Here's a puzzle.
In the applet, select box. One of the two balls it contains will pop up. After which you'll have to drag labels (they will appear at the right moment) into the empty label locations on the boxes.
Solution
Click on the box (incorrectly) labelled white and black. It is bound to contain two balls of the same color, for otherwise the original label would be right. Verify that once the right label for that box is found, there's only one way to place the remaining labels.
Selecting a box labelled by two balls of the same color may or may not give you a clue to the right contents. For example, assume you pick the box labelled black/black. If you are lucky and out pops a black ball, you may be sure of the content: the box contains balls of different colors. But if, per chance, the shown ball is white, you get no useful information about what is inside.
That puzzle is very well known and probably is very old. A novelty could be in increasing the number of balls, boxes, and guesses. With three balls, one can label 4 boxes. (Using only the first letters, the labels are WWW, WWB, WBB, BBB, which naturally corresponds to 4 boxes.) Unfortunately, the puzzle becomes cumbersome as neither 2 nor 3 guesses may suffice to determine the correct labeling. 2 guesses never suffice. 3 guesses and some luck (i.e., given suitable popouts) may lead to a solution in some cases. What about 4 guesses?
(Permutations in which none of the elements remains in its place are called derangements. With 3 boxes, there were only 2 derangements. With 4 boxes, there are 9. Derangements are relevant to the puzzle since when we try to determine the contents of the boxes, we actually seek a derangement of a given (wrong) labelling.)
[Sharygin, p. 82] places a problem in a fanciful context:
