This is a very interesting puzzle and not at all difficult as it may seem. Lets assume that the sages are A, B and C. Let us see the possibilities of conbinations that are possible
A B C ------ W W W - (1) W W B - (2) W B W - (3) W B B - (4) B W W - (5) B W B - (6) B B W - (7) B B B ---> Not possible since only 2 black balls
Combination 1: Sage A sees that other 2 people has white balls so he guesses Black as the probability is 2/3 and fails. Sage B does the same. Sage C now knows that for sure he has white ball.
Combination 2: Sage A sees a White and Black so passes. Sage B does the same. Sage C now knows he has a Black ball.
Combination 3: Sage A sees White and Black so he passes. Sage B sees White and White so he guesses Black and he is correct. Sage C now knows he has White as A passed and B guessed.
Combination 4: Sage A guesses white since there are only 2 Black balls. Sage B now knows he has a Black since A guessed (B may also pass). Whatever B does, C knows he has a Black.
Combination 5: Sage A guesses Black as he can see other two has white and is correct. Sage B might guess based on Sage A or pass. Sage C knows he has white.
Combination 6: Sage A passes. Sage B guesses White. Sage C knows he has Black.
Combination 7: Sage C can see that he has White.