EPTWho is the smartest poster at TL.net? - Page 9
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Liquid`Drone
Norway28784 Posts
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Frits
11782 Posts
The removing of the other doors and leaving 1 doesn't chance a thing, it stays the original chance. How the hell could you just say that after reading the bold part? HAHAHAHAHA didn't read that last part yet, i think he's just smart and fooling us all. At least I hope so for his sake. | ||
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Jamers
Israel1327 Posts
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YoUr_KiLLeR
United States3420 Posts
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Liquid`Drone
Norway28784 Posts
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Jamers
Israel1327 Posts
On May 04 2005 18:31 YoUr_KiLLeR wrote: if the door you picked keeps its original probability, why doesn't the door that is left keep its probability as well? Because the person removing the cards KNOWNINGLY chose NOT to remove it, since the person removing the cards knew none of them were an ace of spades. So, he either chose not to remove it because it IS the ace of spades (99/100 times), or he didn't remove it because you already picked the ace of spades yourself (1/100 times). edit: Ermm you asked about doors, so just same thing but about doors. | ||
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YoUr_KiLLeR
United States3420 Posts
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Frits
11782 Posts
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ihatett
United States2289 Posts
On May 04 2005 18:35 YoUr_KiLLeR wrote: and to get back on topic, i would say cyric is one of the more intelligent posters. if not, at the least he is very articulate. lol edit: at the subject change | ||
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imRadu
1798 Posts
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YoUr_KiLLeR
United States3420 Posts
On May 04 2005 18:33 Liquid`Drone wrote: because you know that one of the two doors is the correct one.. if you didn't know that one of the two doors is the correct one, then the other door would have an equal chance of being correct. (however, a huge majority of the time, they would both be wrong. ) if you know that one of the two doors is correct, nothing gives either door a higher chance of being correct than the other. | ||
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-_-
United States7081 Posts
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XaI)CyRiC
United States4471 Posts
For anyone who is too lazy to look up the link, the scenario is basically three prisoners on deathrow, one of whom will be pardoned at random. Then, one of the prisoners (A) convinces the warden to reveal one of the prisoners who will not be pardoned (C). According to the addendum, the probability of A being the one pardoned is still 1/3, while the other prisoner (B) has his odds go up to 2/3. Is this just a matter of perception, because it would seem that from B's perspective he would be the 1/3 and A would be the 2/3? My brain fails me, please enlighten. (EDIT: Fixed typo) | ||
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ihatett
United States2289 Posts
On May 04 2005 18:39 YoUr_KiLLeR wrote: if you know that one of the two doors is correct, nothing gives either door a higher chance of being correct than the other. holy fucking god Since we have thrown at you all of the logic we can (to the point that only a dumbass wouldn't get it) try it for yourself like someone said. Get a deck of cards, and try to choose the ace of spades. Then, look through the rest of the deck, if the ace of spades is in there, pick it out, otherwise pick any other card. Tell me which card is the ace of spades more often, your original card or the second one. | ||
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imRadu
1798 Posts
there are 100 doors, behind one of them is a prize. You pick a door . The host then opens 98 other doors showing you there is nothing behind them. There are 2 closed doors left. The one that you picked and the one the host left closed. The host now asks you: Are you sure you wanna stay with the door you initially picked or will you pick the other one? Correct answer: Pick the other door. End. | ||
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MoltkeWarding
5195 Posts
On May 04 2005 18:41 XaI)CyRiC wrote: I'm probably going to reveal my lack of intelligence here, but I am curious about the Addendum #5 in the link given for this probability problem. I've been convinced as to the actual Monty Hall problem with the three doors already btw, this picture helped. For anyone who is too lazy to look up the link, the scenario is basically three prisoners on deathrow, one of whom will be pardoned at random. Then, one of the prisoners (A) convinces the warden to reveal one of the prisoners who will not be pardoned (C). According to the addendum, the probability of A being the one pardoned is still 1/3, while the other prisoner (B) has his odds go up to 2/3. Is this just a matter of perception, because it would seem that from C's perspective he would be the 1/3 and A would be the 2/3? My brain fails me, please enlighten. Er...if you're talking about the probability of A being pardoned after its revealed C is not being pardoned, the probability of A being pardoned is now 1/2. | ||
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baal
10678 Posts
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XaI)CyRiC
United States4471 Posts
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FrEaK[S.sIR]
2373 Posts
On May 04 2005 18:41 XaI)CyRiC wrote: I'm probably going to reveal my lack of intelligence here, but I am curious about the Addendum #5 in the link given for this probability problem. I've been convinced as to the actual Monty Hall problem with the three doors already btw, this picture helped. For anyone who is too lazy to look up the link, the scenario is basically three prisoners on deathrow, one of whom will be pardoned at random. Then, one of the prisoners (A) convinces the warden to reveal one of the prisoners who will not be pardoned (C). According to the addendum, the probability of A being the one pardoned is still 1/3, while the other prisoner (B) has his odds go up to 2/3. Is this just a matter of perception, because it would seem that from C's perspective he would be the 1/3 and A would be the 2/3? My brain fails me, please enlighten. Why would C's perspective be different? Nobody was telling C anything, only A. The only perspective that would matter is A because he was the one who was talking to the warden. The other perspectives are non-existant. | ||
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baal
10678 Posts
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