Bob and Jack are standing facing each other. They are told that there are 11 cones divided into 2 colours (black and white). Bob has 6 of those cones behind his back and Jack has 5. None of them can see his own cones but can see the other one's cones. They are also told that the number of the black cones is higher or lower by 1 than the number of the white cones. Each one of them was asked "do you know how many of the cones behind your back are black and how many are white?"
The answers were as follows:
Bob: I don't know.
Jack: I don't know.
Bob: I don't know.
Jack: I don't know.
Bob: I don't know.
Jack: I don't know.
Bob: I know!
How many cones behind Bob's back are black and
how many are white?
Method 1:
After B1: nothing new determined
After J1: Bob knows all his cones are not the same color.
After B2: Jack knows all his cones are not the same color
After J2: Bob knows that at least two differ in color from other four.
After B3: Jack knows that at least two differ in color from other three.
After J3: Bob knows that at his are divided 3 and 3.
Method 2:
Assuming there's only one solution and nothing in the presentation allows us to differentiate white from black, anything that would lead to 4 black, 2 white, could apply equally to 4 white, 2 black. So it must be the same for each color: 3. Confirmed by the fact the puzzle doesn't ask for the colors of the cones behind Jack's back, which are 3 and 2, but we don't know which color is 3 and which is 2.

Posted by Charlie
on 20070309 17:16:16 