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 Black and White (Posted on 2003-02-25)
Imagine that you have three boxes, one containing two black balls, one containing two white balls, and the third containing one black ball and one white ball.

The boxes were originally labelled for their contents (BB - WW - BW) but someone has inadvertently switched the labels so that now every box is incorrectly labelled.

Without looking inside, you are allowed to take one ball at a time out of any box that you wish, and by this process of sampling, you are to determine the contents of all three boxes.

What is the smallest number of drawings needed to do this?

 See The Solution Submitted by Ravi Raja Rating: 3.1250 (8 votes)

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 answer | Comment 23 of 25 |

just one.

Take one ball from the box marked BW ( since this is wrong, its either actually BB or WW, tf; if you pull out Black its BB ... )

Now means the only other Double color is the color opposite of that you have just drawn.  So the one marked with the Double of the opposite color (if you drew a black ball the lable Im talking about would be WW) is then one that actually has the BW, because if it was the other box           [It would make the third remaining box be labled correctly which the problem says is incorrect.]

And the third double of the color you hand drawn is BW.

uh what?

ex:  you picked B, from the BW box, since BW is wrong the box can only be BB.  Now the BB box isnt Actually BW because that would make the box labled WW to actually be WW (which is wrong).

So since BB isn't BW (or BB) it must be WW and that leaves the last one [WW] to be BW.

 Posted by gabbo on 2004-09-15 22:46:12

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