The solution I’ve come to is that the fourth sage has the label WWW, with the contents of his box WWB. My reasoning is as follows:
The first sage has BB* #
The puzzle states:
The first sage takes out two black balls and says, “I know the color of the third ball.”
If the sage has taken out BB, there are only two options for the final ball: W and B. The only way for the first sage to know the answer is if the sage has a label with the opposite. For example, if the label is BBW, then the contents of the box must be BBB since we know the contents of the boxes must not match the label.
Therefore, there are two options for the first sage:
The second sage has BW* #
Onto the second sage, who says:
The second sage takes out one black and one white ball and says, “I know the color of the third ball.”
We know the contents of the box must be BBW or BWW. Similar to the first sage, we can assume the second sage has the opposite pairing:
The third sage has label BB* #
The third sage is interesting:
The third sage takes out two white balls and says, “I don’t know the color of the third ball.”
The third sage has the following information at this point:
- sage 1’s box contains BB* (BBB or BBW)
- sage 2’s box contains BW* (BBW or BWW)
- sage 3’s box contains WW* (BWW or WWW)
- sage 3’s label.
So what label could sage 3 have such that he cannot eliminate either BWW or WWW as a possible candidate?
Among the labels BBB, BBW, BWW, and WWW, the valid candidate are BBW and BBB:
- if 3’s label is WWW, then the box must contain BWW.
- if 3’s label is BWW, then the box must contain WWW.
So the third sage must have the labels BBB or BBW, the opposite of whatever sage 1 has.
So sage 3’s options are:
Sage 4 #
So now we get to sage 4:
The fourth sage says, without taking out any balls, “I know the color of all the balls in my box and also the content of all the other boxes.”
Sage 4 has the same knowledge and reasoning as we do. So we know that:
- Sage 1 and Sage 3’s labels are BBW or BBB.
- Sage 2’s label is one of BBW or BWW.
Since sage 1 and 3 have labels of BBW and BBB, then sage 2’s label must be BWW. Since all other labels are taken, sage 4’s label must be WWW.
If sage 2’s label is BWW and they know the contents of their box, then their box must contain BBW.
This also means that sage 1’s box is clear: it is BBB, which implies that sage 1’s label is BBW:
So we know the following:
Since the sage knows their label is WWW, they know that their box must not contain WWW. Since sage 3 has drawn WW*, it must contain WWW. By deduction the last sage must have the oly remaining combination, which is WWB: