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We have 3 boxes. In every one there are 2 balls. One of them has 2 black balls, second 2 white balls, third black and white ball.

On every box is a right plate(label): BB,WW,BW. Unfortunatelly somebody mixed the plates and now only NONE of the boxes has a right plate on it. You can draw only one random ball from selected box to decode what is in all the boxes. Which box would you choose?

My opininon: I have no clue, because:

  • if I choose box with a plate BB I only know that it is wrong plate if I draw white ball, but I can draw black ball and it still can be BW box.
  • if I choose box with a plate BW I will know nothing about righteousness of a plate on it
Yoda
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4 Answers4

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Now with the edit, this is solvable. Choose the BW tagged box. All boxes are wrongly tagged by definition. Pick a ball, if ball is black, then, this box is not WW for sure, hence it is BB. Now if the BB-tagged box were originally BW box, then the WW-tagged box would be rightly tagged, which would be a contradiction, hence, BB-tagged box is actually a WW box, and WW tagged box is actually a BW box.

Similarly, you can retag boxes if a white ball came out of BW box.

Juanito
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It is impossible, if you pick black and you get a black one it could be correct or it could be the mixed, so you can't decode there.

If you pick white you could also get a white one.

If you picked mixed then you don't get any information because it could be correct and be mixed or it could be incorrect and be the box of whatever color you got. Therefore there is no answer.

This problems does work however if we know all boxes are labelled incorrectly

Here I adress the new problem:

Since none of the boxes has the right label you can just take a ball from the box marked mixed. If the ball obtain is black in reality that must be the black box. that means the box marked white is mixed and the ball marked black is white. If you get a white ball then the box marked mixed is white, the ball marked white is black and the ball marked black is mixed.

Asinomás
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  • I must apologize. I have edited the orignal post. I am very sorry. None of the boxes has right label on it. – Yoda Jun 02 '14 at 22:29
  • No problem, I have added a paragraph addressing the new question, this is a very famous question by the way. – Asinomás Jun 02 '14 at 22:36
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Clearly impossible. You can easily compute all 18 cases and find out noone of them gives enough information alone.

I suggest you check the question to see if your understanding is correct

AnalysisStudent0414
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  • I must apologize. I have edited the orignal post. I am very sorry. None of the boxes has right label on it. – Yoda Jun 02 '14 at 22:30
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Easy. Pick from BW box, what ever color you get (let's say W) that box had to hold 2 of, so it WW. You know that the BB box must hold the the BW balls because it can't hold the BB balls. Only thing left is is the WW box and the BB balls.

Darla Simmons
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