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An updated Monty Show paradox for all you COR brainiacs
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| Jocker |
So, some people probably know about the Monty show phenomenon. I have updated the problem, so now it seems even more paradoxical (is there such a word?:)
So, there are 5 doors. Behind one of them is a car, behind other four - nothing (or, in original, goats:)). So you are allowed to pick 2 (TWO) doors, and if the cars is behind ANY of those doors, you win.
But before you open those doors, the game show host opens two other doors, which have no car behind them (he knows where the car is).
So, you end up with 2 closed doors you originally chose and 1 other closed door remaining. The host gives you an option to switch from your original choice of those 2 doors to this last 1 remaining. What should you do?
(Well, the term "paradox" gives you a big hint at the answer:D:D) |
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| ASFSE |
| your avatar is cool |
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| 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 :( |
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| Jocker |
| 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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| mezzir |
| 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? |
good point :p
cool :toothless |
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| venomX |
| 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 :p. if you change options your probability of winning doesnt really increase. |
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| 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. |
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| venomX |
| 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 :p |
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| Psy-T |
| quote: | Originally posted by venomX
why? its the same as not changing the door :p |
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 :p 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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| venomX |
| 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 :p 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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| Psy-T |
| 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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