The Monty Hall Problem

[Xfaustx]

i will throw one more idea out. just to see what anyone thinks. its prolly wrong but bear with me. what if you broke the problem up into 2 problems. the beginning and end... 1st part would be 1/3 chance we can agree on. suppose now the end is a different problem and there was never this 3rd door and now is just 2 doors. you now have 2/6 and 3/6 answers. would there be anyway to work this?

uhhhh what?


Kids, drugs are bad O.K.?



Unless you buy them from me

how much?

[BarthelloSylph]

I'm glad you see why 2/3 is right. Unfortunately that means that 50% side is illogical, not logical. You have not shown us that the incorrect, illogical answer is correct or logical.

Agreeing with both sides is mutually exclusive and makes us wonder if you really understand the 2/3s thing at all. Or if you are just tired of being ragged on.

PS This is the Monty thread.

DAMNIT thought i was in Nyna thread. whoops.

the math for the 2/3 is simple. theres nothing hard about it. but when people are all answering different questions. theres reasoning behind each answer. someone needs to change the topic title to the specific question that is being answered.

i will throw one more idea out. just to see what anyone thinks. its prolly wrong but bear with me. what if you broke the problem up into 2 problems. the beginning and end... 1st part would be 1/3 chance we can agree on. suppose now the end is a different problem and there was never this 3rd door and now is just 2 doors. you now have 2/6 and 3/6 answers. would there be anyway to work this?

Not sure what the hell is going on at the end of your post.

The thing is, the way the question is worded, there's only one right answer. There is no two problems. There is no first or second step. The problem starts once Monty shows you the goat. The question is "Switch or stay". And the correct answer is switch, cause you'll win 2/3s of the time, and you will lose 1/3 of the time.

That's just the right answer to the Monty Hall question. Not sure if you are trying to make up another one or just type mumbo jumbo but there's no other solution. I see what you are saying - that there are 2 choices in your mind - but there arent. There are three choices, A B C. And by switching, you win 66% of the time (2/3s) and lose 33% (1/3). The only time you lose by switching is if the car was in the original door you picked - a 1/3 chance. 2/3s of the time, the car will be in one of the TWO DOORS you didn't pick. The odds dont change cause you see a goat. It's math baby.

[Xfaustx]

what if i kill this goat?



it was a lil mumbo jumbo, but maybe thinking of a new formula for both parts.

[aurik]

what if you get banned before you can

[Xfaustx]

what if you get banned before you can

banned for what? lol thats like the 5th mod threat of banning from you because apparently you cant take jokes. BG forums is serious buisiness. wow

[Insanecyclone]

i will throw one more idea out. just to see what anyone thinks. its prolly wrong but bear with me. what if you broke the problem up into 2 problems. the beginning and end... 1st part would be 1/3 chance we can agree on. suppose now the end is a different problem and there was never this 3rd door and now is just 2 doors. you now have 2/6 and 3/6 answers. would there be anyway to work this?

uhhhh what?


Kids, drugs are bad O.K.?



Unless you buy them from me

how much?

Drugs are bad..m'mkay?
http://img105.imageshack.us/img105/6236/mrmackeytu7.gif

[Millael]

I really, really hate to quotetrain, but I feel it is necessary at this point in time.

Can we just ban XfaustX and be done with it?

[Jeryhn]

and there can be 2 answers cuz jesus said so.

i never said 2/3 was wrong, nor 1/2, i support both

i was just supporting the 50% side. never did i say 2/3 was wrong

I didn't know Joe Liberman lurked BG.

[GRT]

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

[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.

[BarthelloSylph]

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?

[GRT]

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

[aurik]

Yet ANOTHER way to think of the problem:

You choose a door from N doors.
Then Monty asks you "do you want to stay with your door? Or do you want to switch to ALL of the remaining doors"
If you switch, he then goes and reveals ALL of the remaining doors. If the car is behind ANY one of them you win.

This is exactly the same situation as the original game.

[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

[BarthelloSylph]

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

I highlighted in bold the part of your response that I found the most wrong. But don't feel alone - this is the part most commonly missed.

Remember - this is a math problem - there is only one right answer - and its 2/3s.

[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.

[BarthelloSylph]

GRT - by your logic, once the 98 doors are opened, there is a 50/50 chance that the car will be behind the door I chose initially.

And I am telling you that that's not true. And I'm willing to bet my life savings on the fact that, if you switch, you will find the door you switched to has the car over 2/3s of the time (in my 100 door example).

Every dime.

Do you understand this problem enough to bet that?

[aurik]

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

[GRT]

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

[BarthelloSylph]

aurik your post may have been the easiest to understand of all the posts. if these morons cant get it after reading that, there's no hope.