bonanova Posted May 25, 2014 Report Share Posted May 25, 2014 An urn contains a certain number of black balls and a certain number of white balls. You draw two balls from the urn at random. If they are of opposite color you throw them away and put a white ball into the urn. If they are of the same color you throw them away and put a black ball into the urn. You reduce the number of balls by 1 each time. After some number of repetitions there will be only a single ball left in the urn. If you are told the respective numbers of black and white balls at the outset, can you accurately predict the color of that final ball? Quote Link to comment Share on other sites More sharing options...
0 m00li Posted May 25, 2014 Report Share Posted May 25, 2014 after each drawing the number of white balls in the urn can decrease only by 2 or not decrease at all. After each drawing the number of black balls can either go down by 1 or go up by 1 Hence, at the outset (assuming there is at least one ball in the urn to begin with): if the number of white balls is even (or 0), ultimately all white balls will finish and only 1 black ball can remain. if not, ultimately only one white can remain Quote Link to comment Share on other sites More sharing options...
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bonanova
An urn contains a certain number of black balls and a certain number of white balls.
You draw two balls from the urn at random.
If they are of opposite color you throw them away and put a white ball into the urn.
If they are of the same color you throw them away and put a black ball into the urn.
You reduce the number of balls by 1 each time.
After some number of repetitions there will be only a single ball left in the urn.
If you are told the respective numbers of black and white balls at the outset,
can you accurately predict the color of that final ball?
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