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An updated Monty Show paradox for all you COR brainiacs

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)

your avatar is cool

hehe

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

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?

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

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.

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.

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

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).

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.

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.

door 1 (1/5)
door 2 (1/5)
door 3 (1/5)
door 4 (1/5)
door 5 (1/5)

also viewable as

doors 1&2 (2/5)
doors 3&4&5 (3/5)

you pick doors 1 & 2, giving you a 2/5 chance to win
the doors you havent chosen are more likely to contain the car (3/5)

the host opens doors 3 & 4 which are empty, which leads us to:

doors 1&2 (2/5)
doors 3&4&5 (3/5)

also viewable as:

door 1 (1/5)
door 2 (1/5)
door 3 (0/5)
door 4 (0/5)
door 5 (3/5)


....

numbers and odds are great and all. but you could have the shittiest luck in the world, and get screwed over. sounds like something that would happen to me. haha

quote:

Originally posted by Psy-T door 1 (1/5) door 2 (1/5) door 3 (1/5) door 4 (1/5) door 5 (1/5) you pick doors 1 & 2, giving you a 2/5 chance to win the doors you havent chosen are more likely to contain the car (3/5) the host opens doors 3 & 4 which are empty, which leads us to: door 1 (1/5) door 2 (1/5) door 3 (0/5) door 4 (0/5) door 5 (3/5) ....

i dont believe so, after eliminating door 3 & 4 you would have:

door 1 (1/3)[because there are two doors that are not here anymore]
door 2 (1/3)
door 3 (1/3)

your argument presupposes that somehow you derive some information from the other doors being opened, which you dont really. Also every door has the same probablity, even from the beginning, each has 1/5, no door has a higher probability than the rest. this method is used to trick people usually cuz our brains are not very good at probabilistic reasoning (mine included of course). ill follow your argument a bit more:
you pick 2 doors, chances of winning 2/5, chances of losing 3/5 but that doesnt mean any doors up to now have any more chances of being right than another. you get this information, two doors dont have the car, ie. #3 and #4. now you still have to doors, chances of winning 2/3 (theres only 3 available doors now), chances of losing 1/3. the doors still have the same chances of being right/wrong. so if you change doors you still have the same 1/3 chance for each door which renders the effort of changing doors futile because its all chance and you have no real control over the result.

quote:

Originally posted by venomX your argument presupposes that somehow you derive some information from the other doors being opened, which you dont really.

see, but i do

i can't explain beyond what i already did, so just trust me on this and try to simulate it with 5 cards (preferably automatically via a computer somehow, but manually would do).

lol

quote:

Originally posted by tubularbills numbers and odds are great and all. but you could have the shittiest luck in the world, and get screwed over. sounds like something that would happen to me. haha

there's no such thing as luck

TAKEN FROM A WIKIPEDIA ARTICLE ON THIS PROBLEM, NOTE THAT THIS IS ONLY WHEN DEALING WITH 3 DOORS TOTAL, THOUGH THE SAME LOGIC STILL APPLIES



EDITED FOR EXPLANATION:
SAY FOR EXAMPLE YOU CHOOSE DOOR 1
YOU HAVE A 1/3 CHANCE OF GETTING IT RIGHT, AND 1/3 EACH CHANCE OF IT BEING BEHIND DOOR 2 OR 3.
SO WHEN THE HOST OPENS DOOR 3, DOOR #1'S CHANCES DON'T CHANGE AT ALL, STILL 1/3
HOWEVER STILL DOORS 2 AND 3 COMBINED ACCOUNT FOR 2/3
WE KNOW ITS NOT BEHIND DOOR 3 SO ESSENTIALLY DOOR 2 HAS A 2/3 CHANCE, AND DOOR 1 HAS A 1/3 CHANCE, AS ILLUSTRATED BY THIS TRAGICALLY TRANSPARANT VENN DIAGRAM

quote:

Originally posted by Psy-T see, but i do i can't explain beyond what i already did, so just trust me on this and try to simulate it with 5 cards (preferably automatically via a computer somehow, but manually would do). lol

it doesnt, i just read abt this recently in my scientific methods class, its called gambler's fallacy. my book has this exact example on it

edit: damn wikipedia im gonna have to go read this thing again

quote:

Originally posted by mezzir TAKEN FROM A WIKIPEDIA ARTICLE ON THIS PROBLEM, NOTE THAT THIS IS ONLY WHEN DEALING WITH 3 DOORS TOTAL, THOUGH THE SAME LOGIC STILL APPLIES

the picture without a discreption doesnt say much

quote:

Originally posted by venomX the picture without a discreption doesnt say much

TR00F!

Better problem:

quote:

Guessing Game. The following little-known problem has a similar flavour to the previous puzzle. It was told to me many years ago and, to this day, I still find it astonishing. Alice and Bob each choose a positive integer and reveal their numbers to Xander, but not to each other. Xander then writes two positive integers on a blackboard and tells Alice and Bob that one of them is the sum of their chosen numbers. He then alternately asks Alice and Bob, �Do you know the other person�s number?� until one of them answers, �Yes�. Prove that after finitely many questions, one of them will know the other player�s number. (Of course, it is necessary to assume that Alice and Bob are completely logical as well as completely honest!)

Monty Hall problem has been done to death. Switching doors gives a better chance than not. http://en.wikipedia.org/wiki/Monty_Hall_problem/revised

quote:

Originally posted by Akridrot Better problem:

QUESTION ABOUT THAT...WHEN IT SAYS FINITELY MANY QUESTIONS...ARE ALICE AND BOB ALLOWED TO ASK QUESTIONS OR GUESS? OR DOES JUST JUST GO BACK AND FORTH WITH THEM SAYING NO UNTIL ONE SAYS YES

quote:

Originally posted by venomX it doesnt, i just read abt this recently in my scientific methods class, its called gambler's fallacy. my book has this exact example on it

it doesn't?? meaning you don't receive the information that doors 3 and 4 have a zero probability of holding the car? come on...

also, gambler's fallacy is in fact the fallacy you are falling to here:

quote:

The gambler's fallacy is a logical fallacy that mistakenly believes past events will affect future events when dealing with random activities, such as many gambling games.

you're the one who thinks that past event affect the future event by increasing your choices' probability from 2/5 to 2/3; the elimination of two doors does not increase the probability of your initial choices in being right, however, it does inherently tell us that those two doors have a zero probability of holding the car, and hence the last door has a probability of 3/5 of holding the car.

quote:

Originally posted by mezzir QUESTION ABOUT THAT...WHEN IT SAYS FINITELY MANY QUESTIONS...ARE ALICE AND BOB ALLOWED TO ASK QUESTIONS OR GUESS? OR DOES JUST JUST GO BACK AND FORTH WITH THEM SAYING NO UNTIL ONE SAYS YES

If they were allowed to ask questions and guess, it wouldn't be much of a problem would it?

In all of the variations I've seen, it's simply "BACK AND FORTH WITH THEM SAYING NO UNTIL ONE SAYS YES"

Also prove that the person who says yes KNOWS the other person's number.

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