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Another Riddle

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OK, POW#1 one knew because he could tell the other two couldn't work out what theirs were from what they could see.

POW#3 didn't know what colour his was... so therefore #1+#3 either BOTH had Black, or one had Black, and one had White, because if they both had white he would have to have Black.

POW#2 realises that POW#3 doesn't know, so knows that his could either be black or white because he knows that POW#1's beanie is black, so his could be black or white, so he doesn't know.

POW#1 therefore realises that as neither of the other two know, his must be black.

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Let's see if I can show the answer in a graphical way.

The POWs were lined up so that #3 could see #2 and #1, #2 could see #1 and #1 couldn't see anyone.

3-> 2-> 1->

There were seven possible combinations of beenies:

B   B   B   (a
W   B   B   (b
B   W   B   (c
W   W   B   (d
B   B   W   (e
W   B   W   (f
B   W   W   (g
3-> 2-> 1->

The only way #3 would know for certain the color of his beenie was if both #2 and #1 were wearing white (g which would mean #3 was wearing black. But this wasn't that case, so #3 didn't say anything.

Since #3 stayed silent, #2 realized that #3 could see at least one black beenie. So according to #2 the possible combinations were:

?   B   W   (a
?   W   B   (b
?   B   B   (c
3-> 2-> 1->

#2 could see that #1 was wearing a black beenie and realized that the beenie on his own head could still be either black or white, (b or (c, so he didn't say anything either.

Since #2 stayed silent, #1 realized that #2 didn't know if it was case (b or (c and in both of those cases #1 was wearing a black beenie.

10 points to Licensed Devil for figuring it out.

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