# The Monty Hall Problem (Logic Puzzle)



## AngryReader (Jan 2, 2011)

There is a statistics problem that is throw around the internet, water coolers and dinner tables that asks people to determine the probability of some event occurring. I'll post the problem in a moment for anyone to look at who hasn't already see it.  First I want to establish why I am posting this.


  Almost everyone gets this problem wrong. I had an debate with some family members of Christmas, and it was my contention that one of the main reasons people don't properly identify all the elements of this problem is because it is almost always presented using loaded language. I'm going to rephrase the problem from the way people generally say it and see if anyone who hasn't seen this before can get it correct. This doesn't prove anything conclusively of course about my belief.


  The problem is named for a TV show host in the 1960's named Monty Hall. You are a contestant on his game show, and you are presented with three doors. Behind one of the doors is a new car (this must be a pretty small new car), behind the other two doors are goats.


  Monty Hall asks you pick a door. After you pick your door, he opens one of the other two doors. The door he opens has a goat behind it.


  What is the probability that the door you chose has the new car behind it?


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## Pyan (Jan 2, 2011)

Well, as the problem is stated above, it's a 50/50 chance. But that depends on you having to stick with your original door, or if you can switch to the other one...


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## AngryReader (Jan 2, 2011)

You can switch


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## littlemissattitude (Jan 3, 2011)

AngryReader said:


> The problem is named for a TV show host in the 1960's named Monty Hall. You are a contestant on his game show, and you are presented with three doors. Behind one of the doors is a new car (this must be a pretty small new car), behind the other two doors are goats.


 
I don't do logic problems.  Don't get along with them, something I discovered when I took a logic class in school.  My instructor and I came to an understanding about the issue: I wouldn't do the logic problems he handed out at the beginning of class as brain teasers, and he wouldn't give me a bad time about not doing them.  I still got a B in the class, even without doing them.

Anyway, I just had to comment, for those who never saw the show (which is in a new incarnation with a new host these days): the doors are pretty big, so there is no size issue for the cars.


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## chrispenycate (Jan 3, 2011)

It is dependent on the host knowing which door the car is behind. If he knows, you should switch, if he doesn't know, it is irrelevant. The extra information he holds changes the odds. 
(Sorry if that's not clear, but I didn't feel like writing the odds out for each case)


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## The Judge (Jan 3, 2011)

It surely isn't a logic problem, but a mathematical/statistical one.  I'm usually good at the former, but can never get my head around probabilities -- even when I've been told the answer to these things I still can't grasp them.


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## Ursa major (Jan 3, 2011)

The Wiki article, Monty Hall problem, explains what's going on in relatively simple terms. (There are nice diagrams, ones that inlcude pictures of doors, goats and cars, to help one's understanding.)


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## Parson (Jan 3, 2011)

Ursa, 

The Wiki article is indeed very enlightening. Count me in among those people who would have consistently assumed the wrong answer and had to be hit over the head with the logic in the article -- Thank you Marlyn -- to believe the right answer.


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## J Riff (Jan 4, 2011)

It's one in three, unless you discount the first door.


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## Dave (Jan 4, 2011)

I don't think it is the loaded language that makes any difference. It is because there are only two doors. If you have chosen your door already among the three, and then one of the doors you have not chosen is discounted, that tends to confirm your investment and belief in your chosen door. You were correct once, so why not again? The odds may be slightly in your favour now to switch, but your heart will always tell you to stick with your original choice.

It would be much more obvious with more doors that the odds were changing. Doesn't 'Deal or No Deal' work like that with boxes? I can't watch that show though, I just want them to pick the winning box and get it all over with. My daughter's old school used to play something similar at fund-raising nights called 'backwards bingo' picking out every number until only one number was left - (and everyone had gone to sleep!)


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## AngryReader (Jan 4, 2011)

pyan said:


> Well, as the problem is stated above, it's a 50/50  chance. But that depends on you having to stick with your original door,  or if you can switch to the other one...



After further consideration, the way the problem is stated 50/50 is still "wrong" (see qualification below), and switching is a red herring. I originally  felt that switching was a red herring in this question, and one of the major reasons why so many people guessed 50/50 rather than 66/33. This discussion so far has supported my hypothesis as an inordinately high number of people have gotten the answer correct. Normally more than 90% of people get it wrong. As I stated it, the probability the door you chose has a new car is  1/3, since there is another door now open, and only two doors remaining, the probability the other door has the new car is 2/3. Conversely the proability your door has a goat is 2/3. Once again, see qualification below.



The Judge said:


> It surely isn't a logic problem, but a mathematical/statistical one.


That differentiation is essentially meaningless, but I recognize that most people think that. There are two reasons I identified this as a logic problem as well as a statistical. Firstly, because you can deduce the probability changes even if you can't calculate it, as chrispenycate did. The second reason is below, and has to do with how people perceive the problem. 

Many of you already look at wikipedia, as I expected, to find the answer. I have a slightly different view of this problem than wikipedia however. I am not willing to say that those people who said the probability of 50/50 are wrong. I am saying that the probability that the new car is behind the door you choose is simultaneously 1/2 and 1/3. And you have to use reason to see that one model is more valid than the other. So in some respects, the people who get this "wrong", are still kind of right. It is that one answer is more right than the other.

I can explain more, but it gets a lot more esoteric, and I'm not sure anyone here cares.


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## Ursa major (Jan 4, 2011)

I think _I_ saw the explanation on _Numb3rs_ o), but I don't know how to post links to a TV programme, hence my use of Wiki.



EDIT: In trying to find some sort of link to the episode (which was number 13 of season 1), I came across this: http://education.ti.com/xchange/US/Math/Statistics/7715/Ep113_StickorSwitch_Manhunt.pdf.

It seems, according to the following quote from that paper, that the game hosted by Monty Hall was not played to demonstrate how maths and instinct don't always give the same result but to persuade the contestant to chose a door hiding a goat:


> The reason that this problem is a variation of the game show is that in reality, the host was not required to offer the choice. According to the New York Times (see a link to the article is in the "Extensions"), if the contestant picked a goat, the door was opened and the game was over. In other cases, once the contestant opted to stick or switch, the host would offer them money to change their mind, in order to build excitement and encourage the contestant to ultimately end up with a goat.


It seems to me that _under these conditions_ if a goat wasn't revealed to be behind the door, the car must be there. So in answer to the question, the probability is 100%.


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## Moonbat (Jan 4, 2011)

This problem was explained in the film *21* with Kevin Spacey, a film about some maths kids going to Vegas and counting cards to win. Anyway, I have always wondered about the problem.
Let's see if I get this right.
There are three doors and behind one of them is a car, you choose any door, the odds of getting the car are 1/3, then one of the other doors is opened to reveal not a car (a goat or anything non-car) then you are given the chance to change the door you chose. Normally people would say it doesn't matter, both doors now have odds of 1/2, but because you made the initial choice when the odds were 1/3 if you stick with your first choice the odds are still 1/3, whereas if you swap to the new door they are now 1/2 (or is that 2/3?)
Ok? I think I've explained that right, but what I want to know is, does the knowledge that one of the options (lets say door c) that is now open and revealed a non car affect the odds on your choice?
Even though, when you chose, the odds were 1/3 now that there are only 2 doors, surely your odds have changed to 1/2? But what I find more interesting is, what if you change doors, but then change back? Does that reset your odds to 1/2 for your original door (let's say door A)?
Also there is the part about the odds becoming 2/3, I assume because the open door can figure in the odds against it being non-car, but that's altering the facts before us.


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## AngryReader (Jan 4, 2011)

As chrispenycate pointed out, the operative factor is that the host has knowledge of the contents of the door. The door he selects is not random, and therefore provides you with additional information, as he can't reveal the car. 

When you pick a door, there is a 1/3 probability there is a car behind it. When the host opens another door, the probability that there is a car behind the door you chose remains 1/3. The proabaility that the door you did not pick, and the host did not open, has the car is now 2/3. Since the host will never open the door with the car behind it, you add the probability of the car being behind the door he opened to the unopened door. 

The probability is 50/50 if the host opens one of the two other doors at random, and it happens to have a goat behind it. In a similar situation, you pick a door, and the hosts opens a door he knows to have a goat, you leave, and someone else comes up who didn't see the host open the door and now this new person has to pick a door. What the is probability that he chooses the door with the new car behind it? 1/2.

The first scenario is a dependent trial, the second is an independent trial.


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## The Judge (Jan 4, 2011)

AngryReader said:


> When the host opens another door, the probability that there is a car behind the door you chose remains 1/3.


Why?


> The proabaility that the door you did not pick, and the host did not open, has the car is now 2/3.


Why?


> Since the host will never open the door with the car behind it, you add the probability of the car being behind the door he opened *to the unopened door*.[my bolding]


Both remaining doors are unopened, not simply the one I haven't chosen.  

Three doors: A, B, C.  I pick A, host opens C.  That proves nothing about A or B  except that one has a goat behind it and the other doesn't.  If I've  picked A and it had a car behind it, it still does.  If it had a goat  behind it, it still does.  

If A had a car behind it, then the presenter could have chosen either B or C doors.  If A had a goat behind it, the presenter could have only have chosen C.  That I can accept.  But how does that change the probablility so that it now becomes more likely that B has the car?  Why isn't it just as likely A has the car, and that the presenter chose C on a whim, rather than from necessity?  Esoteric calculations, undoubtedly.

This is why I say it isn't logic.  Logic I can follow.


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## Ursa major (Jan 4, 2011)

I thought it was rather simple:

the _probability_ that the _two_ doors you didn't choose have a car behind them is 66.666...%;
by opening a door on a goat, the game show host has transferred all of those two doors' probability to the door you didn't choose and he didn't open, making it twice as likely that it has a car behind it than the one you chose.


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## The Judge (Jan 4, 2011)

Ursa major said:


> I thought it was rather simple


Careful.  We'll be measuring you for a larger size in hats at this rate... 



> by opening a door on a goat, the game show host has transferred all of  those two doors' probability to the door you didn't choose


And I say again, why?  Why doesn't the opened door's probablility get distributed between the two remaining doors as it would if the presenter had picked a door at random?


I haven't looked at the wiki article and the other things, which may explain it in more detail.  The point I'm making is that it isn't self-evident, so, frankly, I don't think it's unreasonable for 90% of those who come across the puzzle to fail to understand it (ie me ).


As a matter of interest, has anyone ever carried out specific tests to see whether in fact door B ends up with twice as many cars as A in these circumstances?  (That presumably would be the end result if in fact the probability had changed.)


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## HoopyFrood (Jan 4, 2011)

I can remember someone trying to explain this to me (while I was...hem...a little tipsy). I think I kept asking "Why?" a lot, too. In the end I just got too angry to care!


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## Parson (Jan 4, 2011)

Judge,

You really should look at that Wiki article. It has an illustration of the truth that the door you didn't choose has a 66 2/3% likelihood of having the car, while your door remains with only a 33 1/3% likelihood of having the car. 

One of my favorite sayings is "Just because something is logical does not make it true. But it does make it the way to bet."


I believe that Monty Hall did on rare occasions open one of the other doors or boxes in the show to make the contestant agonize even more.


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## Ursa major (Jan 4, 2011)

The Judge said:


> Careful. We'll be measuring you for a larger size in hats at this rate...


I only found it simple because the Wiki article spelt it out. 



The Judge said:


> And I say again, why? Why doesn't the opened door's probablility get distributed between the two remaining doors as it would if the presenter had picked a door at random?


Because the host (in the hypothetical case) _always_ opens a door with a goat behind it; he does not choose the door at random unless _you_ have already selected the door hiding the car. His action has changed the odds (although, obviously, he hasn't changed the position of the car).

If he always selected the door at random, wherever the car was, then one third of the time, he'd reveal the car. He _never_ does, though.


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## PTeppic (Jan 5, 2011)

I always found the maths much simpler by considering it thus: having been told the rules and considering the statistics, the odds are enhanced by always switching. Every time. Which means in your first choice you WANT to pick a goat. The odds of the goat are 2/3, thus you chance of "winning", based on the assumption of a second action, are 2/3.


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## AngryReader (Jan 5, 2011)

The Judge,

I think they key takeaway point here is that if you're ever walking through a dark forest, and some demon offers you three doors, one that leads to your destination, and the other two to certain death, now you know to switch.


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## Moonbat (Jan 5, 2011)

The wiki article does make it quite obvious why the probability is 2/3 and not 1/3.
But it does depend on the 'host' knowing where the car is, and ensuring that they open a door that has a goat behind it. So, knowing that they would always do that, you could be quite certain that switching is in your best interests, but only in that the chance of you choosing the car straight away is 1/3 and the chance of you choosing a goat is 2/3. If the host then opens a door with a goat behind it, then (as there were 2 goats) you are still more likely to have chosen a goat first, so by switching you improve your chances.

I knew about this problem, and the answer prior to this thread, but it always intrigued me, and now (thanks to the helpful wiki article) it has become clear why it works the way it does.

I wonder if it works with political parties? If I give you the choice between 3 political parties, and then once you've placed your vote I show that one of the other parties is corrupt, woudl switching your vote give you a better chance of choosing an honest party?


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## Parson (Jan 5, 2011)

*Moonbat:

*That's the simplest logic question ever!! There is no chance of picking an honest party because there are none.


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## mosaix (Mar 1, 2011)

The Judge said:


> As a matter of interest, has anyone ever carried out specific tests to see whether in fact door B ends up with twice as many cars as A in these circumstances?  (That presumably would be the end result if in fact the probability had changed.)



Hi TJ, you can do it for yourself.

The Monty Hall page

Just look at the stats at the bottom of the screen.


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## HoopyFrood (Mar 1, 2011)

Switched the first time and won the car.

Didn't switch the second time and still won the car.



See. 50/50


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## Parson (Mar 1, 2011)

I am glad to see that the game after 700 plays is showing that switching results in winning 2/3's of the time, while not switching results in winning 1/3 of the time. Before I read through this I never would have guessed it, and the Judge's idea of testing it out made me wonder yet again if it would really work out. But it did!!


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## mosaix (Mar 1, 2011)

HoopyFrood said:


> Switched the first time and won the car.
> 
> Didn't switch the second time and still won the car.
> 
> ...



Don't know what you're complaining about Hoopy - you won two cars!


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## HoopyFrood (Mar 1, 2011)

Haha! Of course! Didn't even think about that. Woo hoo.


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## RJM Corbet (Apr 3, 2011)

Sorry, here's another one: There are two identical giants. One always tells the truth, the other always lies. There are two doors: one leads to freedom, the other to death. You are allowed to ask one giant one question, to which the answer must be either yes, or no. What's the question?


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## mosaix (Apr 11, 2011)

RJM Corbet said:


> Sorry, here's another one: There are two identical giants. One always tells the truth, the other always lies. There are two doors: one leads to freedom, the other to death. You are allowed to ask one giant one question, to which the answer must be either yes, or no. What's the question?



Assuming the doors are labeled A and B ask either giant "If I was to ask the other giant '_does door A lead to death?_', what would he say?". 

The answer always represents the false state of door 'A'.


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## RJM Corbet (Apr 11, 2011)

Sssh! Don't tell anybody. Let them suffer ...


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