Three.
There are six possible orderings, so you need three bits of information.
One marble from the BW box.
If it is B, then WW has BB and BB has BW
If it is W, then WW has BW and BB has WW
One, because you know that every box is incorrectly labelled.
Draw the ball from the BW box. Because “BW” is wrong, both that are inside must be the same colour.
If the one you pick out is white, then both inside are white. This means that the box labelled BB has BW and the box labelled WW has BB.
If the one you pick out is black, then both inside are black. This means that the box labelled WW has BW and the box labelled BB has WW.
If the state of each box’s labelling is “unknown” instead of “incorrect,” I get best case 3, worst case 4. I can’t construct a scenario where you’ll be able to do it in 3 drawings 100% of the time.
One.
Take one out of the BW box. If it’s Black, this must actually be the BB box; then the one labelled BB must be the WW box and the one labelled WW must be BW.
(and while I was distracted by work, two other people posted it already. Pah)
Wow. You should stick to Latin mistakes.
One from any box.
Pick the box you like the most and claim all other boxes are filled with lies.
Nope you know enough to do it in 1 because you know that every box is incorrectly labeled which reduces the possible number of orderings.
You pull one marble from the box labeled BW which tells you what that box is because it has to contain all marbles of one color. Then the box labeled the opposite color the marble you pulled from the BW box is the box that is actually BW and the other box is the remaining one either BB or WW.
Got this exact question in a job interview.
Oh har-de-har I missed that part! You’re hired! (What the hell kind of job?)
Zero is the minimum number, if you’re lucky. I want to say a third of the time, but there’s something in the back of my mind about permutations that says that’s high. You can eliminate the original configuration at least, since it’s clear the labels were switched.
As outlined above, the first step is to determine whether the box labelled BW is actually WW or BB, and the second step is to switch the labels on two boxes. So, pick zero, and you have even chances of beating out all the other job applicants.
If the boxes were completely unlabeled, then the minimum draws * would indeed be three. But since you know you have three incorrect labels, it’s one.
A lot of these types of puzzles work this way, by hoping you miss a key input. Like, “As I was going to St. Ives, &etc.”. In this case, because it’s completely absurd that anyone would know that the labels are wrong without knowing the content, you got fooled!
* unless you are a lucky guesser, like @potato, in which case zero.
Yeah, I didn’t read that before trying my hand at it. That’s a kind assessment, thanks.
very likely some kind of programming job…
I don’t know. I took “career aptitude” tests that specifically asked me to rank how much I enjoy “packing things into boxes”, which seemed to have a pretty clear correct answer, and deepened my skepticism about much of the field of psychology.
I am a programmer though :). I was lucky to find I enjoyed it, and worried about robots making all the jobs obsolete even at a young age.
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?
I had half a lame joke about bringing Donald Trump in and getting his gut feeling to rank the boxes by laziness. Seems like an expansion on your strategy.
@rtkwe had the best answer. I am still puzzling over beating it without “cheating” as I have so far, but I don’t think it’s possible.
Edit: Sorry, it seems @nimelennar and @Omni_Dilletante arrived at it first.
