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Monty Hall Problem

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#1
if you are unfamiliar, just google it..
im just tired of people posting this on facebook, most of them not even undersatanding the crus of the issue instead babbling about 1/2 > 1/3 because math yaaayy.. :D

1. The fundamental issue i have with the problem is of memory,
suppose i initially chose a door, and then he shows me a goat; now i have to choose again..
now i FORGET what was my original choice; now how do i decide; there is no inherent reason to favour one door over another..
so a mathematical problem which varies depending on how much memory you have.. is just absurd..

2. Yes, when i chose door 1 initially it had a probability of 1/3, assuming all doors are equally likely..
Now, again i have been presented with a choice, which is entirely a new situation; which causes the probability of the first door to become 1/2 as well..

P(door 1 | no door information) = 1/3
P(door 1 | with information of 1 false door) = 1/2

the choice isnt about what you chose before, but whether you want door 1 or 2..
or are you thinking of it as a time series...??

3. Now i take it to 4 doors instead of 3...
applying the solution.. (car in door 1 or 2)
I choose door 1, goat revealed in door 4
I change, I choose door 2, goat revealed in door 3
I change, I choose door 1==correct answer

But if there were only 3 doors, the correct answer is door 2...

The absurdity is just beyond absurd.. So basically it all depends on HOW many doors were there??? rather than how many doors are there now>?

4.
Now if the situation is SLIGHTLY different, the host opens 2 DOORS at a time
I choose door 1, goat revealed in door 4 AND 3
I choose door 2==correct answer

So it just depends on number of decision making nodes... !!!

No wonder this stupid stuff has no mention in any real statistics books, or applicative use of statistics as well..
 

ProxyAmenRa

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#2
Each door has a 1/n chance of being the right one. Where n is the number of doors. When you choose a door the chance that it is the right one is 1/n. However, there is a (n-1)/n chance that the right door is in the other group.

When shown a wrong door from the other group, the chance that the other group has the right door is still the same but there are fewer doors. To figure out whether or not you should choose a door from the other group is simple as diving the chance that the other group has the right door by (n-2); [(n-1)/n]/(n-2). This is the chance per door of the other group. If this chance is greater than 1/(n), then you should choose a door from the other group.

Simple!
 

Cognisant

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#3
All doors until opened are equally likely.
If one door is opened the probability distribution is recalculated.

How does anyone fuck this up?
The choice is irrelevant.

Unless I've missed something, is the car being moved?
 
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#4
If this chance is greater than 1/(n), then you should choose a door from the other group.
Simple!
1. It does not really address the issues i raised

2. I have read this solution , i know about priori and prosteriori.
But i feel, you are making a mistake
Why should i take 1/(n).
Given that we take out the wrong door from the WHOLE experiment, rather than from just the other door, 1 should be reduced from n as well,

Hence, the condition is that the chance should be greater than 1/(n-1).
It just sounds absurd that you are willing to apply new information to only some part of the population and not to the whole thing.
 

Jennywocky

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#6
People get confused when the host opens one of the doors in the other group to reveal a goat, and they now think there is a 50/50 chance between either door.

In reality, you are being given an option between choosing ONE door (1/3 chance of winning) or choosing the other SET of doors (2/3 chance of winning). It's already a given that one of those doors in the other set has a goat behind it, whether he shows you the goat or not. When you pick the other door (the ambiguous one), you're still really choosing the other SET of doors -- it's just the 2/3 probability is being symbolized by one door versus both, when he reveals one of them as a dud.

So after you pick one door (1/3 chance), he says, "Do you want to keep your one door, or choose the other set of TWO doors (one of which already has a goat for sure)?" There's a 2/3 chance the car is in the other set of doors, regardless of whether he shows you the goat or not.

It doesn't really matter how many doors there are, except that there is more and more chance the car is behind the other doors (assuming the host opens all the other doors but one to reveal goats, then leaves you the option of trading your door for the other remaining door). In fact, as n -> infinity, there's an infinite chance the car is behind the other remaining door.



All doors until opened are equally likely.
If one door is opened the probability distribution is recalculated.

How does anyone fuck this up?
The choice is irrelevant.

Unless I've missed something, is the car being moved?
You are offered a choice to pick from one of three doors. One door has a car behind it; the other two have goats.

After you choose a door, the host opens one of the other two doors to reveal a goat. He then offers you the chance to switch to the remaining door.

There is no "recalculating odds," and you can see this if you actually create an outcome grid and tally up the outcomes where you win. The way the problem is set up, you're not choosing individual doors -- you are choosing SETS of doors, where one set is the door you originally choose and the other set encompasses all remaining doors. So do you want the set that has a 1/n chance of winning (because it only ever had one door in it), or do you want to choose the set that has all the other remaining doors (n-1 / n chance of winning) in it?
 
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#7
That video just boils my blood each time...
1. The first time, the experiment was done given a condition that there are 3 possible doors, hence the "1/3" only applies to that condition only..
When new information is given the "1/3" is no longer valid; past is irrelevant.

2. Wow it gets more intuitive when you see a 100 doors wooowww....
I just want to take a big hammer and smash that old conservative elementary math teacher's head.. thinking more boxes means woow
I can give a similar woooow application of the logic...
As per Monte Hall problem, you should not only switch doors, but also pay to switch the doors, so if there are 100 doors. Then as per Monte, after opening 98 doors, I should not only switch, but i should be READY TO PAY to switch..
AS per his logic,if the car is worth 10k;;
i should be ready to pay = 10*(1/100-1/2)= 49/100= 4.9k
So as per Monte Hall and the old bag in the video, If im offered to switch doors by paying 4K, I should switch it and pay 4K (cos 4<4.9)

As you can see, 100 doors is even more stupid, just because you show BIG numbers doesnt make it correct; now i showed MONEY and it proves that you are wrong...
Really the 100 box example getin on my nerves
 
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#8
People get confused

In reality, you are being given an option between choosing ONE door (1/3 chance of winning) or choosing the other SET of doors (2/3 chance of winning). It's already a given that one of those doors in the other set has a goat behind it, whether he shows you the goat or not. When you pick the other door (the ambiguous one), you're still really choosing the other SET of doors -- it's just the 2/3 probability is being symbolized by one door versus both, when he reveals one of them as a dud.

So after you pick one door (1/3 chance), he says, "Do you want to keep your one door, or choose the other set of TWO doors (one of which already has a goat for sure)?" There's a 2/3 chance the car is in the other set of doors, regardless of whether he shows you the goat or not.
That is what their theory says, i know what they say as well...

There's a 2/3 chance the car is in the other set of doors, regardless of whether he shows you the goat or not.
Applying similar logic,
I choose door 1.
He shows door 3.

I take the set as (Door 1 + Door 3), there is a 2/3 chance that the car is in set (door1, door 3). But there is a 1/3 chance for (door2).

Hence, i should take door 1 (ie, STAY), because I CHOOSE the set (door 1, door 3) which has a greater probability.....

SEE WHAT I DID THERE< I JUST CHANGED THE PERSPECTIVE of WHICH DOOR I CHOOSE TO PUT IN MY DAMN SET...

this is just... mathematical hypocrisy.. you are ready to choose one door for the set but not choose the other door for the set.

People arent confused, the human brain just skips these 2 steps and goes to the 3rd step because it recognises there are 2 ways of making these sets, and each equally likely, thus both switching and staying are equally preferable,
It is these "hi school math teachers" that are confused
 

Seteleechete

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#9
Say If I did this experiment 1000 times if I switched every time I would win more often than if I didn't thus switching is smarter every time. The point is that the 1/3 is still valid even after you gain the new information as the video demonstrates, it never becomes 1/2. The past is in no way irrelevant it is part of the problem and information you can and should use you shouldn't "forget" about it. If you "forget" or if they switch the car randomly again after opening a door it is no longer the same problem it is an entirely new one, as in it is no longer a monty hall problem since you have decided to ignore part of the given information.
 

Seteleechete

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#10
That is what their theory says, i know what they say as well...



Applying similar logic,
I choose door 1.
He shows door 3.

I take the set as (Door 1 + Door 3), there is a 2/3 chance that the car is in set (door1, door 3). But there is a 1/3 chance for (door2).

Hence, i should take door 1 (ie, STAY), because I CHOOSE the set (door 1, door 3) which has a greater probability.....

SEE WHAT I DID THERE< I JUST CHANGED THE PERSPECTIVE of WHICH DOOR I CHOOSE TO PUT IN MY DAMN SET...

this is just... mathematical hypocrisy.. you are ready to choose one door for the set but not choose the other door for the set.

People arent confused, the human brain just skips these 2 steps and goes to the 3rd step because it recognises there are 2 ways of making these sets, and each equally likely, thus both switching and staying are equally preferable, it is these "hi school math teachers" that are confused
You can't do that door 1 was never part of the set of doors that he could open. If you change to set door 1+ door 3 he could never open door 1 since you choose it thus he has to open door 3 if the car is in door 3 this becomes impossible to do, because this was not the set he was given.
 

Seteleechete

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#11
the sets are {UNCHOSEN DOOR, UNCHOSEN DOOR} and {UNCHOSEN DOOR}

you pick 2 or 1. 2>1
 
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#12
Say If I did this experiment 1000 times if I switched every time I would win more often than if I didn't thus switching is smarter every time. The point is that the 1/3 is still valid even after you gain the new information as the video demonstrates, it never becomes 1/2. The past is in no way irrelevant it is part of the problem and information you can and should use you shouldn't "forget" about it. If you "forget" or if they switch the car randomly again after opening a door it is no longer the same problem it is an entirely new one, as in it is no longer a monty hall problem since you have decided to ignore part of the given information.
1/3 HAPPENS only because you choose to make these 2 sets after host opens a door;
{UNCHOSEN DOOR, OPENED DOOR} and {CHOSEN DOOR}

Now if you make these 2 sets;
{CHOSEN DOOR, OPENED DOOR} and {UNCHOSEN DOOR}
It becomes so that staying IS MORE PREFERABLE than switching

---> As there is no reason to choose one mode of set making over the another; we can hand over equal probability to them

=(PREFER SWITCHING cos Set making choice 1)*0.5 + (PREFER STAYING cos Set making choice 2)*0.5
=EQUALLY PREFERRED BETWEEN SWITCHING OR STAYING
 
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#13
the sets are {UNCHOSEN DOOR, UNCHOSEN DOOR} and {UNCHOSEN DOOR}

you pick 2 or 1. 2>1
What?
I believe you made a typo.
Ofc i chose a door, how could there be 3 unchosen doors...
 

Seteleechete

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#14
There is 1 3rd chance that a door is correct

(correct door, Opened door) (wrong door) 1 3rd chance

(opened door, Correct door) (wrong door) 1 3rd chance

(opened door, wrong door) (correct door) 1 6th chance

(wrong door, opened door) (correct door) 1 6th chance
 
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#15
There is 1 3rd chance every door is correct

(correct door, Opened door) (wrong door) 1 3rd chance

(opened door, Correct door) (wrong door) 1 3rd chance

(opened door, wrong door) (correct door) 1 6th chance

(wrong door, opened door) (correct door) 1 6th chance
Why would the second 2 have lower probability of making sets than the first two??

As you have not included "chosen door" in your sets, do you mean to say that the "chosen door" is irrelevant??

The question just devolves into; whether the correct door is inserted within the set or the wrong one..
The sets devolve into

(correct door, Opened door) (wrong door) 1 4th

(opened door, Correct door) (wrong door) 1 4th

(opened door, wrong door) (correct door) 1 4th

(wrong door, opened door) (correct door) 1 4th
 

ProxyAmenRa

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#16
1. It does not really address the issues i raised

2. I have read this solution , i know about priori and prosteriori.
But i feel, you are making a mistake
Why should i take 1/(n).
Given that we take out the wrong door from the WHOLE experiment, rather than from just the other door, 1 should be reduced from n as well,

Hence, the condition is that the chance should be greater than 1/(n-1).
It just sounds absurd that you are willing to apply new information to only some part of the population and not to the whole thing.
You are incorrect because you focus on single door probability and not group of doors probabilities.
 

Seteleechete

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#17
the set with 1 door is the originally chosen door.

Every door has 1 3rd of chance of being correct

In yours door number 3 has half a chance of being correct so it is impossible

the chosen door is irrelevant(beyond making the groupings) since you don't know what is behind it and you can switch

He will never open the door in the set with 1 door

He will never open a correct door

There are no other possible combinations

He has to open a door

By your set at the start: door 1 has 1 4th chance of being correct door 2 1 4th chance and door 3 1 half in reality they are all 1 3rd of a chance

(possible option 3 and 4 share a correct door)
 
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#18
the chosen door is irrelevant(beyond making the groupings) since you don't know what is behind it and you can switch
But in your set combination, you have never included chosen door in your set grouping, then how can you say it was used for making the groupings.
Thats the flaw in your sets.

Your statements are unclear, especially with your usage of the words like "your"
Maybe could you be a little more clear so i can counter your point you want to make.
 

Seteleechete

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#19
There is 1 3rd chance that a door is correct

(correct door, Opened door) (wrong door(orignally chosen door)) 1 3rd chance - assuming door 1 holds the car

(opened door, Correct door) (wrong door(originally chosen door)) 1 3rd chance -assuming door 2 holds the car

(opened door, wrong door) (correct door(originally chosen door)) 1 6th chance assuming door 3 holds the car and door 1 is opened

(wrong door, opened door) (correct door(originally chosen door)) 1 6th chance assuming door 3 holds the car and door 2 is opened.

Every door has 1 3rd of a chance of being correct. Since the last 2 options both assume that door 3 is correct the chance is lowered to 1 6th for each option. In this set all 3 doors have 1 3rd of a chance of being correct

After the door is opened you get to pick the 2 doors or 1 door. the culimative chance of the 1 door is 1/3 the 2 door is 2/3


(correct door, Opened door) (wrong door( originally chosen)) 1 4th -door 1 correct

(opened door, Correct door) (wrong door(originally chosen)) 1 4th -door 2 correct

(opened door, wrong door) (correct door(originally chosen)) 1 4th -door 3 correct

(wrong door, opened door) (correct door(originally chosen)) 1 4th -door 3 correct

In this set door 1 and 2 has only 1 4th of a chance of being correct while door 3 has 1 half of a chance thus this set is wrong since every door is supposed to have 1 3rd of a chance at the start.
 

Hadoblado

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#20
The dress is blue!

1) You've got a 2/3 chance of selecting a goat at the start, and a 1/3 chance of selecting the car.
2) If you select a goat and switch, you will always end up with the car (do the probability tree).
C) For a 2/3 chance of ultimately getting a car, you should hope to pick a goat and then switch.
 

Jennywocky

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#21
That is what their theory says, i know what they say as well...



Applying similar logic,
I choose door 1.
He shows door 3.

I take the set as (Door 1 + Door 3), there is a 2/3 chance that the car is in set (door1, door 3). But there is a 1/3 chance for (door2).

Hence, i should take door 1 (ie, STAY), because I CHOOSE the set (door 1, door 3) which has a greater probability.....
You don't have that flexibility, and you're becoming confused.

Your initial chance of success is 1/3.
The chance the car is in the other set of doors is 2/3 regardless of whether you are shown what's behind any of the doors (because you know there is ALWAYS at least one goat in that other set, and all the host is doing is showing it to you).

We don't even need to be theoretical about it.
It's been computer simulated, and when they run the test many many times, the odds of winning if you switch approaches 2/3.
 

del

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#22
The crux of the "paradox" is that the host always shows you a goat, so you can infer information based on his behavior.

For 3 doors, the probability your initial guess was incorrect is 2/3. The host shows you the other wrong door. You switch, you win. OTOH, the probability your initial guess was correct is 1/3. He shows you a goat, you switch, you lose.
 

Cognisant

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#23
There's no sets of doors, the car dosen't move, the fact that the host shows you where the car isn't is completely irrelevant because nothing has changed form the initial conditions, the car hasn't move.

It's an illusionary choice.
 

Cognisant

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#24
Assume you're presented three doors but you don't pick any, monty removes one door and now you have two doors to choose, obviously the remaining two have a 1/2 chance of being the correct door but why would choosing one of the doors in the round before change this?

If you haven't opened a door yourself you choice does nothing to alter the probability.
If the probability dosen't change being given the same choice again is meaningless, the probability is equal among all remaining does and MUST remain equal so your choice dosen't change anything and thus is just the illusion of choice.
 

del

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#25
Assume you're presented three doors but you don't pick any, monty removes one door and now you have two doors to choose, obviously the remaining two have a 1/2 chance of being the correct door but why would choosing one of the doors in the round before change this?
Because Monty won't "remove" the door you chose initially.

When you choose a door, it limits what Monty can show you in the next round, even though you did not open it.
 

Hadoblado

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#26
Precisely del.

Cog you have missed something. Nobody would bother with this problem if it was as obvious as it looked. It would be a complete non-event.

When I received this one as homework it took me several hours to solve it. I thought I'd done poorly until the next session where I realised nobody else had gotten close. I then had a panic attack trying to explain it in front of everyone. Good times.
 

Cognisant

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#27
Because Monty won't "remove" the door you chose initially.
But regardless whether you pick the right or wrong door he still has a wrong door to pick, if you pick the right door he has two doors to choose from, even if there's 100 doors and he picks 98 that last door might seem somehow special but it's not, every door has 1/n chance of being the correct door and revealing one merely changes the value of "n", you can't infer anything further from it.

Probability doesn't have memory.
Getting heads three times in a row with a coin toss doesn't mean there's a 1/16 chance of getting heads again, each coin toss is its own separate event so regardless of what nonsense you spout after Monty removes a door there's only two doors so your chance of being right with either one is and can only be 1/2.

If probability had memory the implications would be staggering, it would overturn the world as we know it.
 

Blarraun

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#28
[bimgx=500]http://i.imgur.com/gAyseQ1.png[/bimgx]
Simply:
The door you choose is 1/3 likely.
The remaining doors are 2/3 likely.

He then removes one door and you are left with:
Your door that is 1/3 likely.
The other door that is now 1/3+1/3=2/3 likely to be winning.
Therefore switching wins 2/3 of the time.
 

Seteleechete

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#29
You are forgetting this is not a normal probability case, special rules apply here that normally don't. Instead of doors lets do balls in a bowl. You have 3 balls 1 blue 2 red. You randomly place 1 of these balls in one bowl(Y) and 2 of them in another(X). bowl (X) either has 2 red balls (1 3rd of a chance) or 1 blue ball and 1 red ball (2 3rds of a chance). You now pick up a red ball from bowl (X). The probability of bowl (X) having a blue ball after picking up a red ball remains 2/3.
 

ProxyAmenRa

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#30
If probability had memory the implications would be staggering, it would overturn the world as we know it.
The a priori solution to the Monty Hall Problem is also empirically true. Ergo, you are viewing the problem incorrectly.
 

Pyropyro

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#31
Doesn't matter. had goat.
 

Cognisant

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#32
Alright I get it now.

Selecting one door out of 100 doors has a 1/100 chance but after the 98 are removed the only way the other choice could not be the car would be if you selected it initially, so there's an overwhelming probability that the car is behind that other door.
 
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#33
What if 2 doors were empty and there was a killer goat behind one door. Sticking with my first choice is actually good? To avoid the goat?
 

dark+matters

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#34
Each door has a 1/n chance of being the right one. Where n is the number of doors. When you choose a door the chance that it is the right one is 1/n. However, there is a (n-1)/n chance that the right door is in the other group.

When shown a wrong door from the other group, the chance that the other group has the right door is still the same but there are fewer doors. To figure out whether or not you should choose a door from the other group is simple as diving the chance that the other group has the right door by (n-2); [(n-1)/n]/(n-2). This is the chance per door of the other group. If this chance is greater than 1/(n), then you should choose a door from the other group.

Simple!
Perfect! *lays a freshly plucked sprig of catnip from her window sill on the arm of ProxyAmenRa's reading couch*
 
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#35
I commonly see this presented by statisticians with poor semantics: "You should switch!" Switching never made sense to me because you have to make a choice between the two anyway. Declining to switch merely picks the same one you already chose, albeit with different odds.
 

Jennywocky

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#36
What if 2 doors were empty and there was a killer goat behind one door. Sticking with my first choice is actually good? To avoid the goat?


Killer goat: Changes everything.

I commonly see this presented by statisticians with poor semantics: "You should switch!" Switching never made sense to me because you have to make a choice between the two anyway. Declining to switch merely picks the same one you already chose, albeit with different odds.
Oh, you can be on my game show any time!
 

Jennywocky

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#39
I just think it's funny: The host showing you a goat is entirely a red herring.

Why? You already know that at least one of those doors has a goat, and the only thing new is that you now know which door definitely has a goat behind it; but it's irrelevant to the outcome which of those doors it is. if he didn't show you anything, the odds are still the same.

All that matters is whether the car is behind your door or in the set of unchosen doors.
 
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