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For Spun Silver -- Was Thomas Aquinas wrong?
- Thread starter Mathman
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- Feb 17, 2007
The solution is very subtle. To understand it, let’s change it a little and put yourself in the position on one of the applicants. You are led to a room, blindfolded, and a hat is placed on your head. When the blindfold comes off, you look at your 2 adversaries, and you see that one is wearing a RED hat and the other has a BLACK hat. (This is different than the actual problem, but will help explain the solution.) At this point, you don’t know what you are wearing.
So the boss instructs, “Raise your hand if you see at least one black hat on the other 2 guys.” All 3 of you raise your hand because all of you see at least one black hat.
OK, now stay with me. In our alternate scenario here, one of your adversaries has a red hat, the other has black. All 3 of you see at least one black hat. What color is your hat?
So the boss instructs, “Raise your hand if you see at least one black hat on the other 2 guys.” All 3 of you raise your hand because all of you see at least one black hat.
OK, now stay with me. In our alternate scenario here, one of your adversaries has a red hat, the other has black. All 3 of you see at least one black hat. What color is your hat?
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- Jun 21, 2003
I think a way around the subtlety is just to ask yourself how many red hats there are.
There can't be two or three (and everyone knows this) because then someone would see two red hats (but no one did).
Suppose there is exactly one red hat. Then each of the other two, seeing one red hat and knowing that's all there are, would know that he had on a black hat. But no one spoke up.
Therefore there are zero red hats.
There can't be two or three (and everyone knows this) because then someone would see two red hats (but no one did).
Suppose there is exactly one red hat. Then each of the other two, seeing one red hat and knowing that's all there are, would know that he had on a black hat. But no one spoke up.
Therefore there are zero red hats.
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- Feb 17, 2007
I think a way around the subtlety is just to ask yourself how many red hats there are.
There can't be two or three (and everyone knows this) because then someone would see two red hats (but no one did).
Suppose there is exactly one red hat. Then each of the other two, seeing one red hat and knowing that's all there are, would know that he had on a black hat. But no one spoke up.
Therefore there are zero red hats.
Yes, that's it. 
Basically, if there was one red hat, each guy with a black hat would see that the other guy with a black hat sees a black hat. Once a few minutes go by, one guy sees that the other 2 can't figure out what they are wearing. So he knows he must have on a black hat.
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- Jun 21, 2003
This is a clever problem.The interesting thing is that in order for it to work, the other two guys have to be sort of smart, but not too smart.
First, they have to be smart enough to know that if they saw a red hat then their's is black. But not smart enough to figure out the whole game, in exactly the way that the winner did, beating the winner to the punch.
In fact, if you played this game against two idiots, you would not know which color hat you had on. :scratch:
First, they have to be smart enough to know that if they saw a red hat then their's is black. But not smart enough to figure out the whole game, in exactly the way that the winner did, beating the winner to the punch.
In fact, if you played this game against two idiots, you would not know which color hat you had on. :scratch:
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- Jun 21, 2003
In the spirit of logical conundra which depend not only on the objective facts of the situation but also on what people in the story know about these facts, here is a genuine epistomological “paradox.” This puzzle is usually credited to W. V. Quine, professor of philosophy at Harvard. If you Google “Hangman’s Paradox” you will see all kinds of scholarly opinions (some thought-provoking, some nonsense) about it.
The Paradox of the Surprise Execution
A prisoner is sentenced to death by hanging, with these conditions. He will be hanged at noon on Wednesday, Thursday, or Friday of next week. But if he can figure out, before the hangman shows up at his cell door, which day it is going to be, then he will receive a pardon.
The prisoner reasons as follows.
“I can’t be hanged on Friday, because if Thursday passes and I am still alive then I will know that Friday must be the day, so I will have earned my pardon.
“So it can’t be Friday.
“Can it be Thursday? No, by the same reasoning. I already figured out that it can’t be Friday. So if Wednesday passes and I am still alive, then I will know that it must be Thursday. I will have figured out the day and I’m home free.
“Since it can’t be Thursday and it can’t be Friday, it must be Wednesday, since that’s the only day left.
“But now I have figured out that it must be Wednesday. Having figured this out, now I get to go free.”
Imagine the prisoner’s surprise when the hangman shows up at his cell door at noon on Thursday, rope in hand.
“Hey," he protests. "You can’t hang me on Thursday. I just figured out that you can’t!”
The hangman replies, “That’s why I can.”
The Paradox of the Surprise Execution
A prisoner is sentenced to death by hanging, with these conditions. He will be hanged at noon on Wednesday, Thursday, or Friday of next week. But if he can figure out, before the hangman shows up at his cell door, which day it is going to be, then he will receive a pardon.
The prisoner reasons as follows.
“I can’t be hanged on Friday, because if Thursday passes and I am still alive then I will know that Friday must be the day, so I will have earned my pardon.
“So it can’t be Friday.
“Can it be Thursday? No, by the same reasoning. I already figured out that it can’t be Friday. So if Wednesday passes and I am still alive, then I will know that it must be Thursday. I will have figured out the day and I’m home free.
“Since it can’t be Thursday and it can’t be Friday, it must be Wednesday, since that’s the only day left.
“But now I have figured out that it must be Wednesday. Having figured this out, now I get to go free.”
Imagine the prisoner’s surprise when the hangman shows up at his cell door at noon on Thursday, rope in hand.
“Hey," he protests. "You can’t hang me on Thursday. I just figured out that you can’t!”
The hangman replies, “That’s why I can.”
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- Feb 17, 2007
I haven’t heard of that version, but I heard something similar many years ago. A guy walks into a casino and sees a table set up for an experimental game. He goes to check it out. He is handed 2 envelopes, each containing a voucher for a certain amount of money.
Here are the rules:
He is allowed to open one, and only one, envelope. The other remains sealed until the game is over.
Each envelope contains an amount which in an increment of $100. So it might have $100. $200, $300, $400 etc, with $100 being the minimum.
The difference in the amounts between the 2 vouchers is $100, so one envelope must have exactly $100 more ( or less ) than the other.
After opening one envelope and finding out how much it contains, he has the choice of keeping it or trading it for the other envelope which remains sealed.
So the guy opens one envelope and sees a voucher for $400. He reasons as follows. They would not put a $100 voucher in an envelope, because that is the minimum. The other envelope would have to contain $200. So there’s no game there.
If $100 is disallowed, then $200 can not be used either. If somebody opened a $200 voucher with $100 disallowed, he would know the other envelope must contain a voucher for $300. There would be no game involved.
If $200 is disallowed, then $300 can not be used either. If somebody opened a $300 voucher with $200 disallowed, he would know the other envelope must contain a voucher for $400. Again, there would be no game involved.
So upon opening his $400 voucher, he decided the other envelope can not contain a $300 voucher, it must have a $500 voucher. So he traded. Was he correct?
Here are the rules:
He is allowed to open one, and only one, envelope. The other remains sealed until the game is over.
Each envelope contains an amount which in an increment of $100. So it might have $100. $200, $300, $400 etc, with $100 being the minimum.
The difference in the amounts between the 2 vouchers is $100, so one envelope must have exactly $100 more ( or less ) than the other.
After opening one envelope and finding out how much it contains, he has the choice of keeping it or trading it for the other envelope which remains sealed.
So the guy opens one envelope and sees a voucher for $400. He reasons as follows. They would not put a $100 voucher in an envelope, because that is the minimum. The other envelope would have to contain $200. So there’s no game there.
If $100 is disallowed, then $200 can not be used either. If somebody opened a $200 voucher with $100 disallowed, he would know the other envelope must contain a voucher for $300. There would be no game involved.
If $200 is disallowed, then $300 can not be used either. If somebody opened a $300 voucher with $200 disallowed, he would know the other envelope must contain a voucher for $400. Again, there would be no game involved.
So upon opening his $400 voucher, he decided the other envelope can not contain a $300 voucher, it must have a $500 voucher. So he traded. Was he correct?
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I would like to gamble all my money on the fact that I would like to excute this thread!!!
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I think PolymerBob and Math are having just tooooo much fun.I would like to gamble all my money on the fact that I would like to excute this thread!!!
Dee
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- Oct 27, 2006
I think that polymer and mathman have no idea what they are talking about. I think they are making stuff up and since we don't know what they are talking about either...we have no idea...so they seem really smart.
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