Quote:
Originally Posted by orlok
Once again, thanks for the attempts to clarify. I still stick to my position that the elimination process does not hold water if you start from the premise that they are all sitting in the room, and can all see four red hats from the start. Blue hats just don't come into it. Can someone show how you would work out the colour of their own hat starting from this principle?
Otherwise I'll just go sit in the corner with the big hat with a D on it...
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I'll have one more go, working purely from the perspective of one of the people sitting in the room, seeing four red hats.
OK, I'm in a room with four other people. We're called A, B, C, D and E. (Dull names, but this is a logic puzzle.) I'm A. They are all wearing red hats, but I don't know what colour hat I have. But I know that they can all also see either four red hats (if I also have a red hat) or three red hats (if I have a blue hat).
Now I need to consider what one of them, say B, can see. If my hat is blue, B sees three red hats and one blue hat.
Now B doesn't know that B has a red hat, so if B considers what C can see, B must consider the possibility that C can see either three red hats and one blue hat (if B's hat is red), or two red hats and two blue hats (if B's hat is blue). Now, I, A, know that B's hat is red, but B does not know that, so he can legitimately wonder what C can see if B's hat is blue.
I can think that B might think that C might see two red hats and two blue hats. I can think this, even though I know that C can see at least three red hats. That's because I'm thinking about what B might legitimately think, if my hat were blue.
Now as I can legitimately consider that B can see only three red hats, and so that B might think that C might only see two red hats and two blue hats (mine and B's), I can consider that B can also legitimately consider what C would think that D can see. I think that B could think that C can see two red and two blue, and so I can think that B can think that C might think that D could only see one red hat and three blue ones (mine, B's and C's). I can build a chain of assumptions about what other people might think that other people can see.
In summary:
I can think that B might see only three red hats.
And I can think that B might think that C see only two red hats.
And I can think that B might think that C might think that D can see only one red hat.
And I can think that B might think that C might think that D might think that E can see no red hats, only blue hats.
And I can think that B might think that C might think that D might think that E can't know whether E's hat is red or blue.
Until someone says, at 11:55, that they can see a red hat.
Then I think that B might think that C might think that D might think that E can see no red hats, and so know that E's hat is red, and so would leave at 12:00.
Except, of course, that E does not leave at 12:00.
Then I can no longer think that B might think that C might think that D might think that E sees no red hats.
I must now think that B might think that C might think that D might see one red hat, and I must think that B might think that C might think that D must think that E can also see one red hat. In which case both D & E would leave at 13:00.
Except, of course, that D & E do not leave at 13:00.
I must now think that B might think that C might see two red hats, and I must think that B might think that C must think that D and E can also see two red hats. In which case, C, D & E will leave at 14:00.
Except, of course, that C, D & E do not leave at 14:00.
I must now think that B might see three red hats, and I must think that B must think that C, D, and E also see three red hats. In which B, C, D & E will leave at 15:00.
Except that B, C, D & E do not leave at 15:00.
I can now conclude that B does not see three red hats, but also sees four red hats. My hat must be red, and we all leave at 16:00.
This probably hasn't helped. But I've done my best.