# Can you solve the "scrambled box tops" puzzle?

Originally published at: https://boingboing.net/2020/05/19/can-you-solve-the-scrambled.html

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No.

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Well technically, it’s the bottoms that are scrambled, not the tops. Do I win?

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Okay, every box is mislabeled, so they are all incorrect.

So, if I draw from the box labeled BB, and I’m lucky and get a B, then I know that the contents must be BW, and since all the boxes are wrong, then it means that the contents of BW must then be WW (since if it was BB then WW would be correctly labeled. But if I drew a W, then I wouldn’t be able to tell whether BB should be WW or BW without drawing a second.

Spoiler in case it’s the right answer: it’s unclear if you need to demonstrate that your draws guarantee an answer.

Spoiler: that’s not the answer. Couldn’t get over associating the titles with the contents, even though I knew the contents were necessarily different.

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One.

Since they’re all mislabeled. You take one out of the box marked BW. You know the BW box is not mixed, therefore it contains all whatever color you picked. If it’s white, then you know that the BB box must be the mixed box, since it can’t have all black, and the WW box must contain the black ones. If it’s black, then the BB box is white and the WW box is mixed.

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Best way to solve this is to use the probabilty equation developed by invading a land war in Asia. You see if you…WHAT IS THAT…

Oh, nothing I guess. I totally didn’t look in the box…

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The key is that all 3 boxes are mislabeled.

Sample the box labeled BW. Because all of the labels are incorrect, this box must now contain either BB or WW marbles. Whatever color marble you get will tell you which. Now, whatever box that is, that label now refers to the other solid-color box. That is, if the box labeled BW is filled w/ BB marbles, the box labelled BB will be filled with WW marbles. The remaining box (WW in this case) now has the BW marbles. One sample guarantees correct determination.

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aka “My Drink with Dread Pirate Roberts”

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There are only two ways to arrange the labels to be all wrong. (Google derangements) If the contents are (BB, BW, WW), then the possibilities are (BW, WW, BB) and (WW,BB,BW). Two possibilities can be distinguished by a single binary feature, so it remains to find one which distinguishes them. That feature can be chosen to be which color lies underneath the BW label.

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I went for an interview with VMware many years ago and was asked this very thing. Needles I do not work there today…

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Let’s see. Three boxes, two marbles per box, that’s six draws, but I might pick the same marble twice, so draw multiple times to hopefully get lucky and draw a different marble. I think I can do it in 578 draws.

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I got bored and made things more complicated for myself

Ok, if we remove the condition that the boxes definitely don’t have the correct label (ie the labels are randomised), then I think you can tell with just three draws if you’re lucky.
(If you’re unlucky you could go on sampling the last two boxes forever).
Sample the first box and draw (eg) a B. If the second box also gives you a B, then the third box must be WW. Now resample box 1, if it’s a W then that must be BW, therefore BB is the other one, and you’ve done it in three draws. There’s only a 25% chance of this happening though.
If instead the first two picks are different, then you must sample the third box. You will get two of one colour, and one of the other. If you get BBW then the box that gave you W must be WW, and vice versa.
Then you just have to keep sampling the other two boxes until one of them changes colour.

One out of the box marked BW.

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Of course, that’s the real issue here: can you trust that all labels are false? If the labels were merely randomised, then there is a chance that the right label is on the box “by mistake”, so to say. And that is the real-life assumption most people make, including myself.

Oh yeah, that makes it even more tricky, in that you have to return the marble after drawing. Otherwise you could do it as such:

``````1. B- -- --  (first may be BW or BB)
2. B- B- --  (last is now WW)
3. BB B- -- (It's now known, BB BW WW)

1. B- -- --
2. B- W- -- (crap, now all three are potentially BW)
3. B- WW -- (damn it, the ends are still potential BW)
4. BB WW -- (at least we didn't have to look in the third box, damn it)
``````

All of this assumes the labels are useless, of course. As stated elsewhere, the original puzzle hints that the labels state what is not in the box.

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