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?
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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.
jyoak said,
August 15, 2009 at 11:08 pm
Just what I had in mind, Blaine!
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.