STRANGE MetaSolution
Solution and Meta-Solution
White
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Solution:
The obvious answer is 7.
At the 7th meal each of the 7 reasons as follows:
If there's only the six BEM that I see, at the 6th meal they would have reasoned as follows and not shown up today:
I see 5 BEM. If I'm not Blue Eyed they should not be here today having reasoned as follows at the 5th meal:
I see 4 BEM. If I'm not Blue Eyed they should not be here today having reasoned as follows at the 4th meal:
I see 3 Bem. If I'm not Blue Eyed they should not be here today having reasoned as follows at the 3rd meal:
I see 2 BEM. If I'm not Blued Eyed they should not be here today having reasoned as follows at the 2nd meal:
I see another BEM. If I'm not Blue Eyed he should not be here having concluded he had blue eyes at the first meal.
MetaSolution:
But if there are 7 BEM they already know there are some BEMs before the stranger comes and tells them so. The stranger has provided no additional information, so why didn't the Monks figure this out long ago?
Answer: Before the stranger makes the announcement the string of logic cannot make the last step. A single BEM cannot conclude he has Blue Eyes until the stranger comes and announces the fact.
This is STRANGE because with 7 actual BEM's it's clear to them all that noone could possibly think he's a lone BEM, yet for the string of logic to work the information available to the Monks must be such that IF there were a Lone BEM he could conclude he was blue eyed after one meal.
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End White
I think I need to see more examples of such "Strangeness" before my mind quits rebelling against this notion.
PairTheBoard
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