The Monty Hall Problem

[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.

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

You already agreed to statement A:
A) The chance that your initial choice is correct is 10%

The implication of this statement is that
A.1) The chance that your initial choice is incorrect is 90%.

Disagree or agree?

[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

You already agreed to statement A:
A) The chance that your initial choice is correct is 10%

The implication of this statement is that
A.1) The chance that your initial choice is incorrect is 90%.

Disagree or agree?

disagree with A: my initial choice is never "correct" because my initial choice is never final

disagree with A.1: since my initial choice is never a final decision, you can say 0.99999999999999 never reach 1

[aurik]

[quote=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

You already agreed to statement A:
A) The chance that your initial choice is correct is 10%

The implication of this statement is that
A.1) The chance that your initial choice is incorrect is 90%.

Disagree or agree?

disagree with A: my initial choice is never "correct" because my initial choice is never final

disagree with A.1: since my initial choice is never a final decision, you can say 0.99999999999999 never reach 1 [/quote:19711]

That doesn't even make sense. You pick a door from 10; there is a probability that is correct or incorrect. If it's not 10% then what is it, and why?

[GRT]

[quote=aurik][quote=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

You already agreed to statement A:
A) The chance that your initial choice is correct is 10%

The implication of this statement is that
A.1) The chance that your initial choice is incorrect is 90%.

Disagree or agree?

disagree with A: my initial choice is never "correct" because my initial choice is never final

disagree with A.1: since my initial choice is never a final decision, you can say 0.99999999999999 never reach 1 [/quote:d94d1]

That doesn't even make sense. You pick a door from 10; there is a probability that is correct or incorrect. If it's not 10% then what is it, and why?[/quote:d94d1]

Because since you are never allowed to open the door in step A, you will never get your prize. You can also say that your probability in getting what you want is 0%.

[aurik]

That doesn't even make sense. You pick a door from 10; there is a probability that is correct or incorrect. If it's not 10% then what is it, and why?

Because since you are never allowed to open the door in step A, you will never get your prize. You can also say that your probability in getting what you want is 0%.

You're confused. Just because you don't open the door doesn't mean that the probability that it IS the correct door is undefined. The probability that your initial choice HAPPENS to be the right door exists, and is 10%.

[Tajin]

Because since you are never allowed to open the door in step A, you will never get your prize. You can also say that your probability in getting what you want is 0%.

[GRT]

That doesn't even make sense. You pick a door from 10; there is a probability that is correct or incorrect. If it's not 10% then what is it, and why?

Because since you are never allowed to open the door in step A, you will never get your prize. You can also say that your probability in getting what you want is 0%.

You're confused. Just because you don't open the door doesn't mean that the probability that it IS the correct door is undefined. The probability that your initial choice HAPPENS to be the right door exists, and is 10%.

it's not undefined, it's 0%.

What you are stuck with is that the Monty has no choice but open certain doors. The reality is, to you, the door he opens is the same as the other one.

Let me help you a bit. Imagine Monty is not a person, and the doors are not really doors, and they are not put one next to another. Imagine it's three poker cards, and after you pick out one and lay it on the table, the other two are shuffled a few times by some magic hands and one of them is taken out, the other one is layed on the table next to the one you first picked. Now you choose.

Get it yet?

[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?

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

It's only a "question of behaviors" because it's easier to see the underlying mathematical truth without the mathematical jargon as you increase the number.

[Byrd]

You might find this page interesting:

http://math.ucsd.edu/~crypto/Monty/monty.html

Or this one:

http://www.grand-illusions.com/simulator/montysim.htm

Both of these let you run your own simulations a relatively infinite number of times. You can choose whether to switch or keep your choice.

Guess what both answers come up with?

There's a ton more if you Google "The Monty Hall Problem."

[Tajin]

Let me help you a bit. Imagine Monty is not a person, and the doors are not really doors, and they are not put one next to another. Imagine it's three poker cards, and after you pick out one and lay it on the table, the other two are shuffled a few times by some magic hands and one of them is taken out, the other one is layed on the table next to the one you first picked. Now you choose.

Get it yet?

Keep going, i kind of need a new signature


Bolded part: You just fucked up the probabilities of whatever "stupid" example you we're trying to "help a bit" with.

[GRT]

http://math.ucsd.edu/~crypto/Monty/monty.html

Thanks for helping me with the proof.

--------------------# of Players------Winners----Percent Winners
Switched---------------462------------319-----------69.0
Didn't Switch----------430------------147------------34.2


add total number of players: 892

add total number of winners: 468

total estimate of percent winners: do it yourself

[GRT]

Let me help you a bit. Imagine Monty is not a person, and the doors are not really doors, and they are not put one next to another. Imagine it's three poker cards, and after you pick out one and lay it on the table, the other two are shuffled a few times by some magic hands and one of them is taken out, the other one is layed on the table next to the one you first picked. Now you choose.

Get it yet?

Keep going, i kind of need a new signature


Bolded part: You just fucked up the probabilities of whatever "stupid" example you we're trying to "help a bit" with.

moral of this response in comparison with the result? more people who are stupid/stubborn/conformist than people who aren't

(not really, that's bad math XD)

[Devek]

You might find this page interesting:

http://math.ucsd.edu/~crypto/Monty/monty.html

Or this one:

http://www.grand-illusions.com/simulator/montysim.htm

Both of these let you run your own simulations a relatively infinite number of times. You can choose whether to switch or keep your choice.

Guess what both answers come up with?

There's a ton more if you Google "The Monty Hall Problem."

I just played that.. switched each time and lost the car 10 times in a fucking row before I finally won.

I went like 0/200 on my JSE sachet too..

I fuck fuck fuck fucking hate anything with luck involved.

[Tajin]

OHHHHHHHHHHHHHHHHHH BOY HAHAHAHAHAHAHAHAH


;; i have to leave for home :D if the PC is free i will come back and see how you keep owning yourself


Otherwise i'll be AFK banging hot goats


Keep it up *'.')b

[GRT]

OHHHHHHHHHHHHHHHHHH BOY HAHAHAHAHAHAHAHAH


;; i have to leave for home :D if the PC is free i will come back and see how you keep owning yourself


Otherwise i'll be AFK banging hot goats


Keep it up *'.')b

are you really that stupid? lol

[Enkidu]

http://math.ucsd.edu/~crypto/Monty/monty.html

Thanks for helping me with the proof.

--------------------# of Players------Winners----Percent Winners
Switched---------------462------------319-----------69.0
Didn't Switch----------430------------147------------34.2


add total number of players: 892

add total number of winners: 468

total estimate of percent winners: do it yourself

I don't understand your point here. 69% of the time those who switched doors won a car, 34.2% of the time those who did not switch did not win a car. Adding the two together gives you nothing substantial.

[GRT]

http://math.ucsd.edu/~crypto/Monty/monty.html

Thanks for helping me with the proof.

--------------------# of Players------Winners----Percent Winners
Switched---------------462------------319-----------69.0
Didn't Switch----------430------------147------------34.2


add total number of players: 892

add total number of winners: 468

total estimate of percent winners: do it yourself

I don't understand your point here. 69% of the time those who switched doors won a car, 34.2% of the time those who did not switch did not win a car. Adding the two together gives you nothing substantial.

is it really that hard to understand? i remember doing IQ test as a kid i was only average

anyway, i'll explain. the original question is what's your chance of winning a car if monty eliminates the one goat door after your initial pick... i thought that the fact that he stops you and let you pick all over again nullifies the significance of your first pick so the overall probability of your chance of winning is the same as your second pick, which is 50%

which is proven by the UCSD statistics, 468/892 is roughly 50%


also, to the people who thought that it is significant that monty could not open the door you first picked... i say you are right in the study of behaviors, but not the study of math

because when you don't switch, it doesn't matter that monty opened any of the doors, so your chance remains 33%. but if you do switch, that's the same thing as if you are allowed to open both of the other doors at the same time, which is of course a 66% chance of winning (so it still does not matter which of the two doors monty opens)

[BarthelloSylph]

GRT those are some of the dumbest posts I've ever read on any message board ever.

If you really think you are breaking the boundaries, and discovering some new mathmatical shit, then you are really just a sad excuse for moron.

But all that aside GRT, I think what aurik is getting at is this. And I'll try to speak from the heart here.

-------------------------------------

Begin initial phase. GRT, all I am gonna say is - put yourself in the shoes of the contestant. There are three doors, 2 have a goat behind them and 1 has a car. You pick a door to start off, and then the gameshow host will reveal a goat behind one of the two doors you did not. End initial phase.

Begin decision time. GRT, let's say you have a choice. Stick with your current door, or switch doors. Let's say you have to pick one or the other You can't say "it doesnt matter", you have to pick... which is it?

[aurik]

That doesn't even make sense. You pick a door from 10; there is a probability that is correct or incorrect. If it's not 10% then what is it, and why?

Because since you are never allowed to open the door in step A, you will never get your prize. You can also say that your probability in getting what you want is 0%.

You're confused. Just because you don't open the door doesn't mean that the probability that it IS the correct door is undefined. The probability that your initial choice HAPPENS to be the right door exists, and is 10%.

it's not undefined, it's 0%.

you're saying that the chance to guess the correct door off the start is 0%?

[Enkidu]

is it really that hard to understand? i remember doing IQ test as a kid i was only average

Well, that's nice.

anyway, i'll explain. the original question is what's your chance of winning a car if monty eliminates the one goat door after your initial pick... i thought that the fact that he stops you and let you pick all over again nullifies the significance of your first pick so the overall probability of your chance of winning is the same as your second pick, which is 50%

which is proven by the UCSD statistics, 468/892 is roughly 50%

Like I said, adding the two yields nothing substantial. The chance of winning upon switching is always 66.66...%. The chance of winning without switching is always 33.33...%. There is never a 50% chance of winning.

You're taking a study and changing the results to better suit your argument. Change the number of players who switched to 10,000 and from the winning percent calculate the number of winners, now add the winners from both switching and not switching. Of course now the percentage is higher. Like I said, nothing substantial comes from adding those two quantities.

also, to the people who thought that it is significant that monty could not open the door you first picked... i say you are right in the study of behaviors, but not the study of math

because when you don't switch, it doesn't matter that monty opened any of the doors, so your chance remains 33%. but if you do switch, that's the same thing as if you are allowed to open both of the other doors at the same time, which is of course a 66% chance of winning (so it still does not matter which of the two doors monty opens)

No. If Monty opened the door you chose, you would be left with less information then if he opens another door. (EDIT: assuming the door you chose was a goat, of course)