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Server Time: 2/10/2012 10:31:13 PM PACIFIC |
 One more hat problem, vanagon40, 8. May 2003 22:58 | ||
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| Seven guys in prison. The warden offers a chance at release, if the prisoners can correctly guess the color of the hat on his own head. The hat is either black or white, and prisoners cannot see their own hats. The prisoners are lined up front to back, single file. The prisoner in back can see the other six hats, the next only five, etc., and the guy in front can see no hats. The prisoner in back guesses first, then down the line, with the prisoner in front guessing last. Each prisoner that guesses his own hat color is released. What is the optimum system? | ||
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| Re: One more hat problem, vanagon40, 8. May 2003 23:01 | ||
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| The prisoners can devise a plan prior to the "contest." However, once in line, the prisoners cannot signal in any manner and must state only "black" or "white." | ||
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| Re: One more hat problem, stdioh, 9. May 2003 08:37 | ||
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| The game is about partial information. The guy at the back has a 50/50 chance nomatter what colour he picks, so he uses his guess to tell the guy in front what colour his hat is. Now, prisoners in even numbered spaces are all 100% correct and prisoners in odd numbered spaces each have a 50% chance. Thus 5/7 should get released. I'm not sure if there is a better solution to this one yet. | ||
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| Re: One more hat problem, Wren, 9. May 2003 10:28 | ||
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| I can't figure out anything better than this so far. Should we keep on trying, vanagon? :O) | ||
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| Re: One more hat problem, stdioh, 9. May 2003 08:44 | ||
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| Oh, I see now that the guy in the back can see all of the other hats...that changes things. ok... The guy in the back says white if there are an uneven number of hats. There will be either 2 white and 4 black, 1 white and 5 black, 0 white and 6 black, or without loss of generallity, the other combinations. Now he has a 50% chance nomatter what he guesses. Now the guy in seat 6 looks and sees an uneven number of hats. If number 7 had told him even (black) then he would know which hat he was wearing. Likewise, if number seven had told him uneven he also has enough information to know what colour his hat is. He always goes free. Now, based on his decision, everybody knows what colour his hat was, so the next guy in line looks at the hats remaining and includes the hat of the man who went free in his tally, then does exactly what the man who went free did and also gets his colour right. Thus everybody goes free except the first to speak, who has a 50% chance and 6.5/7 are saved. | ||
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| Re: One more hat problem, Wren, 9. May 2003 10:22 | ||
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| "Likewise, if number seven had told him uneven he also has enough information to know what colour his hat is." No he doesn't. Say seat 7 looks and sees 2 white and 4 black hats. He says "white", because he sees an uneven number. Now seat 6, wearing a white hat, looks and sees 1 white and 4 black hats. He doesn't know whether he's wearing a black hat (to make 1 white and 5 black) or a white hat (to make 2 white and 4 black). I'm still stumbling over this one. | ||
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| Re: One more hat problem, stdioh, 12. May 2003 10:00 | ||
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| I should clarify that. The information he gives isn't whether there are the same number of white and black. He gives the parity of white vs. black....if there is an even number of each or an odd number of each. | ||
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| Re: One more hat problem, shorn, 9. May 2003 11:23 | ||
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| No enough info. to determine it because there was no stipulation as to the distribution of hats. No matter what each guy sees, each has the same 50/50 shot of getting it right. | ||
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| Re: One more hat problem, vanagon40, 12. May 2003 09:16 | ||
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| Sorry, almost forgot about the answer. Stdioh had it right. The last prisoner sees all hats except his and states Black if the number of Black hats is odd, and White if the number of Black hats is even (or zero). He has a 50% chance himself, but saves everyone else. The next prisoner knows whether the number of Black hats is even or odd (expect for the prisoner who already guessed, who is now irrelevant), and can see all the remaining hats except his. He can correctly determine the color of his hat. The next prisoner also knows whether the number of Black hats is even or odd, he knows the color of the hat of the prisoner directly behind him and in front of him, and can correctly determine the color of his hat. Etc.,etc., ..... | ||
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| Re: One more hat problem, shorn, 12. May 2003 09:26 | ||
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| Why does the second prisoner know the number of Black hats? There was no discussion as to the distribution of hat color. I don't see how you can state your solution as such with the information given. | ||
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| Re: One more hat problem, stdioh, 12. May 2003 09:59 | ||
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| The guy in the back has a 50/50 chance...the then can see how many of the other hats are black. He indicates the parity by his choice. Now the next guy looks and sees how many hats are black in front of him and if the parity has changed then he knows his hat is black...if it has not then he knows his hat is white. Now if he choses black, everybody changes their parity that they are storing (as was indicated by the last guy) and if he says white then they do not change their parity. It continues until everybody is free. | ||
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