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