If we change it to state “[…] so that it’s unclear whether any box is correctly labelled,” then I see two strategies, both solved in 3 draws if you’re lucky, 4 if you’re not, though the first has a much higher chance of being lucky.
Tactic 1: Draw one ball from Box 1 and one from Box 2.
If they’re the same colour, then the third box is two of the other colour. Draw one ball from either Box 1 or Box 2, and you will know what the one you picked from contains, which will give you the other by elimination.
If the two drawn from Box 1 and Box 2 are different colours, then draw one more from Box 1. If Box 1 is BW, then you know all three boxes. (Box 2 is two of whatever colour is in Box 2, and Box 3 is two of the other colour).
If Box 1 is not BW, you have to draw one more from Box 2. You know know what is in Box 1 and what is in Box 2, and, by process of elimination, what is in Box 3.
With Tactic 2, I calculate that you’ll have a 3/4 chance that you’ll get your answer in 3 balls, and a fourth ball will guarantee that you know the answer.
Tactic 2: Draw two balls from Box 1…
If you draw BW, then draw one from Box 2. Since both of the other two will have two of the same colour, that’s all you’ll need.
If the two balls in Box 1 are the same colour, you’re boned. Draw 2 balls from Box 2, which will tell you what’s in Box 3 by elimination.
With Tactic 2, I calculate that you’ll have a 1/3 chance that you’ll get your answer in 3 balls, and a fourth ball will guarantee that you know the answer.
I know that this is not the puzzle we’ve been given (see my previous post, above), but I wanted to know: can anyone improve upon that, so that the answer is always known within three balls?