Yoak: Lewis Carroll, Colored Stones

I’ve enjoyed several books by and about Lewis Carroll with puzzles, games and neat observations.  I’m going to post a few here.  Here’s a simple one with which to get started.

Suppose that I secretly flip a coin and place either a blank or white stone in a bag based on the result.  I then put a white stone in the bag for two stones in total.  I invite you to pull one stone out and it turns out that it is white.  What is the chance that the other stone in the bag is also white?

 

3 Comments »

  1. Blaine said,

    August 14, 2009 at 2:50 pm

    Unbiased coin, 50:50 chances, so the answer sounds like it should be…
    [spoiler]Well, let’s not jump into that trap. Instead, think for a second.

    Initially the bag could have one of two states (with equal probability)
    BLANK-WHITE
    or
    WHITE-WHITE

    There are *three* ways you could have pulled out a white stone:
    1) You could have pulled the second stone from the BLANK-WHITE bag.
    2) You could have pulled out the first stone from the WHITE-WHITE bag.
    3) You could have pulled out the second stone from the WHITE-WHITE bag.

    In two of the three cases, the other stone is also white, so the chance is:
    2/3. Conditional probability can be tricky, can’t it?[/spoiler]
    Not convinced? Try the experiment yourself.

  2. jyoak said,

    August 15, 2009 at 11:08 pm

    Just what I had in mind, Blaine!

  3. Andy said,

    August 19, 2009 at 10:18 am

    Bayes’ theorem:
    P(both white|draw one white) = P(draw one white|both white) * P(both white) / P(draw one white)
    P(draw one white) = 1 – P(draw one black) = 1 – (1/2 * 1/2) = 3/4
    So:
    P(both white|draw one white) = 1 * 1/2 * 4/3 = 2/3
    Yay.

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