From: Eliezer S. Yudkowsky (sentience@pobox.com)
Date: Thu Sep 11 2003 - 09:15:04 MDT
Emlyn O'regan wrote:
> I think it's 4/9, but I brute forced it.
4/9 is correct.
Eliezer S. Yudkowsky wrote:
>
> 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.
To see why you didn't need to brute-force it, try calculating the
probability that the third chip was white given that the first chip was
white, or the probability that the first chip was white given that the
second chip was white. Also 4/9.
In fact, for any j != k, p(Wj|Wk) == 4/9, provided that the only other
information given is as described above. So if we knew only that the
sixty-seventh chip was white, we would estimate a 4/9 probability that the
twenty-third chip was white.
Since all bowls of this kind possess this symmetry, we can find the
constant for any particular bowl by estimating the probability that the
second chip was white, given that the first chip was white. If we got a
white chip on the first round, we would add one white chip for the second
round, making the probability of a white chip 4/9. This is then the same
as the probability that a white chip was drawn on the second round, given
that a white chip was drawn on the fourth round.
-- Eliezer S. Yudkowsky http://singinst.org/ Research Fellow, Singularity Institute for Artificial Intelligence
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