Re: Today's evil Bayesian math problem

From: Brett Paatsch (bpaatsch@bigpond.net.au)
Date: Thu Sep 11 2003 - 02:23:39 MDT

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    Feeble second attempt 5/9. 5/(4+5)

    = 55.5'%

    > Eliezer writes:
    >
    > > > Suppose that a bowl has 5 red chips and 3 white chips.
    > > > We sample chips from the bowl using the following procedure:
    > > > On each round we draw a random chip, replace it, and
    > > > then add another chip of the same color to the bowl. For
    > > > example, if on the first round we happen to draw a red
    > > > chip, there would then be 6 red chips and 3 white chips to
    > > > draw from on the second round.
    > > >
    > > > Given that a white chip was drawn on the fourth round,
    > > > what is the probability that a white chip was drawn on the
    > > > second round?
    > > >
    > > > (This problem is extra bonus evil because it's so easy if you
    > > > know the rules.)
    > >
    > > Evil Hint #1: Not only is it possible for you to do this problem
    > > in your head, the answer can be obtained in ONE step.
    >
    > Aaagh! ;-) But this (my effort below) ISN'T Bayes !
    >
    > R1 R2 R3 R4 (w drawn is given!)
    >
    >
    avg
    > w2 5r3w 5r4w 5r5w 5r6w 6/11 ) 5/11
    > 6r5w 5/11 )
    > 6r4w 6r5w 5/11 )
    > 7r4w 4/11 )
    >
    >
    avg
    > r2(~w2) 6r3w 6r4w 6r5w 5/11 4/11
    > 7r4w 4/11
    > 7r3w 7r4w 4/11
    > 8r3w 3/11
    > (3/9 or 4/9)
    > 5 + 4 = 100% 9 = 100%

    p(w2) = 5/9
    p(-w2) = 4/9

    Brett

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