Re: Today's evil Bayesian math problem

From: Brett Paatsch (bpaatsch@bigpond.net.au)
Date: Thu Sep 11 2003 - 02:00:43 MDT

  • Next message: Brett Paatsch: "Re: Today's evil Bayesian math problem"

    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)

    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
                  (4/9 or 3/9)
          (3/8)

    So 100% = 11
           5 + 4 = 100%

    w is 25% more likely to come from a (r2 w chip) group
    than the others.

    P= 3/8 + 25% = 3/8 + 2/8 = 5/8

     = 62.5%

    But this is *not* one step, I didn't do it in my head,
    and I would not like to bet that 62.5% is right either ;-)

    What's the Bayesian way?

    Regards,
    Brett



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