The "Three Doors" mathematical problem

[Batman Jr.]

I'm glad I can help!

 

[T'ai Chi]

I've tried to make this clear as possible. Let me know if it works.

Assume, without loss of generality, that the prize is behind door 1.

An outcome from the game can be kept tract of by the 4-tuple (a,b,c,d). The letter a is the door you initially choose, b the door Monty opens, c the door you stay with, and d will be the outcome, that is, WIN or LOSE.

NOT SWITCHING DOORS

Our sample space, that is, the set of all possible outcomes, is:

S = {(1,2,1,WIN), (1,3,1,WIN), (2,3,2,LOSE), (3,2,3,LOSE)}

Therefore, P(LOSE) = P(2,3,2,LOSE) + P(3,2,3,LOSE).
(where P(grapefruit) means 'the probability of grapefruit')

The probability associated with initially choosing door 2 is the same as the probability of initially choosing door 3, and that is 1/3. However, when you do this, you lose, and you can lose in two distinct ways, so P(LOSE) = 2/3.

So P(WIN) = 1 - P(LOSE) = 1 - 2/3 = 1/3.

SWITCHING DOORS

S = {(2,3,1,WIN), (3,2,1,WIN), (1,2,3,LOSE), (1,3,2,LOSE)}

Because we only win when we switch to door 1, it means we only win if we initially choose door 2 or door 3. When we initially choose a door, we have a 1/3 probability of doing such. Therefore, our probability of winning is in part based on the probability of choosing door 2 or door 3.

So P(WIN) = P(2,3,1,WIN) + P(3,2,1,WIN).

P(WIN) = 2/3.

 

[Humphreys]

Right conclusion, but dead wrong analysis, Humphreys.

I don't think so.

First off is the evidence from the game show, which I have to guess you've never seen.

I'll admit I've never seen the show, but I've seen the puzzle described lots of times.

Monty always revealed a goat door. Why? Because there is no drama in doing otherwise. There's nothing left to choose. The contestant knows she's lost. By revealing a goat door, the drama is increased, and the contestant still has a chance and a seemingly difficult choice to make.

Yes. But it wouldn't make any difference once this stage was reached whether he originally knew or not.

Now for the analysis:
(Assume your explanation is right...)
o There is only one correct door.
o The contestant chooses one to start the game. 1/3 probability she is right.
o The chance one of the other doors is correct is 2/3.
o Monty chooses one of these at random.
o The chance he chose the prize is 1/3, but the game is now over, isn't it?
o The chance he didn't choose the prize is also 1/3, but, given that, the chance you should switch to the other door is 1/2.
o No advantage in switching!

Monty's choice is still restricted to the doors you didn't choose. So, how can it make any difference whether he knows what door contains the prize and chooses it deliberately, or he doesn't know what door contains the prize but picks it by chance anyway!

There is no difference. Remember, he doesn't have a free choice of doors - he can't pick the same door you picked.

When we are given the option to swich, we are already past the stage where Monty chooses a door. We presume he must have picked a goat door, otherwise the game would be over already.
That's why it makes no difference at this stage whether Monty has knowledge or not.

How can it make a difference whether he picks a goat door deliberately, or by chance? The result is the same. Still switch!

(Assume now that Monty knows...)
o There is only one correct door.
o The contestant chooses one to start the game. 1/3 probability she is right.
o The chance one of the other doors is correct is 2/3.
o Monty selects (with knowledge) a goat.
o That selection, being done with certainty, is not part of the probability picture. Effectively, the 2/.3 probability has now jumped to the door the contestant did not originally choose.
o Switch, d***it, switch!

But what if Monty selected a goat without knowledge? It makes no difference once this stage is reached! (if it is indeed reached).

But I could be wrong about all this (although this has never happened before)...

 

[Humphreys]

Actually, I'm starting to realize why I may be wrong about this now. I can't quite understand yet why it would make a difference whether Monty picked a goat door deliberately, or by chance.

But I can see that if we had 100 raffle tickets, and you picked one, then we removed 98 of the others by chance, and they were all losers - it would do us no good switching.

So you are right Bill, I think, but can you explain better why?

I'll think about it more, anyway.

 

[Humphreys]

Ha, don't worry, I understand it now. I was wrong - it does matter whether Monty has knowledge or not. The whole puzzle only works if we make that assumption.

Thanks for the correction, Bill.

 

[metacristi]

De Bunk

Im sure some of you have come across this before..but i was fascinated by the debate surrounding this.

Somehow, both sides of the discussion seem to make sense....

Heres the question.

You've got three doors..

Behind two of them is a goat, the other a car

You choose one door to open....but you dont open it yet.

Then, someone opens one of the other doors to reveal a goat.

You are left with two closed doors

But before you open your chosen door, you are given the choice to switch to the other unopened door and open that one instead.

Do you stay with your original choice of door or do you switch

Would switching from your original choice increase the odds of you finding the car...

( i know this is probably old maths problem...but it done my head in..)

If you want the link to the simulation...ask..

DB

My solution to the Monty Hall problem:

Let

A,B,C =doors
'X'=1=the car is behind door 'X'
'X'=0=there is nothing [or the goat] behind door 'X' ; where 'X'=A,B or C


There are possible 4 distinct events after choosing our door,let it be A,(where the underlined doors are those shown later as having the goat [or nothing in some variants of the problem] behind):

event1: A=1 AND B=0 AND C=0 ;P1=1/6

event2.: A=1 AND B=0 AND C=0 ;P2=1/6

event3: A=0 AND B=1 AND C=0 ;P3=1/3

event4: A=0 AND B=0 AND C=1 ;P4=1/3

We have:

P[event1]=P1=(1/3)*(1/2)*1=1/6

P[event2]=P2=(1/3)*(1/2)*1=1/6

P[event3]=P3=(1/3)*1*1=1/3

P[event4]=P4=(1/3)*1*1=1/3


Therefore the probabilities to have chosen a door having or not the car behind are respectively:

P[the car behind]=P1+P2=2*(1/6)=1/3

P[nothing behind]=P3+P4=2*(1/3)=2/3

From here it's clear why switching increase the chances to find something behind the door from 1/3 to 2/3.

 

[Paul C. Anagnostopoulos]

Man, all this math! By opening one of the other doors and showing you a goat, Monty has effectively given you both other doors. Easy pie.

~~ Paul

 

[CurtC]

BillHoyt, you're right that it comes down to whether Monty Hall always shows you a goat-door. But I watched the show lots of times, and the fact is, he didn't. Sometimes, he would just reveal the door that the contestant had picked, right or wrong. Offering the switch was *not* something he did every time.

So now, you have to figure in his motivation in giving you this choice. Is it done to keep you from winning? In that case, never switch. Is it done to make the show more exciting? Then it probably is 50-50. Is it done randomly? Switch. Is it done to give away more merchandise? Switch.

If anyone denies that the motivation of the host is what makes the difference, consider this scenario. You're walking down a city street and come upon a street hustler offering you a game of three-card monte. You put your money down, and the hustler moves the cards around thoroughly, at a speed you can't keep up with. You pick your card. The hustler, instead of revealing the one you picked, turns over a non-winning card, and offers you the choice to switch your guess to the remaining one. Should you switch? If you do, you're a fool.

 

[aerosolben]

I'm glad to see no one here is insisting on the p = 1/2 solution. Everytime this problem comes up, there are always people who adamantly insist the probability is 1/2. Trying to explain the idea is infuriating, as the straight dope link demonstrates.

I dislike the two child problem even more, because it depends so much on phrasing: "the first child is a girl" gives 1/2 chance of a male sibling, while "at least one child is a girl" gives 2/3 chance of a male sibling.

 

[Doc Dish]

Originally posted by De_Bunk Heres the link to one of the better simulators...

You can let it run, at hi-speed, a defined number of times to see the odds...

http://www.grand-illusions.com/simulator/montysim.htm

DB

I just let it run 20000 times! The results are (drumroll please):

Keep choice: 10000 times Wins: 3319 cars (33%) Losses: 6681 goats (67%)
Change choice: 10000 times Wins: 6714 cars (67%) Losses: 3286 goats (33%)

 

[athon]

What if I prefer the goat?

Athon

 

[Doc Dish]

Originally posted by athon: What if I prefer the goat?

Athon

You should seek professional assistance

 

[T'ai Chi]

I personally had a hard time thinking about it until I actually wrote out the sample space.

 

[Number Six]

The assumption is that the probability of him revealing a false door is independant of which door you chose. The way the puzzle is phrased, we don't know whether he does this everytime or if he intentionally tries to throw people off.

IIRC correctly, on the game show Monty always revealed one. So in that particular case, always switch choices.

Walt

That is it. It is best to switch _only if_ Monty's choice to reveal a door to you is independent of whether or not you picked the right door. If you know beforehand that Monty is going to ask you if you want to switch, then you switch. But if you don't know that he's going to ask and then he asks, you can't say whether it's best to switch or not because you don't have enough information to come to a conclusion.

 

[Badger]

What if I prefer the goat?

Athon

Then it's a win-win situation. Imagine how many goats you could get for a car!

 

[athon]

Then it's a win-win situation. Imagine how many goats you could get for a car!

Why would I want goats for my car? That's not where I want to keep them.



Athon

 

[Underemployed]

I am going to apply for government funding to let me sit and watch all the recordings of this monty hall show to provide a rigourous analysis of real-world outcomes. Anyone want to co-author who has a big-screen TV?

 

[Rolfe]

But if you don't know that he's going to ask and then he asks, you can't say whether it's best to switch or not because you don't have enough information to come to a conclusion.

If you don't have that information, then overall it's still best to switch. If he isn't deliberately avoiding revealing the car, then you still sit on the 50/50 odds, so it's not going to make your chances any worse.

Given that there is some chance that he is deliberately avoiding the car, and in that case your odds are 2/3 in favour of switching, then always switch. You won't decrease your chances, and depending on the game he's playing (which you don't know for sure), you may increase your chances.

Rolfe.

 

[Matabiri]

My brain came out of my ears on this one some years ago. In the process I tried the following "reductio ad absurdum".

I came up with another reductio ad absurdum a while ago which I think works better:

Imagine you're playing 100 simultaneous games. You pick 100 doors. If you opened those doors now, you'd win, on average, 33 cars, right.

The host opens 100 doors which you didn't pick, revealing 100 goats. How would the act of opening those doors raise the number of cars you win from 33 to 50? It doesn't.

 

[Tanja]

I came up with another reductio ad absurdum a while ago which I think works better:

Imagine you're playing 100 simultaneous games. You pick 100 doors. If you opened those doors now, you'd win, on average, 33 cars, right.

The host opens 100 doors which you didn't pick, revealing 100 goats. How would the act of opening those doors raise the number of cars you win from 33 to 50? It doesn't.

If the host reveals 100 doors with goats, then by changing your choice, instead of getting the 33 cars you chose right the first time, you get the 66 (67) cars you DID NOT choose right the first time.

Edited to add: I am not sure what you actually meant by your example, that the odds go up or that they don't?