess. In the remaining 200 trials, it's in your door 100 of them and in the other door 100 of them: therefore it's a 50/50 choice.

Remember, the problem statement says that you're already in the situation where Monty has revealed a goat, therefore those 100 where he revealed the prize don't count for this analysis.
 
GreyICE said:
I recanted because I was explicitly mathematically wrong, and can prove it.

It very much matters WHY Monty opened the door.
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OK. I tried quite hard to see why you said DUH, told me abruptly that it was me who wasn't getting it, and announced that I lacked understanding. I thought, silly me, that you wouldn't have been quite so obnoxious unless you were sure you were right.

So, you agree that I was right all along? Thanks for mentioning it.

I repeat, unless Monty is rigging the game by only offering the switch if he knows the correct door has already been picked, then one should always switch. In one scenario, the switch will not confer any benefit, but neither will it reduce the chance of winning. In the other scenario, the switch will increase the chance of winning.

Thus, if we can exclude the rigging scenario, switching makes sense whatever, because the worst it will do is make no difference, and it might improve your chances.

Rolfe.
 
Diogenes said:
You can't add that one, and have the game satisfy the original puzzle ..

The puzzle calls for the car not to be revealed when Monty opens the door ...
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One of the main problems here is that people don't carefully state the problem, and then go on to disagree because they have different versions in mind. For example, I think most people think the answer is 50/50 because they have something like my version in mind, plus the additional information that the door which was opened had a goat behind it (so CurtC's item B).
 
sol invictus said:
Agreed on all counts.

I might add one:

D) Monty doesn't know where the car is, but will always open one of the two doors you didn't pick and will always offer you the chance to switch. In that case you want to switch to the door he opens if it's the car (obviously!), and it doesn't matter whether you stay or switch (to the door he didn't open) if he opens a door to a goat.
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Do you really think they are going to have a game where you can switch to the door Monty opened if he reveals a car?
 
Rolfe said:
OK. I tried quite hard to see why you said DUH, told me abruptly that it was me who wasn't getting it, and announced that I lacked understanding. I thought, silly me, that you wouldn't have been quite so obnoxious unless you were sure you were right.

So, you agree that I was right all along? Thanks for mentioning it.

I repeat, unless Monty is rigging the game by only offering the switch if he knows the correct door has already been picked, then one should always switch. In one scenario, the switch will not confer any benefit, but neither will it reduce the chance of winning. In the other scenario, the switch will increase the chance of winning.

Thus, if we can exclude the rigging scenario, switching makes sense whatever, because the worst it will do is make no difference, and it might improve your chances.

Rolfe.
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Yes, I'm sorry. Have I eaten enough crow for you, oh impenetrable text wall?

I got it conflated with the baby problem, where in fact it doesn't matter, as long as you know whether the person has a son or not. That's a variant on the coin-box problem, rather than on the forced-choice problem like Monty Haul.
 
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CurtC said:
This is exactly wrong. There's all kinds of misinformation swirling around here, so
Everyone please pause and read the following.

Sometimes the problem is stated such that Monty always shows you a goat before giving you the choice to switch. If it's stated that way, there is no argument that you'll win 2/3 of the time by switching.

However, usually when I see it stated, it's not explicit in constraining Monty to always offer the switch, all you know is that in this case, your only data point, he did offer the switch.

IN THIS CASE THERE IS NOT ENOUGH INFORMATION TO ARRIVE AT AN ANSWER WITHOUT KNOWING MONTY'S REASONS FOR OFFERING THE SWITCH.

Let's take three assumptions that we can make:

A: He knows where the prize is, will always reveal a goat door, and offer you the choice to switch. We've already determined that you'll win 2/3 of the time by switching.

B. Monty doesn't know where the prize is, he just happened to pick one door which was a goat, and now he's giving you the choice. In this case, if you simulate it yourself with pen and paper, you'll quickly see that your chances are not improved by switching: it's a 50/50 proposition.

C. Monty knows where the prize is, but wants you to lose the game, therefore he offers the choice to switch only for contestants who picked right the first time. In this case, obviously, you need to stick with your original guess (you'll lose 100% of the time by switching).

I hope it's clear that the host's motivation DOES matter, and that simply finding yourself in the situation where he has offered a switch is not enough information without knowing more about how Monty plays the game.
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No, you are wrong. You don't even have to have Monty in the equation. All you need to know is that there are 3 doors, one of which has a prize. You pick one. Your odds of it being behind that door are 1 in 3. Assume none of the other doors are revealed and you are given the chance to pick both the other doors instead (which you are essentially doing even if an empty door is revealed on that side). Your odds are 2 in 3. Or simpy draw an imaginary line between your door and the other two door - the odds are better if you pick the side on which there are two doors. This become more than obvious if you have a million doors. The one you pick on one side and the 999,999 on the other side. The prize is almost certainly on the side with the 999,999 doors. Revealing 999,998 doors on the other side that are empty virtually guarantees that the prize is behind the remaining door. The odds of it be being the door you picked remain 1 in a million. Monty is completely irrelevant to the puzzle. All we know is that a second door is opened revealing no prize and that is all we need to know.

I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.
 
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billydkid said:
I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door
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Unfortunately, a lot of times this stipulation is not made. You are only told that 1) he does open a door, and 2) there is a goat behind it. You are not told that he always does this when this game is played, nor that when he does it, that he knows that he will reveal a goat.

and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.
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As noted, you are making an assertion that is not always part of the equation: that he always does this and that the door WILL NOT contain the prize. If that is part of the problem, fine. But it is not always the case. What is often (usually?) given is that a) he HAS opened the door (nothing about it being always) and that b) it DID contain a goat (not that it always contains a goat).

If he doesn't always open a door, then you do not know that he only opened it because you guessed right. Therefore, you need to know his motivation.

Second, whether he knows which is the right door or not also matters. If he does not, and is only guessing, then there is a 1/3 chance that he could reveal one with the car. The problem with that is that, as pointed out above, he has just ruined your chance of winning because it means that you did not select the car initially. This is a case where you would have been wise to switch, but because he revealed the car accidentally, you aren't able to make that switch. Thus, it hurts your chances of winning if you switch.

You can convince yourself of this by using cards (ok, I did it with excel, but it becomes pretty obvious pretty quick - if the second door is selected randomly, then the odds of winning in those events when the car is not revealed is 50/50)

So you need to know 2 things: 1) that he is not acting maliciously, and 2) that he knows what he is doing and only reveals a losing door. If you can establish those two things, than you can assuredly say the odds are 2/3.

ETA: Billy, below you will see the problem as presented in the first post of this thread
Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?
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You can see that the assumptions you have made (he always does this, and he always reveals a goat) are not explicitly noted. This is a common occurence.
 
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billydkid said:
No, you are wrong. You don't even have to have Monty in the equation. All you need to know is that there are 3 doors, one of which has a prize. You pick one. Your odds of it being behind that door are 1 in 3. Assume none of the other doors are revealed and you are given the chance to pick both the other doors instead (which you are essentially doing even if an empty door is revealed on that side). Your odds are 2 in 3. Or simpy draw an imaginary line between your door and the other two door - the odds are better if you pick the side on which there are two doors. This become more than obvious if you have a million doors. The one you pick on one side and the 999,999 on the other side. The prize is almost certainly on the side with the 999,999 doors. Revealing 999,998 doors on the other side that are empty virtually guarantees that the prize is behind the remaining door. The odds of it be being the door you picked remain 1 in a million. Monty is completely irrelevant to the puzzle. All we know is that a second door is opened revealing no prize and that is all we need to know.

I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.
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The fallacy of this is revealed if you imagine the following scenario.

You have three doors. You choose one. You have a 1/3 chance of choosing the right door.

You flip a coin, and draw a dot based on a random door, based on that coinflip. Your claim is that switching to the door without the dot improves your chances of winning to 2/3rds.

This is obviously fallacious, because we've gained no new information.

Yet it's exactly the scenario you're proposing that switching is always helpful, except that we don't know if we have a car behind that door or not.

Imagine your million doors. You eliminate 999,998 of them through a process of utterly random chance.

Your chance of having picked the right door in the first place is now 1/2.

What you missed is that by RANDOMLY eliminating 999,998 doors, you're already sitting in an immensely unlikely scenario, of which there are two possibilities remaining - you picked the door with the car, or the door with the car is the other door.

It's when Monty is forced to eliminate those 999,998 doors without revealing the car that the chances become 999,999 in a million that it's behind the other door.

As I said, the oddity only occurs when Monty is forced to change the random car choice into a goat choice, which gives you additional information. A piece of paper, a pen, and 10 minutes will satisfy you on this (it did me).
 
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Diogenes said:
The problem was stated in the OP ..

If you want to propose a different problem, then you should do so ..
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Here is the problem statement from the OP:

"Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?"

Note that this does not say that Monty knows where the car is, only that in this particular case, he revealed a goat. Maybe he doesn't know and sometimes reveals the car which you didn't choose, game over. You just got lucky and he revealed a goat so you get another chance. That is completely compatible with the problem statement in the OP.
 
GreyICE said:
Yes, I'm sorry. Have I eaten enough crow for you, oh impenetrable text wall?

I got it conflated with the baby problem, where in fact it doesn't matter, as long as you know whether the person has a son or not. That's a variant on the coin-box problem, rather than on the forced-choice problem like Monty Haul.
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OK, OK, don't mention it! I just thought since you were so sure, I ought to try to figure out whether you might be right.

Maybe this isn't the place. Tell me about the baby problem (or the coin-box problem). How does it differ from the present example? (Isn't it "Monty Hall"? - there was no name attached to it at all when I first heard of it though, in the mid-1990s.)

Rolfe.
 
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billydkid said:
I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.
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Just try it! It will take you less time than it took to write your last post and you'll quickly see why you're wrong. Take CurtC's case A to begin with since that is the problem as it's usually, if not always, intended to be. You can do it with three cards, or three pieces of paper, or whatever you like.

If you want an explanation there are probably hundreds in this thread. I've never understood why this is confusing, so I'm probably not the one to try to give yet another.
 
Diogenes said:
My reply was that the OP was plainly ( carefully ) stated ..
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If the OP was carefully stated, then the correct answer is "I'm sorry, there is not enough information provided."

If he wanted the "2/3 by switching" answer, he should have more carefully constrained Monty to always revealing a goat.
 
sol invictus said:
Just try it! It will take you less time than it took to write your last post and you'll quickly see why you're wrong. Take CurtC's case A to begin with since that is the problem as it's usually, if not always, intended to be. You can do it with three cards, or three pieces of paper, or whatever you like.

If you want an explanation there are probably hundreds in this thread. I've never understood why this is confusing, so I'm probably not the one to try to give yet another.
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CurtC's case A is not the problem presented in the OP ..

BillyKid's solution is valid for the OP .. Whether the problem is usually presented some other way, or regardless of what assumptions someone chooses to make.

With three closed doors, the chances of picking the car are 1/3 .. ( out of a sufficiently number of tries, the car will be picked 33% of the time )

Once a door is opened, without the car behind it, the chances go to 2/3 that it is behind the remaining door . ( out of a sufficiently number of tries, the car will be picked 66% of the time )

It is very easy to run a computer simulation to show this ..

Or you can do it by blindly pulling marbles out of boxes .. You just have to do it long enough to see the result converge toward 33 or 66% ..

The computer makes it a lot easier ..
 
Rolfe said:
OK, OK, don't mention it! I just thought since you were so sure, I ought to try to figure out whether you might be right.

Maybe this isn't the place. Tell be about the baby problem (or the coi-box problem). How does it differ from the present example? (Isn't it "Monty Hall"? - there was no name attache to it at all when I first heard of it though, in the mid-1990s.)

Rolfe.
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The coin box is weird, conceptually.

You have three boxes, one with two gold coins, one with two silver coins, one with a gold coin and a silver coin.

If you take a random coin out of a random box, and it's gold, what chance is there that the other coin in the box is silver?

Works similar to Monty Hall (yes, I can't spell) only with randomness. Conceptually similar, yet the con man in this case doesn't have to influence the outcome of the event to change the odds of the color of the other coin in the box.
 
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CurtC said:
Here is the problem statement from the OP:

"Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?"

Note that this does not say that Monty knows where the car is, only that in this particular case, he revealed a goat. Maybe he doesn't know and sometimes reveals the car which you didn't choose, game over. You just got lucky and he revealed a goat so you get another chance. That is completely compatible with the problem statement in the OP.
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No. The experiment is
-1- you pick a door
-2- Monty opens another door with goat behind
-3- you may switch or not.

Revealing a car in step -2- is not compliant to the rules of the experiment. It's totally irrelevant why Monty opens a goat door, he just does per definition.

If you want to simulate this experiment, say, 200 times by random choice, then you have to re-iterate experiments such that Monty opens a goat door 200 times in a row by pure chance. If he opens a car door, he invalidates the whole simulation and you have to re-start it from the beginning. I recommend to reveal to Monty the location of the car to shorten experimental effort.
 
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