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An updated Monty Show paradox for all you COR brainiacs (pg. 2)
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| Psy-T |
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)
.... |
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| tubularbills |
| numbers and odds are great and all. but you could have the tiest luck in the world, and get screwed over. sounds like something that would happen to me. haha |
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| venomX |
| 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. |
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| Psy-T |
| 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 :p
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). :p
lol |
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| Psy-T |
| quote: | Originally posted by tubularbills
numbers and odds are great and all. but you could have the tiest luck in the world, and get screwed over. sounds like something that would happen to me. haha |
there's no such thing as luck :p |
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| 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
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 |
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| venomX |
| quote: | Originally posted by Psy-T
see, but i do :p
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). :p
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 :wtf:
edit: damn wikipedia im gonna have to go read this thing again |
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| venomX |
| 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
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the picture without a discreption doesnt say much :p |
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| mezzir |
| quote: | Originally posted by venomX
the picture without a discreption doesnt say much :p |
TR00F! |
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| Akridrot |
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!)
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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 |
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| mezzir |
| quote: | Originally posted by Akridrot
Better problem:
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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 |
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| Psy-T |
| 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 :wtf: |
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. |
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