The stupid 3-doors problem

[DWarrior]

Someone threw this at me a year ago, and I just rememberd it now again.

You're on a show and you have to pick one door out of three. Behind 1 of the prize, behind the other 2 is nothing. The agreement is that you pick your door, then the show host reveals one empty door and you have the option to switch to another.

Should you switch?

Obviously the odds of picking the first time are 1/3, but my friend and the CS professor claimed that the second pick is actually 2/3, because the remaining door somehow "combines" probability with the door that the host opened.

I argued that there is no way to know what's behind the second door, so as soon as the host opens the empty door, you have the option of choosing one door out of 2, so your odds are 1/2, which means that there is really no difference between switching and staying the on the same door.

So, 3 doors, A B and C, you pick A. Game host reveals door B as the empty one. Are the odds of door C having the prize 1/2 or 2/3 (B and C combined)?

Back then, I was awed by my CS professor, and considered him a wise man, so I did not argue (though I kept my theory), but a year later, I still think the odds are 1/2 on both doors.

[BruceZ]

I'm tempted to write a script to remove any posts with the words "3 doors".

Archives, search function, Monty Hall.

[GTSamIAm]

I think there is some confusion with the sequence of events. If you pick the right door the first time, if you automatically win, the chance before you pick the first time is 2/3. The 1/3 chance the first time plus the 1/2 chance the second time. If you don't automatically win, it looks like the chances are 1/2. You always have two doors to choose from. If he never opens the door you have picked when it is empty then obviously you have a 100% chance to win, as you can switch.

[BruceZ]

[ QUOTE ]
I think there is some confusion with the sequence of events. If you pick the right door the first time, if you automatically win, the chance before you pick the first time is 2/3. The 1/3 chance the first time plus the 1/2 chance the second time. If you don't automatically win, it looks like the chances are 1/2.

[/ QUOTE ]

That's wrong. You don't automatically win, and it is 1/3 if you don't switch, so it is 2/3 if you switch.

Here is another way to solve problems like this without having to invoke the equations of Bayes' theorem directly. Say you pick door A, and the host, who must reveal an empty door from one that you did not pick, reveals that door B is empty. There are 2 things that could have happened. Either the prize is behind door C, and the host was forced to open door B (since he can't open your door or door C), or the prize is behind door A which you selected, and the host made a random choice of door B from doors B and C, which are both empty. The first case is twice as likely as the second case, since in the second case the host will choose door B only 1/2 of the time, while in the first case he will choose door B all of the time. So it is twice as likely that the prize is in C than it is in A, so the probability is 2/3 for C and 1/3 for A. So we switch to C. Note that we also need to know that the host chooses randomly when he has a choice of 2 doors. This assumption is almost never stated in the problem, and the lack of this makes the problem ill-posed and subject to debate.

[GTSamIAm]

It never said the host opens a door that you don't pick.

[BruceZ]

[ QUOTE ]
It never said the host opens a door that you don't pick.

[/ QUOTE ]

Right, the problem was not worded correctly, which is the cause of most debate about these problems. You have to know how Monty Hall was normally played. If the host can open your door, then the answer indeed becomes 1/2.

[BruceZ]

BTW, the prisoner problem is worded correctly, but you got it wrong anyway.

[PowerHouse]

If you have a problem understanding why it is 2/3 probability then think if you had to choose between 100 doors you choose Door#1 and then the host opens all but your choice and Door#73. There would most likely be a reason he didn't open Door#73, so with switching you have 99% of being right because all the odds for Door#2-100 are 1% each they all combine into Door#73. Would you still now think that your odds are 50/50 between 1 and 73?

[GTSamIAm]

[ QUOTE ]
BTW, the prisoner problem is worded correctly, but you got it wrong anyway.

[/ QUOTE ]

Yeah, rub it in. But I don't understand exactly how Monty Hall applies to that problem.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
BTW, the prisoner problem is worded correctly, but you got it wrong anyway.

[/ QUOTE ]

Yeah, rub it in.

[/ QUOTE ]

OK. /images/graemlins/smile.gif

[ QUOTE ]
But I don't understand exactly how Monty Hall applies to that problem.

[/ QUOTE ]

Replace being being killed with winning a prize, and they are exactly the same. Sheesh.

[KJL]

I didn't read the other posts so I am most likly repeating another poster. When you pick your door you have a 1/3 chance of picking the right door. The host then shows you a door with a O% chance of being right. Basic math tells you the remaining door as a 2/3 chance of being correct.

Another way to look at it is logically. For these examples say that the prize is behind door 3.
If you pick Door 1 the host shows you Door 2, if you switch you will pick the correct door.
If you pick Door 2 the host shows you Door 1, if you switch you will pick the correct door.
If you pick Door 3 the host shows you either door, if you switch you will be wrong.
So if you switch you are correct 2/3 times.
If you do not switch you will only be correct when you pick the right door, which is 1/3 times.

[SheetWise]

I thought this question was put to rest 15 years ago by Marilyn Vos Savant -- google "Marilyn Vos Savant 3 door problem", or look at http://barryispuzzled.com/zmonty.htm

What WAS interesting is that several professors of mathematics at leading universities disagreed with her, and publicly stated so (after politely being advised not to), the pissing contest was raised to a level where Marilyn posted their letters WITH names and affiliation. MIT finally stepped in and confirmed Marilyns calculations.

SheetWise

[MickeyHoldem]

[ QUOTE ]
I'm tempted to write a script to remove any posts with the words "3 doors".

[/ QUOTE ]

I like this idea; but consider...

"What is the probability that said script will be overwhelmed, crashing forum servers?"

[Snoogins47]

[ QUOTE ]
I thought this question was put to rest 15 years ago by Marilyn Vos Savant -- google "Marilyn Vos Savant 3 door problem", or look at http://barryispuzzled.com/zmonty.htm

What WAS interesting is that several professors of mathematics at leading universities disagreed with her, and publicly stated so (after politely being advised not to), the pissing contest was raised to a level where Marilyn posted their letters WITH names and affiliation. MIT finally stepped in and confirmed Marilyns calculations.

SheetWise

[/ QUOTE ]

It stuns me that people doubted her. Years ago when I first ran into this problem I didn't want to believe that switching was optimal. Simulations weren't good enough, because I wanted to understand why it worked.

I made a little diagram of the possible outcomes, and it became obvious in minutes what the truth was.

This coming from a guy who almost failed his senior math class in high school. This statement is misleading, because the only reason I got poor marks was because I didn't turn in any homework, but it makes my point seem cooler so I'm leaving it in.

[Stephen H]

[ QUOTE ]
[ QUOTE ]
I thought this question was put to rest 15 years ago by Marilyn Vos Savant -- google "Marilyn Vos Savant 3 door problem", or look at http://barryispuzzled.com/zmonty.htm

What WAS interesting is that several professors of mathematics at leading universities disagreed with her, and publicly stated so (after politely being advised not to), the pissing contest was raised to a level where Marilyn posted their letters WITH names and affiliation. MIT finally stepped in and confirmed Marilyns calculations.

SheetWise

[/ QUOTE ]

It stuns me that people doubted her. Years ago when I first ran into this problem I didn't want to believe that switching was optimal. Simulations weren't good enough, because I wanted to understand why it worked.

I made a little diagram of the possible outcomes, and it became obvious in minutes what the truth was.

This coming from a guy who almost failed his senior math class in high school. This statement is misleading, because the only reason I got poor marks was because I didn't turn in any homework, but it makes my point seem cooler so I'm leaving it in.

[/ QUOTE ]

I remember when the Marilyn Vos Savant column came out - I used to read them back in school. I wasn't convinced of the answer at first, especially after seeing so many conflicting replies. So I wrote a simple computer program to simulate it - and I didn't even need to run it to understand. Just in writing the program, it became clear that there were 3 possible cases (prize behind door a,b, or c) and that in 2 of the cases, you need to switch to win the prize. I was then befuddled as to why math PhDs were writing in to argue the point.

[DWarrior]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I thought this question was put to rest 15 years ago by Marilyn Vos Savant -- google "Marilyn Vos Savant 3 door problem", or look at http://barryispuzzled.com/zmonty.htm

What WAS interesting is that several professors of mathematics at leading universities disagreed with her, and publicly stated so (after politely being advised not to), the pissing contest was raised to a level where Marilyn posted their letters WITH names and affiliation. MIT finally stepped in and confirmed Marilyns calculations.

SheetWise

[/ QUOTE ]

It stuns me that people doubted her. Years ago when I first ran into this problem I didn't want to believe that switching was optimal. Simulations weren't good enough, because I wanted to understand why it worked.

I made a little diagram of the possible outcomes, and it became obvious in minutes what the truth was.

This coming from a guy who almost failed his senior math class in high school. This statement is misleading, because the only reason I got poor marks was because I didn't turn in any homework, but it makes my point seem cooler so I'm leaving it in.

[/ QUOTE ]

I remember when the Marilyn Vos Savant column came out - I used to read them back in school. I wasn't convinced of the answer at first, especially after seeing so many conflicting replies. So I wrote a simple computer program to simulate it - and I didn't even need to run it to understand. Just in writing the program, it became clear that there were 3 possible cases (prize behind door a,b, or c) and that in 2 of the cases, you need to switch to win the prize. I was then befuddled as to why math PhDs were writing in to argue the point.

[/ QUOTE ]

Yeah, that's what happened to me. As I was writing the program, I realized that to check if the switch was correct, I only had to write pickedDoor != prizeDoor

[Pokerlogist]

Here's another way to look at this. I have never seen it explained this way:

By chance alone, you will initally choose an empty door 2 of every 3 times. At every one of those "empty door" times, Monty is being forced to choose the only remaining empty door. He must indicate the prize door by default with 100% certainity every one of those times. He has got to indicate the correct prize door 2/3 of the time. So by always switching you will have to be right 2/3 of the time.