• We have updated the guidelines regarding posting political content: please see the stickied thread on Website Issues.

The Monty Hall Problem & Other Apparent Paradoxes

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


(quote)
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.

the reason we discarded those cases was because we had already "won" the car.
and also because the original question was "does switching increase your chance of winning" and since the switch was between the 2 remaining doors we only counted those attempts.
 
nikoteen : I cannot get my head round where you are coming from with your methodology at all.

However

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

33/100 ... you just don't get closer to 1/3 than that ... with whole number anyway ;) ... thus proving the theory that you only have a 1/3 chance of picking the damn car the first damn time, and so the chances of the other remaining door hiding the car in the 2 door state are 2/3. :)
 
lizard23 said:
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.
Click to expand...

I thought this thread would die after two posts as well!

OKay lets try it this way:

There are two goats and only one car.

Therefore if you pick a door you have more likely chosen a goat.

The host shows you another goat behind one of the other doors. So you know that there is a goat behind that door. And you also know that you have probably picked a goat with your first choice because that is where the odds lie.

The best option is therefore to switch as the car is then most likely to be behind the other door. :)

Fabulous book by the way.:yeay:
 
Originally posted by ginoide
well
i understand it's some kind of a mathematician's cavil, but:

first you have a goat, another goat, a car
Click to expand...

(then you choose a door)

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


then what is left is a goat and a car.
Click to expand...

But you are more likely to have already chosen a goat. You make the choice before a door is excluded, and the host's choice of excluded door is not independent of your choice.

David
 
nikoteen said:
the reason we discarded those cases was because we had already "won" the car.
and also because the original question was "does switching increase your chance of winning" and since the switch was between the 2 remaining doors we only counted those attempts.
Click to expand...

But in the real game, Monty didn't just say "Oh, you picked the car straight away, you win." He still forces you to make the choice between switching or not.

David
 
So... basically [any typos duly accredited]:

[A]WIN LOSE [C]LOSE

I choose A and door B is opened.
I don't switch to C and win the car.
I switch to C and lose the car.

I choose A and door C is opened.
I don't switch to B and win the car.
I switch to B and lose the car.


I choose B and door C must be opened.
I don't switch to A and lose the car.
I switch to A and win the car.


I choose C and door B must be opened.
I don't switch to A and lose the car.
I switch to A and win the car.



[A]LOSE WIN [C]LOSE

I choose A and door C must be opened.
I don't switch to B and lose the car.
I switch to B and win the car.


I choose B and door A is opened.
I don't switch to C and win the car.
I switch to C and lose the car.

I choose B and door C is opened.
I don't switch to A and win the car.
I switch to A and lose the car.


I choose C and door A must be opened.
I don't switch to B and lose the car.
I switch to B and win the car.



[A]LOSE LOSE [C]WIN

I choose A and door B must be opened.
I don't switch to C and lose the car.
I switch to C and win the car.


I choose B and door A must be opened.
I don't switch to C and lose the car.
I switch to C and win the car.


I choose C and door A is opened.
I don't switch to B and win the car.
I switch to B and lose the car.

I choose C and door B is opened.
I don't switch to A and win the car.
I switch to A and lose the car.



In all instances, twice I don't switch and win, twice I
do and win.

No advantage - the odds remain 50/50.


However, whilst this scenario separates WIN [A] from LOSE
, it amalgamates the LOSE possibilities, essentially:

ABB
BAB
BBA


The true situation is:

ABC
ACB
BAC
BCA
CAB
CBA


Thus:

CAR GOAT1 GOAT2
CAR GOAT2 GOAT1
GOAT1 CAR GOAT2
GOAT1 GOAT2 CAR
GOAT2 CAR GOAT1
GOAT2 GOAT1 CAR


Does separately distinguishing GOAT1 from GOAT2 make any
difference whatsoever?


Consequently:

CAR GOAT1 GOAT2

I choose CAR and GOAT1 is opened.
I don't switch to GOAT2 and win CAR.
I switch to GOAT2 and lose CAR.

I choose CAR and GOAT2 is opened.
I don't switch to GOAT1 and win CAR.
I switch to GOAT1 and lose CAR.


I choose GOAT1 and GOAT2 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.


I choose GOAT2 and GOAT1 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.



CAR GOAT2 GOAT1

I choose CAR and GOAT1 is opened.
I don't switch to GOAT2 and win CAR.
I switch to GOAT2 and lose CAR.

I choose CAR and GOAT2 is opened.
I don't switch to GOAT1 and win CAR.
I switch to GOAT1 and lose CAR.


I choose GOAT1 and GOAT2 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.


I choose GOAT2 and GOAT1 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.



GOAT1 CAR GOAT2

I choose GOAT1 and GOAT2 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.


I choose CAR and GOAT1 is opened.
I don't switch to GOAT2 and win CAR.
I switch to GOAT2 and lose CAR.

I choose CAR and GOAT2 is opened.
I don't switch to GOAT1 and win CAR.
I switch to GOAT1 and lose CAR.


I choose GOAT2 and GOAT1 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.



GOAT1 GOAT2 CAR

I choose GOAT1 and GOAT2 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.


I choose GOAT2 and GOAT1 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.


I choose CAR and GOAT1 is opened.
I don't switch to GOAT2 and win CAR.
I switch to GOAT2 and lose CAR.

I choose CAR and GOAT2 is opened.
I don't switch to GOAT1 and win CAR.
I switch to GOAT1 and lose CAR.



GOAT2 CAR GOAT1

I choose GOAT2 and GOAT1 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.


I choose CAR and GOAT1 is opened.
I don't switch to GOAT2 and win CAR.
I switch to GOAT2 and lose CAR.

I choose CAR and GOAT2 is opened.
I don't switch to GOAT1 and win CAR.
I switch to GOAT1 and lose CAR.


I choose GOAT1 and GOAT2 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.



GOAT2 GOAT1 CAR

I choose GOAT2 and GOAT1 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.


I choose GOAT1 and GOAT2 must be opened.
I don't switch to CAR and lose CAR.
I switch to CAR and win CAR.


I choose CAR and GOAT1 is opened.
I don't switch to GOAT2 and win CAR.
I switch to GOAT2 and lose CAR.

I choose CAR and GOAT2 is opened.
I don't switch to GOAT1 and win CAR.
I switch to GOAT1 and lose CAR.



A summary of those six possibilities:

LOSE/LOSE/WIN/WIN/LOSE/LOSE/WIN/WIN/WIN/LOSE/LOSE/WIN/WIN
/WIN/LOSE/LOSE/WIN/LOSE/LOSE/WIN/WIN/WIN/LOSE/LOSE


I both win and lose the car on 12 of 24 outcomes.

Strictly 50/50; so why should I switch?


In answer to my query, 'Does separately distinguishing
GOAT1 from GOAT2 make any difference whatsoever?', I note
that any differentiation seems to be irrelevant.


Doubtless, as an acknowledged layman, I'm unaware of
inherent, fundamental, statistical criteria which
substantiates accepted proof why we should always change
our initial selection - two-thirds of the time we would
win the car.

An explanation _why_ the above cited data is spurious
would be much appreciated.


James Easton.
 
scotmedia said:
So... basically [any typos duly accredited]:

[A]WIN LOSE [C]LOSE

I choose A and door B is opened.
I don't switch to C and win the car.
I switch to C and lose the car.

I choose A and door C is opened.
I don't switch to B and win the car.
I switch to B and lose the car.


I choose B and door C must be opened.
I don't switch to A and lose the car.
I switch to A and win the car.


I choose C and door B must be opened.
I don't switch to A and lose the car.
I switch to A and win the car.
Click to expand...


You've made the same mistake as Scabbydog. I've only quoted your first case since the others are identical. You've listed four outcomes and given them all the same probability. But note that you've listed chosing door A twice. The probability of choosing door A is 1/3. The probability of choosing door B is 1/3. The probability of choosing door C is 1/3. Since there are two equally likely cases under choosing door A, each must have a probability of 1/6. The other cases have a probability of 1/3.

What is the probability that the contestant will pick the car to begin with? 1/3. After a losing door is opened, what is the probability that he picked the car to begin with? It's still 1/3 - the event has already occured and nothing will change the probability. Therefore there must be a 2/3 probability that the car is behind the other door.

David
 
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.
Click to expand...

No we don't, because unless we know wether we picked the car door or a goat door the first time, we don't know wether the host has picked a goat door or the only remaining goat door.
 
But we know we ARE *MORE LIKELY* TO HAVE PICKED A GOAT DOOR.
It is a question of probabilities afterall!
Therefore we know it is *more likely* that the host has shown us the one remaining goat door than one of two remaining goat doors.
 
Yes
We would all agree that where there are two goats and only one car, your chances of picking a goat are 2/3

So say you pick door 1 and the host opens door 2 to show you a goat.

So:

DOOR 1: 2/3 chance of a goat-there is more likely to be a goat than a car yes? You know this because there are two goats available.

DOOR 2: Definatly a goat, the host has just shown you, yes?

DOOR 3:If DOOR 2 definatly has a goat and DOOR 1 has a 2/3 chance of having a goat then this door is more likely to have the car. Yes?
 
Congratulations!! (<---- genuinely not sarcastic).
Isn't the sensation of that moment of understanding intoxicating?!
 
OK lets take this one stage further and consider...

Multi-Stage Monty Hall Dilemma

Suppose there are four doors, one of which is a winner. The Monty says:

"You point to one of the doors, and then I will open one of the other non-winners. Then you decide whether to stick with your original pick or switch to one of the remaining doors. Then I will open another (other than the current pick) non-winner. You will then make your final decision by sticking with the door picked on the previous decision or by switching to the only other remaining door."

What is the best strategy now?
 
lizard:
honestly: no, it's not. i feel bad for having wasted lots of time, and having been indirectly called <thick> and <fool> for a result that will not change my life in any way. i hope that if other illuminations await me in my mortal days that they are more gratifying.
 
Right I'm confused now, because I understand that as a mathematical question of probabilities you can say it is best to switch, as an actual "real world" problem, I still recon you are down to a 50:50 situation- the third choice (one goat) has been removed. This means that you are left with the door you chose (you still have no way of knowing what is behind unless you have a very sensitive nose) or the other door each having the same chance of housing a thoeretical goat or a metaphorical car.

Qoute
"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?"

That lottry ticket example stated that you did not know what the numbers were, if it is monty with the other ticket then he is a non malicious fictional character to aid the story, then his ticket has a 50% chance of being a winner (in the real world you would never offer to swap a winning ticket). This agian correlates to the 50:50 chance being correct - only two options to choose from!
 
Oh.
Well, I'm sorry to hear that, ginoide. As I said earlier I don't think anyone is 'thick' for not understanding this problem as it is pretty counter-intuitive, although I do admit to finding it somewhat amusing and allegorical, even, that people are capable of sticking to their insinctive beliefs in the face of a multitude of very reasonable argument to the contrary.

It seems to me that the experienced sensation of 'getting it', whether 'it' be a joke, a puzzle or a spiritual revelation, that "ah-ha!" moment is pretty much identical.

Pi23 *you* are just a trouble maker! ;)
 
I don't understand it, but I can see that it is correct, if you see what I mean. Even though I have a fair understanding of maths, this is as mysterious to me as a rain of fish.

Why chances are better if I start with three doors than if I had come in when there are only two doors?

What if I come into the room and there are three doors, but one of them is already open and has a goat behind it and I havent yet made a decision?

How can the intangible choice I make change the hard maths that controls the game?
 
Well when I played it *according to the instructions* it proved the 'better to switch' theory ... how freaky is that? The java applet must be *psychic*!

:headbutt:
 
lizard23 said:
Well when I played it *according to the instructions* it proved the 'better to switch' theory ... how freaky is that? The java applet must be *psychic*!

:headbutt:
Click to expand...

Ok Ok that is correct, but I was only being controversial for my own pitiful entertainment!

The results were more successful when swapping (85% success against 50% if I stuck) But I'm still of the opinion that this only works because mathematicians are cleverer than what I is!!
:D
 
Breakfast said:
What if I come into the room and there are three doors, but one of them is already open and has a goat behind it and I havent yet made a decision?

How can the intangible choice I make change the hard maths that controls the game?
Click to expand...

Ah well that is a different situation because remember that when you are making your choice when all three doors are closed you have no way of knowing which has a goat behind it. You just know there is a 2/3 chance of it being behind each of the doors.

If one door is already open you can effectively remove that goat from the equation because it enters the picture before you have even made your choice.

So there are only two doors and your chances are 50/50.
 
The first choice isn't a choice that influences the outcome. You could pick any door and the compere shows you a goat, then you have the chance to choose. the choice is do I stick with the already selected door, or move. The chances of the first selection doesn't influence the odds of the actually choice that wins the car or the goat.
 
In my first post (page 2) I went with 6 scenarios where the initial outcome would be (the player choice is always door 1 for the purpose of the table and the bold words are the host selection):

door 1 | door 2 | door 3 win | lose
car goat goat 1 0
car goat goat 1 0
goat car goat 0 1
goat goat car 0 1
goat car goat 0 1
goat goat car 0 1


if you add up the wins versus the loses then you get 4/6 or 2/3 in favour of swapping (with hindsight on the part of the contestant)

however the host cannot select a door that you have selected as that is just plain daft so we have to get rid of the last two scenarios and this gives us 2 wins and 2 loses which is obviously 1/2.

All the web sites i've found say otherwise to the 50/50 answer and various people on this thread have tried computer programs to show it. Therefore I thought about it from the perspective of a computer running an instance of the game and got this:

initially we have this set of possible scenarios so we have three games played (we still select door 1 bytheway)

door 1 | door 2 | door 3
car goat goat
goat car goat
goat goat car

now in the real world the host can THINK of two possible doors he can select but he can ACTUALLY only use one of them per instance of the game, so:

door 1 | door 2 | door 3 win | lose
car goat goat 1 0 (if you ran the game again the host could choose the other one)
goat car goat 0 1
goat goat car 0 1

which gives us 2/3 in favour of swapping

Now this either suggest that the 2/3 gang are correct or that simulating the problem on a computer is fundamentally flawed.
 
however the host cannot select a door that you have selected as that is just plain daft so we have to get rid of the last two scenarios and this gives us 2 wins and 2 loses which is obviously 1/2.
Click to expand...

You present the 6 possibilities - each has a probability of 1/6. When you remove the two impossible ones, you can't redistribute the probabilities. The four scenarios are not equally likely!

car goat goat
car goat goat
goat car goat
goat goat car

The first two scenarios have the car behind door 1. These two scenarios together have the same probability as each of the other two.

As Cecil explained in his Straight Dope column, what if you were asked to pick a door, and then asked if you would like to swap it for the contents of both the remaining doors? This is equivalent to the stated problem, and of course you'd rather try the contents of two doors than one.

door 1 | door 2 | door 3 win | lose
car goat goat 1 0 (if you ran the game again the host could choose the other one)
Click to expand...

Yes, he could choose the other one, but if and only if the car is behind door 1, and that only happens 1/3 of the time.

David
 
For any permutation of initial choice there is always at least one goat for the host show. Therefore the host Will show either the one goat he has, or one of his goats. To me, this doesn't affect the out come. The important choice is between the contestant's chosen door or the unchosen door, the first choice having no affect on the outcome, and is just confusing.

Example

Door 1 = Goat
Door 2 = Car
Door 3 = Goat

If the contestant picks 1 or 3 then the Host can only open door 3 or 1.
If the contestant picks Door 2 then the Host has 1 and 3 to choose from.

This is a limitation not on the Contestant's chances of winning but on the Host's selection of goat.

The Host opens a goat door, if the Contestant picked door 1 then he opens door 3, but if the Contestant had chosen Door 2 then the Host may have opened Door 3. All we know at this stage is that the door opened will have a goat behind it. It has no import in the outcome.

The choice is now between Contestant's choice and the other door. This is now the issue, not what has gone before. We know the Host will always show a goat. Since the contestant can now change his mind or stick, its not a 1 in 3 chance but simply the toss of a coin.

We are making an assumption that somehow the choices are cumulative, which they are not. The choice is your initial choice or the other door, not which of the three doors is the car behind.
 
On 1/3/2004, David wrote:

>But note that you've listed chosing door A twice. The
>probability of choosing door A is 1/3. The probability
>of choosing door B is 1/3. The probability of choosing
>door C is 1/3.

Yes, that's the culprit!

The logic expressed:

[A]WIN LOSE [C]LOSE

I choose A and door B is opened.
I don't switch to C and win the car.
I switch to C and lose the car.

I choose A and door C is opened.
I don't switch to B and win the car.
I switch to B and lose the car.


Should read:

I choose A and EITHER door B OR door C is opened.
I don't switch and win the car.
I switch and lose the car.


Guessing the WIN door and sticking with it, can only ever
be a single event, not two.


In short:

WIN LOSE LOSE

Doesn't matter in what sequence they exist.

On average, we initially select LOSE 2/3 of the time.

As a direct consequence (and this is the significant
point), on 2 out of every 3 plays we must then be shown
what's behind the other LOSE door.

Therefore, if we switch we WIN and should switch every
time.


The odds are in our favour because we know that 'Monty'
can never select the WIN door.


James Easton.
 
The choice is now between Contestant's choice and the other door. This is now the issue, not what has gone before.
Click to expand...

What went before is central to what happens next! Specifically, whether or not the contestant picked the car in the first place!

Would you agree that these two statements will always be true?

1. If (and only if) the contestant picked the car, he should stick.

2. If (and only if) the contestant did not pick the car, he should switch.

Or to put it more formally:

1.
Condition: the contestant picked the car.
Best action: Stick

2.
Condition: the contestant did not pick the car.
Best action: Switch

What are the respective probabilities for those two conditions?

The choice is your initial choice or the other door, not which of the three doors is the car behind.
Click to expand...

Right - and what's the probability that your initial choice was correct? 1/3. So what's the probability that the other door would be the correct choice? 1-1/3=2/3.

David
 
The question then is does the initial choice matter? In the real world possibly, in statistics no. The choice is stick or move. 50-50
 
The question then is does the initial choice matter? In the real world possibly, in statistics no. The choice is stick or move. 50-50
Click to expand...

The existence of two choices doesn't automatically dictate a 0.5 probability for each (aka 50-50). If you have a roulette wheel with 30 black holes and 1 red, which would you bet on?

The statistics are an accurate model of the real world situation. See the myriad simulations available on the Internet.

In one sense it doesn't matter which door you pick, because you don't know which holds the prize. But it does matter if you stick or switch, since the latter doubles your chances of winning over the former.

Consider this: I ask you to pick one of three doors at random. You do so, then I open one of the other doors and reveal a goat. Now here's the twist - I don't give you the chance to switch. You're stuck with your original choice. What's the probability that you picked the car at the start?

David
 
You must log in or register to reply here.
Back
Top