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Title: SINGLE-FILE HAT EXECUTION 2 Post by alien2 on Nov 25th, 2011, 1:35pm Everything is the same except that each prisoner will get a hat of a different color. The ten hats will be randomly chosen from eleven different hats. The eleven colors of these hats are mentioned to the prisoners. They are white, black, blue, yellow, green, red, orange, pink, brown, gray and purple. So what’s the optimal plan? How many lives can be saved for certain? Instead of eleven colors, all hats could be white, each with a different black number ranging from 1 to 11. |
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Title: Re: SINGLE-FILE HAT EXECUTION 2 Post by rmsgrey on Nov 25th, 2011, 3:07pm on 11/25/11 at 13:35:04, alien2 wrote:
So the guy at the back knows his hat is one of just two colours/numbers? I'm not sure whether or not there's a system that would allow the guy at the back to keep his 50% chance of success and still save everyone else, but it's definitely possible to guarantee saving everyone else. |
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Title: Re: SINGLE-FILE HAT EXECUTION 2 Post by alien2 on Nov 27th, 2011, 4:49am on 11/25/11 at 15:07:27, rmsgrey wrote:
That’s right. on 11/25/11 at 15:07:27, rmsgrey wrote:
How? Explain. |
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Title: Re: SINGLE-FILE HAT EXECUTION 2 Post by towr on Nov 27th, 2011, 6:59am on 11/27/11 at 04:49:09, alien2 wrote:
*) for summing modulo 11 it's more convenient if the numbering goes from 0 to 10 |
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Title: Re: SINGLE-FILE HAT EXECUTION 2 Post by alien2 on Nov 27th, 2011, 7:33am Man. I was sure that the answer is 6 out of 10. Delete this topic if you think that the riddle is too simple. I don’t mind. |
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Title: Re: SINGLE-FILE HAT EXECUTION 2 Post by rmsgrey on Nov 27th, 2011, 8:41am The question I haven't resolved is whether there's a suitable coding so that the number/colour the guy at the back should say according to the system is always going to be one of the two he can't see in front of him. It looks like it could be possible - with no agreed system, each person has a choice of two hat colours, one of which is always their own - no-one's going to name a colour they can see, so no-one's going to repeat an earlier colour, so each person will have seen or heard nine different colours by the time they need to guess. I haven't found a system that will work, nor a construction to produce one, but I haven't found a proof that it's impossible either. |
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