The Monty Hall Problem

[dietvanillapepsi]

"Like others have said" as in the posts where they say that the host already knows which door is the correct one. If you pick one of three doors then at least one of the two doors you did not pick has a goat. The host is selectively showing you a door that he knows for sure does not have a car. It is not new information. You're fooling yourself if you think so.

[Seyrr]

"Like others have said" as in the posts where they say that the host already knows which door is the correct one. If you pick one of three doors then at least one of the two doors you did not pick has a goat. The host is selectively showing you a door that he knows for sure does not have a car. It is not new information. You're fooling yourself if you think so.

Yes, that's what we're saying too. So it's still 1/3 that you made the right choice to start. How can you say something like that, and then turn right around and claim it's 50% now? There's no new information, so you're not doing a new probability calculation. You still use the old one, where it's 1/3 that you picked right, and 2/3 that you picked wrong. Staying is saying you picked right to begin with. Switching is saying you picked wrong.

[dietvanillapepsi]

Your math is faulty. If 98 doors are revealed then both doors have a 1/2 chance of being the right one. The "first choice" does not remain 1/100 chance while the "switch" has a dynamic chance. The odds for both change when new information is presented but the odds still remain equal.

You really think there is a 50% chance you picked the right door the first time out of 100 doors? Because that's what you're saying if you think that you still have a 50% chance after he reveals the 98 other doors as being incorrect.

I never said that there is a 50% chance that someone picked the right door the first time out of 100 doors. I said that AFTER 98 doors are revealed that your door has an EQUAL chance of being the correct door as the other door left.

[Epincus]

Once the host has opened a door, the car must be behind one of the two remaining doors. The player has no way to know which of these doors is the winning door, leading many people to assume that each door has an equal probability and to conclude that switching does not matter (Mueser and Granberg, 1999). This "equal probability" assumption, while being intuitively seductive, is incorrect. The player's chances of winning the car actually double by switching to the door the host offers.

The chance of initially choosing the car is one in three, which is the chance of winning the car by sticking with this choice. By contrast, the chance of initially choosing a door with a goat is two in three, and a player originally choosing a door with a goat wins by switching. In both cases the host must reveal a goat. In the 2/3 case where the player initially chooses a goat, the host must reveal the other goat making the only remaining door the one with the car.

More formally, when the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with probability 1/3:

* The player originally picked the door hiding goat number 1. The game host has shown the other goat.
* The player originally picked the door hiding goat number 2. The game host has shown the other goat.
* The player originally picked the door hiding the car. The game host has shown either of the two goats.

If the player chooses to switch, the player wins the car in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three.

The solution would be different if the host did not know what was behind each door, or if the host sometimes had the option of not offering the player the chance to switch. Some statements of the problem, notably the one in Parade Magazine, do not explicitly exclude these possibilities. For example, if the game host only offers the opportunity to switch if the contestant originally chooses the car, the probability of winning by switching is 0%. In the problem as stated by Mueser and Granberg, it is because the host must reveal a goat and must make the offer to switch that the player has a 2/3 chance of winning by switching.

I found this explanation to make far more sense than what was posted above (just me, maybe?) At any rate, I see their reasoning now, and voted accordingly.

Edit: Setting this to the extreme 100 door example works too. Consider them opening 98 doors and showing you a *lot* of goats. Is the chance that you picked the right door the first time now 50/50? NO! It's 1%, and therefore there's a 99% chance the other door is the right one.

Now had he opened 98 doors THEN asked you to pick one...

[evilbau]

The people defending 1/2 look quite silly, as it is provable mathematically (and has been linked and explained ad nauseum) and it is also verifiable anecdotally through cards (as mentioned) and running tests (statistics).

I don't know how you can deny its 2/3

[Byrd]

The host is selectively showing you a door that he knows for sure does not have a car. It is not new information. You're fooling yourself if you think so.

Exactly. Since you knew at least one of the two you didn't know had a goat to start with. Him showing you a goat doesn't provide you any new information at all. So your odds of picking the right door at the beginning were 1/3, you have no new information, so why are you saying that odds change?

[layoneil]

The people defending 1/2 look quite silly, as it is provable mathematically (and has been linked and explained ad nauseum) and it is also verifiable anecdotally through cards (as mentioned) and running tests (statistics).

I don't know how you can deny its 2/3

because the question on the poll itself really makes no sense.

[dietvanillapepsi]

"Like others have said" as in the posts where they say that the host already knows which door is the correct one. If you pick one of three doors then at least one of the two doors you did not pick has a goat. The host is selectively showing you a door that he knows for sure does not have a car. It is not new information. You're fooling yourself if you think so.

Yes, that's what we're saying too. So it's still 1/3 that you made the right choice to start. How can you say something like that, and then turn right around and claim it's 50% now? There's no new information, so you're not doing a new probability calculation. You still use the old one, where it's 1/3 that you picked right, and 2/3 that you picked wrong. Staying is saying you picked right to begin with. Switching is saying you picked wrong.

OK, fine. Let me break it down into math for you guys.

The reason why all 3 doors have an equal chance is because you (not the host) do not know what is in them. The odds change each time you open a door and discover more of the truth.

Odds for each door prior to choice:
A: 33% (1/3)
B: 33% (1/3)
C: 33% (1/3)

If you choose door A then your odds are 1/3 or 33%. The rest of the doors are a combined 2/3 or 66%. In math terms this would be:

B + C = 1/3 + 1/3 = 2/3

A + (B + C) = 100%

OK? Everyone agree with this right?


Odds after door B is opened:
A: 50% (1/2)
B: 0% (0/2)
C: 50% (1/2)

Note that the odds of door B is now zero percent because we now know the door is empty. We cannot ignore this fact. The door is empty so we know that door B has zero percent chance of holding the money.

So now there are only two doors whose contents are still unknown. As you can see, the range for the odds of each door changed from three to two. Now each remaining door has 1/2 chance of holding the money. In math terms this would be:

A = 1/2
C = 1/2

A + (C) = 100%



A and C have an EQUAL chance of being correct. The only reason why the odds change is because the range of possiblities changed when B was confirmed to be one of the two incorrect doors.

[Seyrr]

Well yes, the poll's question is poorly phrased. But still.

[Devek]

He couldn't phrase the question properly for the poll because he didn't understand the problem.

If he was able to phrase the question properly he would of understood that the answer was 2/3 and we wouldn't have this thread at all

[Millael]

If you choose door A then your odds are 1/3 or 33%. The rest of the doors are a combined 2/3 or 66%. In math terms this would be:

B + C = 1/3 + 1/3 = 2/3

A + (B + C) = 100%

OK? Everyone agree with this right?


Odds after door B is opened:
A: 50% (1/2)
B: 0% (0/2)
C: 50% (1/2)

Note that the odds of door B is now zero percent because we know know the door is empty. We cannot ignore this fact. The door is empty so we know that door B has zero percent chance of holding the money.

So now there are only two doors whose contents are still unknown. As you can see, the range for the odds of each door changed from three to two. Now each remaining door has 1/2 chance of holding the money. In math terms this would be:

C = 1/2

A + (C) = 100%



A and C have an EQUAL chance of being correct. The only reason why the odds change is because the range of possiblities changed when B was confirmed to be one of the two incorrect doors.

No no no no no no no. You can't just eliminate B from the equation. Like you said, no new information was given, so you're still dealing with an A + (B+C) situation. That's where you're getting confused.

[evilbau]

a lot of stuff saying 50% chance after one is taken away

that is true, but only in the case where you didn't choose a door to begin with and the door revealed by the host has nothing to do with what you picked.

With the conditions of the problem, you cannot say it is only a 50% chance after one is revealed.

[Seyrr]

"Like others have said" as in the posts where they say that the host already knows which door is the correct one. If you pick one of three doors then at least one of the two doors you did not pick has a goat. The host is selectively showing you a door that he knows for sure does not have a car. It is not new information. You're fooling yourself if you think so.

Yes, that's what we're saying too. So it's still 1/3 that you made the right choice to start. How can you say something like that, and then turn right around and claim it's 50% now? There's no new information, so you're not doing a new probability calculation. You still use the old one, where it's 1/3 that you picked right, and 2/3 that you picked wrong. Staying is saying you picked right to begin with. Switching is saying you picked wrong.

OK, fine. Let me break it down into math for you guys.

The reason why all 3 doors have an equal chance is because you (not the host) do not know what is in them. The odds change each time you open a door and discover more of the truth.

Odds for each door prior to choice:
A: 33% (1/3)
B: 33% (1/3)
C: 33% (1/3)

If you choose door A then your odds are 1/3 or 33%. The rest of the doors are a combined 2/3 or 66%. In math terms this would be:

B + C = 1/3 + 1/3 = 2/3

A + (B + C) = 100%

OK? Everyone agree with this right?


Odds after door B is opened:
A: 50% (1/2)
B: 0% (0/2)
C: 50% (1/2)

Note that the odds of door B is now zero percent because we know know the door is empty. We cannot ignore this fact. The door is empty so we know that door B has zero percent chance of holding the money.

So now there are only two doors whose contents are still unknown. As you can see, the range for the odds of each door changed from three to two. Now each remaining door has 1/2 chance of holding the money. In math terms this would be:

C = 1/2

A + (C) = 100%



A and C have an EQUAL chance of being correct. The only reason why the odds change is because the range of possiblities changed when B was confirmed to be one of the two incorrect doors.

So wrong, I don't know where to begin.

What justification do you have that B's probability was divided evenly between A and C?

We're saying B's probability went entirely into C. The "doors you didn't choose" probability is still 66%, and there's no reason for it to have changed.

Your reasoning is like people who say the likelihood of any event is 50% because "it either happens or it doesn't, and we have to divide the chances equally."

[Byrd]

A + (B + C) = 100%
OK? Everyone so far agree with this right?

So far, so good.

Odds after door B is opened:
A: 50% (1/2)
B: 0% (0/2)
C: 50% (1/2)

Wrong.

A and C have an EQUAL chance of being correct. The only reason why the odds change is because the range of possiblities changed when B was confirmed to be one of the two incorrect doors.

You already knew ahead of time that at least one of B and C was going to be incorrect. Knowing which one doesn't really matter, by sticking with your original choice (whether it be three doors or one hundred) you are effectively saying you picked the right door the first time. By switching, you are saying you picked the wrong door and are picking the best of the alternatives, because the host is eliminating all the bad alternatives. The only way you lose by switching is if you picked the correct one in the first place.

Don't worry, I've known plenty of very intelligent people who have gotten this wrong and simply don't grasp it until they try it for themselves. You won't be the first or the last.

[Jeryhn]

I kinda see both sides of the argument, so let me try to get this down to layman's terms:

You got 3 doors. The host knows everything.
You are first told to pick a door, let's say you pick Door 1.
Then the host reveals what's behind Door 2. It is a goat.
He offers you the chance to stick with Door 1, or to switch to Door 3.

In the grand scheme of the problem, there is a 1/3 chance you picked the right door, and a 2/3rd chance that you didn't.

If you work it in your mind that Door 2 is no longer a possible choice because you now know what's behind it, and its not a favorable outcome, there is a 50% chance that switching to Door 3 will end in an unfavorable outcome. However, that does not change your original odds because there's still a 2/3 chance you were wrong to begin with, and those 2/3 odds still apply regardless of knowing what's behind one of the doors that you did not originally pick.

Did I get the gist of it?

[quannum]

He couldn't phrase the question properly for the poll because he didn't understand the problem.

If he was able to phrase the question properly he would of understood that the answer was 2/3 and we wouldn't have this thread at all

we'd still have this thread. people don't understand math. i posted a link with the correct answer as the second reply and people are still picking the wrong choice. .. you don't have to watch all 20 minutes of it. this classic scenario is all done by 6 minutes in. the rest is alterations on it, monty hall is in collusion with the contestant for example.

edit: here's the link again for people who skipped it http://www.youtube.com/watch?v=o2L_2psS9uI

[Millael]

Here's an AIM chat to help out anyone else who still doesn't get it.

Crovax1234: Actually, here's the easiest way to do it.
Crovax1234: Pretend the doors aren't named or anything.
Crovax1234: It's just 2 subsets of doors.
Crovax1234: Doors you chose
Crovax1234: Doors you did not choose.
Crovax1234: So far so good?
DrkSphere: yep. one has one, one has two.
Crovax1234: Ok
Crovax1234: Now, since "Doors you did not choose" has 2 doors in it, one is guaranteed to be wrong, right?
DrkSphere: Right.
Crovax1234: Even so, there's still a 66% chance that this is the right subset, correct?
DrkSphere: you mean that one of those is the door with the car?
Crovax1234: Yes.
DrkSphere: OH.
DrkSphere: I see.
Crovax1234: So if you show one of these two doors, no new info is given.
DrkSphere: So then even though he removes one, that subset *still* has a 66% chance.
Crovax1234: There you go.
Crovax1234: Now you get it?
DrkSphere: Okay, thanks. Yeah.
Crovax1234: Thank god.

[BarthelloSylph]

Yup Jerry.

The people who go for the 50/50 answer make one crucial mistake. Once the choice comes between the two unopened doors, they ignore the initial round of "Pick door. Host opens door."

We aren't flipping a coin here. The past DOES matter and it does have a bearing on what's going to happen next.

They initially had a 1/3 chance of guessing right. By revealing a WRONG choice, those odds DON'T CHANGE. Each door still exists. Each outcome still exists. There is still two doors with goats, there is still one door with a car. You just know one of the doors with the goats. This is totally different from flipping a coin, where knowing what the previous outcome is DOES NOT help you guess the next one.

[Byrd]

Did I get the gist of it?

You nailed it.

It's an interesting problem in part because, despite the fact there is a right/wrong answer (i.e. not an opinion poll), perfectly intelligent people can pick the wrong answer, and can get really worked up over their opinion of the wrong answer. "Trust me, I know how numbers work," one math major told me during a conversation about this. (He was wrong).

It kind of depends on how you initially view the problem, and it's really hard to change one's opinion. I always find this question interesting because there IS a right answer, and there is still boundless discussion on what is right and wrong. Some people take hours of convincing, many having to resort to a deck of cards to convince them it's in your best interest to switch.

[dietvanillapepsi]

Once the host has opened a door, the car must be behind one of the two remaining doors. The player has no way to know which of these doors is the winning door, leading many people to assume that each door has an equal probability and to conclude that switching does not matter (Mueser and Granberg, 1999). This "equal probability" assumption, while being intuitively seductive, is incorrect. The player's chances of winning the car actually double by switching to the door the host offers.

The chance of initially choosing the car is one in three, which is the chance of winning the car by sticking with this choice. By contrast, the chance of initially choosing a door with a goat is two in three, and a player originally choosing a door with a goat wins by switching. In both cases the host must reveal a goat. In the 2/3 case where the player initially chooses a goat, the host must reveal the other goat making the only remaining door the one with the car.

More formally, when the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with probability 1/3:

* The player originally picked the door hiding goat number 1. The game host has shown the other goat.
* The player originally picked the door hiding goat number 2. The game host has shown the other goat.
* The player originally picked the door hiding the car. The game host has shown either of the two goats.

If the player chooses to switch, the player wins the car in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three.

The solution would be different if the host did not know what was behind each door, or if the host sometimes had the option of not offering the player the chance to switch. Some statements of the problem, notably the one in Parade Magazine, do not explicitly exclude these possibilities. For example, if the game host only offers the opportunity to switch if the contestant originally chooses the car, the probability of winning by switching is 0%. In the problem as stated by Mueser and Granberg, it is because the host must reveal a goat and must make the offer to switch that the player has a 2/3 chance of winning by switching.

I found this explanation to make far more sense than what was posted above (just me, maybe?) At any rate, I see their reasoning now, and voted accordingly.

Edit: Setting this to the extreme 100 door example works too. Consider them opening 98 doors and showing you a *lot* of goats. Is the chance that you picked the right door the first time now 50/50? NO! It's 1%, and therefore there's a 99% chance the other door is the right one.

Now had he opened 98 doors THEN asked you to pick one...

I just saw this post and I see now where our disconnect is. I think everyone (yes, even me) is agreeing that the odds when you first choose a door is 1/3 (33%).

The disconnect is after a door is revealed and you are deciding to switch or not. When you are left with two doors (regardless of how you got there) and you are deciding to switch or not then you are re-choosing. Therefore your odds at that point are recalculated. This is the point where we were disconnecting.

Yes, the odds at the start are 1/3 but those odds do not remain stagnant. Also, Door C does not suddenly gain both the odds of B and C after B is revealed.

The math is not:

A = 1/3
B = 0/3
C = 2/3

In this scenario, you argue that Door C has increasing odds but then contradict yourselves when you say A's odds remain the same. The odds for both A and C increase, not just one of them.