The Monty Hall Problem

[aurik]

hm, i said 1/3 earlier but i was thinking about this today at work and changed my mind

1/2 should be right

an easy way to think about this is to change the question a bit... pretend it's not 3 doors to begin with, but rather 10 doors, and the host said after you pick one door that he will eliminate 8 goat doors, leaving you with 2

so the basic logic here is that the illusion is before he eliminates the wrong answers: because he stops you after you make a first decision and let you begin the entire process of picking out an answer again, the whole question is really just picking one door between two, so the chance should be 50/50

No. Your first choice always has a 1/N chance of being right.

Just because he shows you N-2 "incorrect" choices after you have chosen at a 1/N rate doesn't change the fact that your originally chosen door still has a 1/N chance of being right--mainly, because you already knew that of the remaining N-1 doors, at least N-2 of them are wrong. He's just showing you which ones.

(N-1)/N of the time, he has no choice which door to leave closed. That NECESSARILY means that (N-1)/N of the time, the winning door is the one he leaves closed.

my main point is not necessarily a calculation, but math logic: the evidence of the illusion is that he stops you after you make your first choice and then give you a fresh start to pick one out of two, which nullifies everything that's been done before that (clearifying the illusion)

so the only problem is the two door problem

Except by making a choice and limiting which doors Monty can open, you derive additional information about the state of the puzzle -- namely that there is a higher-than-50% chance that the door he doesn't show you he CANNOT show you because it contains the prize.

that's exactly the illusion: you do not know what he does, and he will always open the goat door, and it doesn't matter which of the two goat doors he does open

it's not a choice between A, B, and C

it's a choice between A, B, and B

Except Monty has no choice what door to open if you did not choose the car the first time (2/3 chance).

[GRT]

GRT, why stop at 10? Let's do 100....

OK I just want you to picture this situation...

Let's say you have to pick 1 door out of 100. 1 has a car, 99 have goats. You pick a door.

I open 98 doors for you to show you 98 goats. That's all doors except the one you randomly choose, and another random door.

Do you really fucking believe there was a 50/50 chance you picked the car door right on the first try? Really?

There is a better chance the car is behind the OTHER door, than the first you picked. It's not 50/50, its better than that.

Think about it dude. Your first pick was 1 in 100. Then I eliminate all the wrong answers except for one - either the door you picked, or another random door.

Do you really think its 50/50 that the FIRST DOOR YOU PICKED is the car? Would you really, ever even CONSIDER staying with your first pick?

heh, i don't "fucking" believe anything, i do believe, however, in the mathmatical world, if the result doesn't matter, neither does the process

the result of your first choice does NOT matter in the math problem of this situation is all i'm saying

But it does matter. Tell me which of the following steps in a logical progression you disagree with
a) In a 10-door situation, 90% of the time your first choice is wrong.
b) Therefore, 90% of the time Monty has no choice on what door NOT open
c) Therefore, 90% of the time the door he does not open contains the car.
d) Therefore, 90% of the time if you switch to the door he does not open, you win the car

i disagree with the very first step: there's no continuation (in the math world) between your first and second choice, so it doesn't matter if you were right or wrong the first time

[GRT]

[quote=aurik]

hm, i said 1/3 earlier but i was thinking about this today at work and changed my mind

1/2 should be right

an easy way to think about this is to change the question a bit... pretend it's not 3 doors to begin with, but rather 10 doors, and the host said after you pick one door that he will eliminate 8 goat doors, leaving you with 2

so the basic logic here is that the illusion is before he eliminates the wrong answers: because he stops you after you make a first decision and let you begin the entire process of picking out an answer again, the whole question is really just picking one door between two, so the chance should be 50/50

No. Your first choice always has a 1/N chance of being right.

Just because he shows you N-2 "incorrect" choices after you have chosen at a 1/N rate doesn't change the fact that your originally chosen door still has a 1/N chance of being right--mainly, because you already knew that of the remaining N-1 doors, at least N-2 of them are wrong. He's just showing you which ones.

(N-1)/N of the time, he has no choice which door to leave closed. That NECESSARILY means that (N-1)/N of the time, the winning door is the one he leaves closed.

my main point is not necessarily a calculation, but math logic: the evidence of the illusion is that he stops you after you make your first choice and then give you a fresh start to pick one out of two, which nullifies everything that's been done before that (clearifying the illusion)

so the only problem is the two door problem

Except by making a choice and limiting which doors Monty can open, you derive additional information about the state of the puzzle -- namely that there is a higher-than-50% chance that the door he doesn't show you he CANNOT show you because it contains the prize.

that's exactly the illusion: you do not know what he does, and he will always open the goat door, and it doesn't matter which of the two goat doors he does open

it's not a choice between A, B, and C

it's a choice between A, B, and B

Except Monty has no choice what door to open if you did not choose the car the first time (2/3 chance).[/quote:a14be]

but you don't know that: it does not matter to you which of the two doors he opens

[aurik]

But it does matter. Tell me which of the following steps in a logical progression you disagree with
a) In a 10-door situation, 90% of the time your first choice is wrong.
b) Therefore, 90% of the time Monty has no choice on what door NOT open
c) Therefore, 90% of the time the door he does not open contains the car.
d) Therefore, 90% of the time if you switch to the door he does not open, you win the car

i disagree with the very first step: there's no continuation (in the math world) between your first and second choice, so it doesn't matter if you were right or wrong the first time

You disagree that if you choose a door at random from 10, the chance that it contains the car is NOT 10%? What chance of winning does it have then? And why?

[aurik]

Except Monty has no choice what door to open if you did not choose the car the first time (2/3 chance).

but you don't know that: it does not matter to you which of the two doors he opens

Ignore this line of thinking. It's more confusing than the other (although still correct...you are 66% certain that your initial choice has forced him to leave a particular door closed)

[GRT]

But it does matter. Tell me which of the following steps in a logical progression you disagree with
a) In a 10-door situation, 90% of the time your first choice is wrong.
b) Therefore, 90% of the time Monty has no choice on what door NOT open
c) Therefore, 90% of the time the door he does not open contains the car.
d) Therefore, 90% of the time if you switch to the door he does not open, you win the car

i disagree with the very first step: there's no continuation (in the math world) between your first and second choice, so it doesn't matter if you were right or wrong the first time

You disagree that if you choose a door at random from 10, the chance that it contains the car is NOT 10%? What chance of winning does it have then? And why?

i don't disagree with the logic itself, i disagree with this step being any relevent to the problem as a whole to start with

[Bagel]

I can't believe this thread is 7 pages long. You're telling me in a billion door scenario you wouldn't switch doors after 999,999,998 were opened and you were left with your original choice or the one that he "happened" not to open?

[GRT]

I can't believe this thread is 7 pages long. You're telling me in a billion door scenario you wouldn't switch doors after 999,999,998 were opened and you were left with your original choice or the one that he "happened" not to open?

hehe, read it again: when you put it like that, it's no longer a math question, but a question of behaviors

[aurik]

But it does matter. Tell me which of the following steps in a logical progression you disagree with
a) In a 10-door situation, 90% of the time your first choice is wrong.
b) Therefore, 90% of the time Monty has no choice on what door NOT open
c) Therefore, 90% of the time the door he does not open contains the car.
d) Therefore, 90% of the time if you switch to the door he does not open, you win the car

i disagree with the very first step: there's no continuation (in the math world) between your first and second choice, so it doesn't matter if you were right or wrong the first time

You disagree that if you choose a door at random from 10, the chance that it contains the car is NOT 10%? What chance of winning does it have then? And why?

i don't disagree with the logic itself, i disagree with this step being any relevent to the problem as a whole to start with

OK, so you agree with step A. The chance that your initial choice is correct is 10%.

B) If your initial choice was incorrect (90%), Monty will open 8 of the remaining 9 to reveal goats. Since only 8 contain goats, he must open exactly those 8.

Disagree?

[GRT]

But it does matter. Tell me which of the following steps in a logical progression you disagree with
a) In a 10-door situation, 90% of the time your first choice is wrong.
b) Therefore, 90% of the time Monty has no choice on what door NOT open
c) Therefore, 90% of the time the door he does not open contains the car.
d) Therefore, 90% of the time if you switch to the door he does not open, you win the car

i disagree with the very first step: there's no continuation (in the math world) between your first and second choice, so it doesn't matter if you were right or wrong the first time

You disagree that if you choose a door at random from 10, the chance that it contains the car is NOT 10%? What chance of winning does it have then? And why?

i don't disagree with the logic itself, i disagree with this step being any relevent to the problem as a whole to start with

OK, so you agree with step A. The chance that your initial choice is correct is 10%.

B) If your initial choice was incorrect (90%), Monty will open 8 of the remaining 9 to reveal goats. Since only 8 contain goats, he must open exactly those 8.

Disagree?

you are too involved with what the host does, and as a math problem, there's only two sides of an equation, and he's not on your side

to you, the two doors you did not pick the first time look exactly the same, you do not know if you picked the right one or wrong one the first time around, so to you, it looks exactly the same if the host opened either of the other two doors

if you understand this then we solve our differences: the host is always picking the goat to open, he always picks B in the choice of A, B, B no matter if you picked A or B the first time, therefore what he has in mind is irrelevent

[aurik]

OK, so you agree with step A. The chance that your initial choice is correct is 10%.

B) If your initial choice was incorrect (90%), Monty will open 8 of the remaining 9 to reveal goats. Since only 8 contain goats, he must open exactly those 8.

Disagree?

you are too involved with what the host does, and as a math problem, there's only two sides of an equation, and he's not on your side

to you, the two doors you did not pick the first time look exactly the same, you do not know if you picked the right one or wrong one the first time around, so to you, it looks exactly the same if the host opened either of the other two doors

Are you saying you disagree with point B or not?

[GRT]

OK, so you agree with step A. The chance that your initial choice is correct is 10%.

B) If your initial choice was incorrect (90%), Monty will open 8 of the remaining 9 to reveal goats. Since only 8 contain goats, he must open exactly those 8.

Disagree?

you are too involved with what the host does, and as a math problem, there's only two sides of an equation, and he's not on your side

to you, the two doors you did not pick the first time look exactly the same, you do not know if you picked the right one or wrong one the first time around, so to you, it looks exactly the same if the host opened either of the other two doors

Are you saying you disagree with point B or not?

you are listening to me, but not hearing me

i'm saying all 4 of your steps are irrelevent to the problem

[aurik]

OK, so you agree with step A. The chance that your initial choice is correct is 10%.

B) If your initial choice was incorrect (90%), Monty will open 8 of the remaining 9 to reveal goats. Since only 8 contain goats, he must open exactly those 8.

Disagree?

you are too involved with what the host does, and as a math problem, there's only two sides of an equation, and he's not on your side

to you, the two doors you did not pick the first time look exactly the same, you do not know if you picked the right one or wrong one the first time around, so to you, it looks exactly the same if the host opened either of the other two doors

Are you saying you disagree with point B or not?

you are listening to me, but not hearing me

i'm saying all 4 of your steps are irrelevent to the problem

They're not irrelevant, and I'll show you why if you simply follow along.

Now do you disagree or agree with point B?

[GRT]

OK, so you agree with step A. The chance that your initial choice is correct is 10%.

B) If your initial choice was incorrect (90%), Monty will open 8 of the remaining 9 to reveal goats. Since only 8 contain goats, he must open exactly those 8.

Disagree?

you are too involved with what the host does, and as a math problem, there's only two sides of an equation, and he's not on your side

to you, the two doors you did not pick the first time look exactly the same, you do not know if you picked the right one or wrong one the first time around, so to you, it looks exactly the same if the host opened either of the other two doors

Are you saying you disagree with point B or not?

you are listening to me, but not hearing me

i'm saying all 4 of your steps are irrelevent to the problem

They're not irrelevant, and I'll show you why if you simply follow along.

Now do you disagree or agree with point B?

why does it matter if i disagree or agree if i don't think it's relevent?

here's the scenario that's simplified:

you are given a choice of X, Y, Y

it does not matter if you pick X or Y, one of the two Ys will be eliminated and you are given the chance to pick all over again

now what's your chance of getting an X?

[Blarg]

Isn't the fact that monty is NOT able to reveal the door you chose in step 1 (because he has to reveal one of the other 2) proof that step one still matters when it comes time to choose to change or not. Therefore it isnt simply a 50/50 shot anymore because the door thats revealed has been affected by step 1.

[GRT]

Isn't the fact that monty is NOT able to reveal the door you chose in step 1 (because he has to reveal one of the other 2) proof that step one still matters when it comes time to choose to change or not. Therefore it isnt simply a 50/50 shot anymore because the choice has been affected by step 1.

and i'm saying that is the illusion, and it doesn't matter in the math world

He will be able to take away a Y in the lineup of X, Y, Y no matter what you picked the first time

[aurik]

They're not irrelevant, and I'll show you why if you simply follow along.

Now do you disagree or agree with point B?

why does it matter if i disagree or agree if i don't think it's relevent?

It does matter, because I'm going to prove that your "proof" is necessarily flawed. Now follow along. Do you agree or disagree with statement B:

B) If your initial choice was incorrect (90%), Monty will open 8 of the remaining 9 to reveal goats. Since only 8 contain goats, he must open exactly those 8.

[BarthelloSylph]

GRT, what if I said you were wrong, and all the mathmaticians who have calculated this problem before were right. How would you feel about that?

[GRT]

They're not irrelevant, and I'll show you why if you simply follow along.

Now do you disagree or agree with point B?

why does it matter if i disagree or agree if i don't think it's relevent?

It does matter, because I'm going to prove that your "proof" is necessarily flawed. Now follow along. Do you agree or disagree with statement B:

B) If your initial choice was incorrect (90%), Monty will open 8 of the remaining 9 to reveal goats. Since only 8 contain goats, he must open exactly those 8.

disagree, because my initial choice was not "incorrect" because the result is yet to be revealed

[GRT]

GRT, what if I said you were wrong, and all the mathmaticians who have calculated this problem before were right. How would you feel about that?

I would feel proud that I'm the only mathmatician who got it right, because only one person said earth was a globe at one time in history.