The Monty Hall Problem: Help!

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Anonymous

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#1
Hello all.....

I have recently read Mark Haddons exceptional book, 'The Curious Incident of the Dog in the Night-Time' and in the process I came across the 'Monty Hall Problem'.

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"This problem has rapidly become part of the mathematical folklore.

The American Mathematical Monthly, in its issue of January 1992, explains this problem carefully. The following are excerpted from that article.

Problem:

A TV host shows you three numbered doors (all three equally likely), one hiding a car and the other two hiding goats. You get to pick a door, winning whatever is behind it. Regardless of the door you choose, the host, who knows where the car is, then opens one of the other two doors to reveal a goat, and invites you to switch your choice if you so wish. Does switching increases your chances of winning the car?

If the host always opens one of the two other doors, you should switch. Notice that 1/3 of the time you choose the right door (i.e. the one with the car) and switching is wrong, while 2/3 of the time you choose the wrong door and switching gets you the car.

Thus the expected return of switching is 2/3 which improves over your original expected gain of 1/3.

Even if the hosts offers you to switch only part of the time, it pays to switch. Only in the case where we assume a malicious host (i.e. a host who entices you to switch based in the knowledge that you have the right door) would it pay not to switch.

There are several ways to convince yourself about why it pays to switch. Here's one. You select a door. At this time assume the host asks you if you want to switch before he opens any doors. Even though the odds that the door you selected is empty are high (2/3), there is no advantage on switching as there are two doors, and you don't know thich one to switch to. This means the 2/3 are evenly distributed, which as good as you are doing already. However, once Monty opens one of the two doors you selected, the chances that you selected the right door are still 1/3 and now you only have one door to choose from if you switch. So it pays to switch.


References


L. Gillman The Car and the Goats American Mathematical Monthly, January 1992, pp. 3-7. "

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Another link: http://www.grand-illusions.com/monty.htm


Now: call me stupid if you like (and god knows maths was never my strong point), but I cannot FOR THE LIFE OF ME get my head around why changing your mind is a good idea! The chances are equal, surely?!!

I think this counts as Fortean as it seems to defy logic!

I am hoping that one of you good and brainy people can explain this to me (pref. in words of one syllable) in such a way that I get it.

How? HOW??? For the LOVE OF GOD!!! *gnashes teeth*



Thanks in anticipation!
 

_Lizard23_

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#2
It's a mind-bender and no mistake.
Because with only 2 doors left common sense dictates that it should be a 1 in 2 probability aka 50 50 chance

*However*, the probability is higher that you chose a goat initially because there are twice as many goats as cars. Thus the probability of having chosen a goat is 2 in 3.

That means that 2 times out of 3 you will have chosen a goat.

And then he's just shown you the other goat.

So 2 times out of 3 the other door is the car.

So it is better to switch.

It's difficult to conceptualize and difficult to explain.
I like it! ;)
 
A

Anonymous

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#3
Damn good book and an intrigueing problem.

If you choose the car door and change your mind you lose.
If you choose the goat A door and change your mind then you win.
If you choose the goat B door and change your mind then you win.

Chance of winning not changing mind = 1/3
Chance of winning changing mind = 2/3

therefore it's better to change your mind.

Is that correct? Probability wasn't my strong point.:D
 

_Lizard23_

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#4
Yes.

It does sort of help if you think of there being 100 doors and 99 goats and 1 car.

And then the guy shows you 98 goats.

That way it is only a 1 in 100 chance you chose the car the first time.

Which means it's a 99 in 100 chance you chose a goat.

Which means there's a 99 in 100 chance that the other door is the car.

And who would argue with that kind of odds?
 
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Anonymous

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#5
WOO-HOO!! I geddit!!

Thanks Lizard23, this was the bit that made the penny drop...

"*However*, the probability is higher that you chose a goat initially because there are twice as many goats as cars. Thus the probability of having chosen a goat is 2 in 3.

That means that 2 times out of 3 you will have chosen a goat."

I am really looking forward to driving everyone mad with this in the pub tonight!!


:D
 
A

Anonymous

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#10
Personally I'd rather have a goat! Goats are vegetarian, relatively placid, and make good pets.
Cars are ugly, loud, ecologically-damaging and very destructive to human life. Give me a nice goat any time!
Great thread. :)

Big Bill Robinson

(That was a party political broadcast by the Green Party) :)
 

OneWingedBird

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#11
Perhaps I've missed something, but I disagree with this.

The only way your chances improve is if you pick a goat door and the host opens the door you picked and lets you rechoose (then your chances go from 1/3 to 1/2).

If the host always opens a door which has a goat behind it and isn't the one you picked, then the odds are the same (1/2) for the remaining two doors. Switching won't make any difference.
 

ginoide

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#12
i agree, BRF.
if the guy opens a door and shows you one goat, at that point of course you will not choose that door. which means you will have to choose between the other two doors, and at that point your chances will be 50-50.
(unless of course you want to win a goat, in which case you'll just choose the door that the guy has just opened and have a 100% chance of success...)
 

_Lizard23_

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#13
The host never opens the door you picked. until the end of the game where you get to see what you (could have) won.

We are assuming the host is not trying to trick you in any way, he's an emotionless construct created for the purposes of a mathematical plaything. We are also assuming you would rather win a car than a goat.

You pick a door. The host always opens one of the other two doors which has a goat behind it. You may now choose either of the remaining 2 doors.

As there are more goats than cars when you picked probability dictates the door you picked initially is more likely to hide a goat than a car.

As I say, consider a scenario where there are 100 doors, 99 goats and 1 car. You choose a door. What is most likely to be behind the door you choose? Now the host opens 98 doors he knows have goats behind them. So now there's two doors, the one you originally chose and one other. You know one of them has a car behind it. Which door is more likely to have a car behind it?

Or if you prefer imagine you and however many thousands of other fools have bought a weekly lottery ticket. Then imagine the draw is done and you don't know the results. Then someone puts your lottery ticket and one other lottery ticket on the table in front of you and tells you one of those tickets has won the jackpot.
Of course "it could be you". But probability dictates quite strongly that it isn't ..... and as you *know* *one* of the tickets is worth millions ........ well of course .... you don't *have* to swap ...... but I'll gladly take the other one if you don't want it.
 

Breakfastologist

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#14
It seems to me that what is strange about this problem is that one of the doors is not a "real" door - it is a condition that never changes.

So when you have to choose door A, B or C and both B and C have a goat behind them, then they effectively counts as a single door. Although the probabilities are correct the choice that matters is not the one in three decision between the doors, but the one in two chance that the thing behind the door you chose is a car.
 

ginoide

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#15
there are thrree doors, two goats, one car

we learn that one door surely has a goat behind it.
so that door doesn't coun't anymore. two doors are left. at this point i'm reasoning in terms of "halves", not in terms of "thirds". talking of "2/3 chances" when there are TWO doors makes absolutely no sense. and at this point there is absolutely no reason why my change-of-mind should increase my probabilities. i have exactly 50% on one door and 50% on the other one.
 
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Anonymous

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#16
I read this much clearer explanation recently.

For the purposes of this explanation, we state that the car is behind door number 1.

Three choices:

a) The contestant picks door number 1. The host opens either 2 or 3, revealing a goat. In this case the contestant should stick with his original choice.

b) The contestant picks door number 2. The host opens 3 (the only remaining goat door). In this case the contestant should switch.

c) The contestant picks door numebr 3. This is similar to b) - the host opens door 2. In this case the contestant should switch.

Not knowing which door holds the prize, the contestant will reason that in 2/3 cases, he will win by switching.

One reason it gets confusing is that sometimes people don't realise that the host's choice is not independent of your original choice.

David
 
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Anonymous

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#17
Goats are vegetarian, relatively placid, and make good pets.
Now I'm even *more* confused. I'm vegetarian, relatively placid, and make a good pet. Does that mean I'm a goat?





Nurse... NURSE!


:goof:
 
A

Anonymous

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#18
Charlotte said:
Now I'm even *more* confused. I'm vegetarian, relatively placid, and make a good pet. Does that mean I'm a goat?



Nurse... NURSE!


:goof:
No, you're not a goat and I'm not a nurse. This problem confused me for a long time as well. So I'll take my shot at explaining it.

You have 3 doors, A B C there is a 1/3 chance the car is behind any one of the doors,and let's say you choose A.

So we have the car has a 1/3 chance of being behind A and there is a 2/3 chance it is behind B or C.

1/3 A or 2/3(B or C)

Now the host comes and opens one of the doors which is never the door you have chosen or the one with the car. Let's say its C.
Now we get:

1/3 A or 2/3 B

I hope that helps understand how the 2/3 odds carry over to the one door. I think one of the confusing things about this problem is the 2nd choice isn't really 50/50 since the odds were 1/3 when you initially chose.
 

nikoteen1

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#19
the answer to the original question "Does switching increases your chances of winning the car?". is without doubt a huge and resounding "it dont matter". whatever goat / car options you made in the past have gone into history and you are left with a 50 50 chance. and to prove my point me and my mate just did an experiment,

we had 100 attempts, if i chose the door with the car behind it we disregarded that attempt.

each attempt i would switch, or stick to my first choice door and make a note of my choice until i had 50 of each.

my mate did not know prior to the attempt if i would stick or switch or did he see my "tally sheet"

the results were..

switch doors = 41
stick to original choice = 59

disregarded choices where i chose the car first time = 33.

(no goats were harmed in the execution of this experiment)

(but i did eat a penguin)
 

Stormkhan

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#20
Knock on each door. If you get a goat-noise in reply, pick the one that goes "beep-beep!"
 

OneWingedBird

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#21
I think one of the confusing things about this problem is the 2nd choice isn't really 50/50 since the odds were 1/3 when you initially chose.
The 2nd choice is 1/2. I don't say how the first choice being 1/3 has any bearing on that. The probability changed when it went from being a question of 3 doors to 2.

It's also completely irrelevant which door you choose first, or indeed second as the host's elimination of one goat door doesn't give any information about which of the two remaining doors the car is behind.
 

phi23

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#22
OK here's the deal, its all about "conservation of probability".

You have to remember to take into account the actions of the host, Monty. Here's why:

1) You pick a door. It is either the right door (1 in 3 chance) or the wrong door (2 in three chance).

2) Monty has to remove a door. More importantly, he has to remove a wrong door. Let's look at what can happen.

a) You picked the right door to start off with. Monty now has the choice of two wrong doors. His picking a door tells you nothing.

b) You picked the wrong door. Now Monty has left one wrong door and one right one. Since he cannot pick the right one, he has to pick the wrong one. In this case, him taking away the wrong door tells you that the other one must be right.

Now the trick is that a) happens 1 in 3 times, and b) happens 2 in 3 times. So- 2 out of 3 times Monty has no choice in which door he removes, hence 2 out of 3 times the door he has left is the one with the prize.

And since your original chance of getting the right door = 1 in 3,
and chances that Monty has "told you" that the other door contains the prize = 2 in 3,

... you should always swap to the other door.
 
A

Anonymous

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#23
The answer is 50/50.

I'll use this truth table to demonstrate why it is so (note that this table uses the contestant's choice as door 1) :

door 1 | door 2 | door 3

true false false
true false false
false true false
false true false
false false true
false false true

if we mark the doors that are opened by the quiz master in BOLD then we have:

door 1 | door 2 | door 3

true false false
true false false
false true false
false true false
false false true
false false true

we can eliminate the following paths because, bearing in mind the initial door choice is door 1, the host has to select one of the remaining alternatives. He cannot select the one you have selected

door 1 | door 2 | door 3

false true false
false false true

this leaves us with the following and we see we have 2 winning instances and 2 losing instances

door 1 | door 2 | door 3

true false false
true false false
false true false
false false true

therefore there are 2 instances where it would better to swap and 2 where it isn't, hence the chance is 50/50.
 

Breakfastologist

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#24
I still maintain it is a sleight of hand question, misdirecting you to the doors when the fact that one door will always be opened renders it entirely contingent.
 
A

Anonymous

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#25
nicoteen:

we had 100 attempts, if i chose the door with the car behind it we disregarded that attempt.
I don't quite understand your post. If you're going to discard all cases where you originally pick the car, then you'd know you've got a goat so you should always switch.

I've simulated this in BASIC and the answer really, truly, honestly, is 2/3.

Scabbydog, by eliminating those cases where the host would open door 1, you've screwed up the probabilities. Or at least I think that's your mistake - as you can tell, this is a tricky one.

Instead of looking at the set of cases with the prize behind different doors, just assume the prize is always behind door 1.

Then look at the three cases that come up, based on the three choices the contestant could make:

1. Car Goat Goat
2. Car Goat Goat
3. Car Goat Goat

Now the host opens a door (not the one you picked) with a goat behind it (marked italic):

1a. Car Goat Goat
1b. Car Goat Goat
2. Car Goat Goat
3. Car Goat Goat

Now, just as in your example, there are four possibilities, and in 2 out of 4 you should switch. But cases 1a and 1b come out of case 1 above, so they have a combined weighting of 1/3, not 1/4 each.

There are a few simulations out there; I'll Google for some later, but they are out there and they show the answer to be 2/3.

David
 
A

Anonymous

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#26
Originally posted by ScabbyDog
The answer is 50/50.

I'll use this truth table to demonstrate why it is so (note that this table uses the contestant's choice as door 1) :

door 1 | door 2 | door 3

true false false
true false false
false true false
false true false
false false true
false false true
Each case a 1/6 probability of occuring (because you've doubled each one).

if we mark the doors that are opened by the quiz master in BOLD then we have:

door 1 | door 2 | door 3

true false false
true false false
false true false
false true false
false false true
false false true

we can eliminate the following paths because, bearing in mind the initial door choice is door 1, the host has to select one of the remaining alternatives. He cannot select the one you have selected

door 1 | door 2 | door 3

false true false
false false true

this leaves us with the following and we see we have 2 winning instances and 2 losing instances

door 1 | door 2 | door 3

true false false
true false false
false true false
false false true
All correct, but you can't just redistribute the 1/6 (for each case) probabilities from the start to 1/4 for each case. The top two cases still have 1/6 probability - the bottom two have sort of "absorbed" their impossible counterpart and therefore have a probability of 1/3 each.

therefore there are 2 instances where it would better to swap and 2 where it isn't, hence the chance is 50/50.
But the two instances where it is better to switch are twice as likely to occur, making it 2/3 in favour of switching.

David
 

ginoide

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#27
well
i understand it's some kind of a mathematician's cavil, but:

first you have a goat, another goat, a car

then the guy shows you a goat, which is therefore EXCLUDED, given the rules-


then what is left is a goat and a car.

a goat. a car.

one. two.

fifty. fifty.

not 33.333, 33.333, 33.333.

no.

just: 50 -50.

that gives you a 50% chance on either bet. keep it, switch it: you have the same chances.
 

JamesWhitehead

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#28
genoide is right. You are being asked to make a new bet with new
odds. Your earlier bets with different odds are irrelevant.

NB: For the purposes of this post I am excluding magic, psi and morphic
resonance and I am pretending to be rational. :rolleyes:
 
A

Anonymous

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#29
The simplest way of thinking about this problem is the following.

I pick a door at random.
The probability of guessing correctly is 1/3.
The probability of guessing incorrectly is 2/3

This latter number is the key. Regardless of you choosing correctly or incorrectly, he will always open a door with a goat behind it, and hence the probability of the selected door being incorrect is still 2/3.

This means that sticking with the door gives you a 2/3 probability of getting it wrong, whereas changing gives you a 2/3 probability of being right. :)
 

_Lizard23_

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#30
This is hilarious!

I really thought I had explained it in a way that anyone could understand ... and I am absolutely no mathematician but do have a fair bit of pride in my communication skills.

I'm not calling anyone thick or anything, I am just dead proud I worked it out, and I completely understand the arguments of the 1/2 camp.

In a way it is a microcosm of the issues of belief and scepticism that we so frequently discuss. Initially I too thought "2 doors, choose again, 50:50, obvious" and I could not get past that for a quite a while. But if you are just prepared to abandon that belief and look at the maths (evidence) open-mindedly a most wonderous revelation may occur, that things are not quite what they seem!

50:50 people ... what about my lottery ticket example? It's just the same thing scaled up ... you're not telling me that your original ticket is equally likely to be the winner as the other? Seriously?

As Pi23 says, they key words here are "conservation of probability". Once we get to the 2 door state it is not the same as if we *started* in a two door state. We have more information that if that were the case. We know that it was less likely for you to have chosen the car when you originally guessed.

I looked into this a bit on the net in an all-too-characteristic moment of self-doubt, and there are loads of simulations available online. Perhaps using one of these might convince you you are possibly not right, that perhaps it is not that simple, and then you will be able to think about the problem rather than applying common sense to it, which in matters of probability and indeed forteanism is often not the most useful tool :)
 
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