Six logicians stand one behind the other facing an opaque wall such that 2 are on one side and 4 are on the other. None of the logicians can turn around or see beyond the wall.
Each wears a black or white hat as shown below; "|" represents the wall, capital letters are used to identify the logicians, and "b" and "w" refer to black and white respectively.
b w | b w b w
A B | C D E F
Each knows the location of the others and the quantity of each colour of hat. Who will be first to declare having which colour?
Hi<o:p></o:p>
I am new here (so go easy on me – “NOVICE”) found the site by accident only a few days ago ( Btw: great site!) Almost got a solution relatively quickly by meditation, a day or two ago, but could not reconstruct the logic that happened at brain speed, needed longer time to reconstruct and test. However I noticed the frustration of others with no solution being declared, if that is what happens here? )<o:p></o:p>
My solution is either that A and D declare colour simultaneously ( Black and White respectively)or A beats D to the call and declares his hat Black<o:p></o:p>
He knows that there will be on the other side of the wall either<o:p></o:p>
1. Three of black and only one of white, in such case his hat is White<o:p></o:p>
2. Only two of each Black and white colour hats, in that case he is Black<o:p></o:p>
Of the possible configurations he will not be in a position to decide whether it is option 1 or 2 until both F and E fail to declare. At the point where D is in position to call the configuration must be<o:p></o:p>
BWBW, WBWB,BWBB he is the only one to know it cannot be WBWW so he discounts it<o:p></o:p>
I suggest A discusses their unfortunate predicament with position B along the lines “ My colleague B we might be stuck here forever! if we cannot hear them call on the other side and given you know that you cannot possibly call and are closest to the opaque wall could you call out to F and see whether we can hear them Please!. Also given that we know that the quickest way to stop staring at the wall is to accept that logically progressive calls should be made by F then E and so on then ask the other side to declare in their own time when they have concluded they cannot make any call”<o:p></o:p>
If this was acceptable then the moment E declared he cannot call both A and D can call whoever is quickest is the answer<o:p></o:p>
Regards<o:p></o:p>
Freddo (NOVICE)<o:p></o:p>
It is not logical to silently wait forever !
