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?



  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)

    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
    P(both white|draw one white) = 1 * 1/2 * 4/3 = 2/3

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