your avatar is cool

hehe

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writing lyrics aint what it used to be

after only a small amount of thinking i'd say you'd stay with the 2
trying to reason this out
at the start you have a 2/5 chance of guessing it correctly (40%)
upon selecting those two, he removes two incorrect answers
so you can either stay with your two or change to the other one
staying with your two keeps your odds, while changing to the 1 gives you 33%, right?
i feel like i'm doing something wrong here

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http://www.youtube.com/mezzir

quote:

Originally posted by mezzir after only a small amount of thinking i'd say you'd stay with the 2 trying to reason this out at the start you have a 2/5 chance of guessing it correctly (40%) upon selecting those two, he removes two incorrect answers so you can either stay with your two or change to the other one staying with your two keeps your odds, while changing to the 1 gives you 33%, right? i feel like i'm doing something wrong here

you are on the right way (about 2/5 and that staying with your first choice still has the same odds). however, you know that sum of odds always has to equal to just one number, which is...

think of it like that: what if you had 1 million doors, and you chose 2 of them and then the host opens 999 997 other ones, and leaves 1 remaining?

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writing lyrics aint what it used to be

quote:

Originally posted by Jocker you are on the right way (about 2/5 and that staying with your first choice still has the same odds). however, you know that sum of odds always has to equal to just one number, which is... think of it like that: what if you had 1 million doors, and you chose 2 of them and then the host opens 999 997 other ones, and leaves 1 remaining?

fuck good point
cool

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you should stay with the two you chose because he opening the two other doors has nothing to do with you getting it right, they're not connected, its whats called the gambler's fallacy . if you change options your probability of winning doesnt really increase.

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quote:

Orbax At that point you kind of crossed the rubicon and you might as well lay siege to Rome

the probability of choosing one of the doors with the car in the begining is 2/5, the probability of choosing one of the doors with the car after the 2 door reveal is 2/3, obviously, you should switch one of your picks to the untouched door.

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quote:

Originally posted by Psy-T the probability of choosing one of the doors with the car in the begining is 2/5, the probability of choosing one of the doors with the car after the 2 door reveal is 2/3, obviously, you should switch one of your picks to the untouched door.

why? its the same as not changing the door

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Poetry>Byron//Blog>TheMean

quote:

Orbax At that point you kind of crossed the rubicon and you might as well lay siege to Rome

quote:

Originally posted by venomX why? its the same as not changing the door

let's blow up the numbers a bit, and do a pretend simulation

we got 52 cards, the ace of diamonds is considered the car
the computer deals you two cards, these are the doors you picked (it doesn't matter whether you actually choose in the begining or are 'granted' two doors randomly, does it?).
the probability that you're holding the ace of diamonds is 2/52 at this point. the probability that the dealer is holding the ace of diamonds is 50/52.
the computer proceeds to reveal 49 of its cards.

now suppose you were a bystander to this game, the main player gets kicked out and you get the choice of the final two cards, one of which could earn you a car. if you pick the same two cards the original player held, you're clearly a dolt since they still have the same probability they had earlier, 2/52. pick a combination of the last remaining card in the computer's deck + an arbitrary one of the two the original player held and you have a probability of 51/52 to win the car.

makes sense?


edit: on a second look it appears i made a bit of a mess in my explanation, but since i originally thought it'd be conductive to understanding it, i'll let it stay (it doesn't obscure the answer really, it's just a convulted way to explain it).

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People who own my ass: Citric Acid, Boomer187, Tribu, Sand Leaper,
Jackson, venomX, jamie, Renegade, Konjin, Akridrot, Miss Bliss.
Psy-T - Down The Rabbit Hole (400minute long acid set)

Last edited by Psy-T on Oct-23-2006 at 22:05

quote:

Originally posted by Psy-T let's blow up the numbers a bit, and do a pretend simulation we got 52 cards, the ace of diamonds is considered the car the computer deals you two cards, these are the doors you picked (it doesn't matter whether you actually choose in the begining or are 'granted' two doors randomly, does it?). the probability that you're holding the ace of diamonds is 2/52 at this point. the probability that the dealer is holding the ace of diamonds is 50/52. the computer proceeds to reveal 49 of its cards. now suppose you were a bystander to this game, the main player gets kicked out and you get the choice of the final two cards, one of which could earn you a car. if you pick the same two cards the original player held, you're clearly a dolt since they still have the same probability they had earlier, 2/52. pick a combination of the last remaining card in the computer's deck + an arbitrary one of the two the original player held and you have a probability of 51/52 to win the car. makes sense? edit: on a second look it appears i made a bit of a mess in my explanation, but since i originally thought it'd be conductive to understanding it, i'll let it stay (it doesn't obscure the answer really, it's just a convulted way to explain it).

yeah but to translate that to this example you would need to be able to tell that the two you choose at the beginning were wrong, which you dont get to know, so if you dont know that the two ones you chose were wrong you still have the same posibilities of winning if you remained with them or if you changed.

edit: i understand what youre getting at, i just think this example is different because you get not further information abt the 2 doors you already chose, just abt the ones you didnt choose.

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Poetry>Byron//Blog>TheMean

quote:

Orbax At that point you kind of crossed the rubicon and you might as well lay siege to Rome

quote:

Originally posted by venomX yeah but to translate that to this example you would need to be able to tell that the two you choose at the beginning were wrong, which you dont get to know, so if you dont know that the two ones you chose were wrong you still have the same posibilities of winning if you remained with them or if you changed. edit: i understand what youre getting at, i just think this example is different

5 doors, you choose 2.
the doors you chose have a probability of 2/5 to be holding the car.
the 3 doors you haven't chosen have a probability of 3/5 to be holding the car.
the host reveals 2 out of the 3 doors you haven't chosen to be empty.
your doors still have a probability of 1/5 each of holding the car, or 2/5 in total.
the remaining door you have not chosen now has a greater probability (3/5) of holding the car.

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People who own my ass: Citric Acid, Boomer187, Tribu, Sand Leaper,
Jackson, venomX, jamie, Renegade, Konjin, Akridrot, Miss Bliss.
Psy-T - Down The Rabbit Hole (400minute long acid set)