Sunday, 17 July 2011

Survivor... Black or White?

A really nice probability problem to talk with students about.  What approaches do they use? 

An urn contains 75 white balls and 150 black ones. A pile of black ones is also available.
The following two-step operation occurs repeatedly. First we withdraw two balls at random from the urn, then:
  • If both are black, we put one of them back in the urn and throw the other away.
  • If one is black and the other white, we put the white one back and throw the black one away.
  • If both are white, we throw both away and put a black ball from the pile into the urn.
Because the urn loses a ball at each step, eventually it will contain a single ball. What color is that ball?

This is another nice problem I found at Futility Closet.  Really an entertaining blog.  Greg cites this as from the 1983 Australian Mathematical Olympiad, via Ross Honsberger, From Erdös to Kiev, 1996.  Another nice problem for teachers to add to their files.

If you can't figure it out,Alexander Bogomolny has a solution for the problem at his "Cut-the-knot" web site... a great source for students (and teachers) to explore. 
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7 comments:

  1. Sue VanHattum17 July 2011 at 17:42

    The probabilities looked so complicated, I wanted to program a simulation. I tried first in Scratch, and then used Excel.

    What I saw there made me think I ought to be able to use logic to figure this out (starting with end results and working backward), but so far, no luck.

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  2. Pat's Blog17 July 2011 at 18:12

    Sue V,

    Logic is the key here, think Parity... reminds me a little of covering a checkerboard with dominoes if the two opposite corners are cut off... that kind of thing... It would make no difference if there were 975 white and 2000 black.... Good luck, I imageine you will see a solution here soon if you are not the first... (surely at least ONE of my ex-students will have a go...)

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  3. Sue VanHattum17 July 2011 at 18:18

    Ahh, but it is possible to change the result.

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  4. Þórsteinn17 July 2011 at 18:39

    It looks like it would be impossible to run out of white balls, so I would have to say white.

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  5. Jeffo18 July 2011 at 13:09

    Every move preserves the parity of white balls (i.e., keeps it an odd number) and changes the parity of black balls. Since the total number of balls reduces with every move, eventually there will be just one (the least odd number) white ball, and some number (possibly zero) black balls. All possible moves at that point reduce the number of black balls, and do not change the number of white balls. Therefore, eventually there will be no black balls and one white ball remaining.

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  6. Jeffo18 July 2011 at 13:11

    Incidentally, the "Cut the Knot" link was broken for me. Can you update?

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  7. Pat's Blog19 July 2011 at 12:32

    Jeffo,
    Thanks, I believe the link is corrected now, and thanks to Alexander Bogomolny, too.

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