Suppose there are two people. The first person sees color 1, so there are two colors left: 2 and 3. Out of two permutations 231 and 321, the first one is even. So the first person assigns color 2 to the ghost, and announces 3. Now the from person heard 3, so there are two permutations to choose from: 132 and 231, The second one is even, so he announces 1.

]]>Pardon my poor English.

Congratulations on your excelent and highly entertaining (and thus EDUCATIONAL) blog!

It is really a joy for me to read your thoughts.

There is something I don’t really get about the hat puzzle, regarding the even permutation solution at the end. (of course ,this is just poor understanding by myself, and has nothing to do with your wonderfull paper!)

Is the permutation (1,2,…,101) not always even? I mean whatever number from (1 to 101)let be asigned to the ghost-sage,thus changing the pair:ghost-“heroe” , what difference does it make for the futurous decisions of the other sages?

Perhaps, you could elaborate (numerically) the method with let’s say 10 (or 8) sages (+ 1 ghost=11)?

(Or anybody else ,if she could devote some time to a stupid’s (that’s me!:-) )question)

Thanks,and keep up the good work! ]]>

They guess blindly.

]]>I like this puzzle very much! Thank you for sharing it!

But I have a question:

Where, in the “black/white” version of the puzzle, does it state that there is close to 1 : 1 proportion of black to white hats?

“The sultan puts either a black or a white hat on each sage.”

Odd or even counts do not mean anything if I have a much higher number of white hats as compared to black hats, for instance. Isn’t this so?

Considering this, I like the “color” version better, because there you have this information:

“The sultan has 101 hats, each of a different color, and the sages know all the colors.”

Kind regards,

Sara

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