From: Eliezer S. Yudkowsky (sentience@pobox.com)
Date: Thu Sep 11 2003 - 16:37:20 MDT
Hal Finney wrote:
> Eliezer writes:
>
>>In fact, for any j != k, p(Wj|Wk) == 4/9, provided that the only other
>>information given is as described above.
>
> That's pretty surprising. In fact, it seems that an even more surprising
> fact is true: the probability of a white chip on every round j is always
> 3/8, the same as it is on round 1. And the probability of white chips
> on both of any two different rounds j and k is always 1/6. From this,
> Eliezer's formula follows, as p(Wj|Wk) = p(Wj&Wk)/p(Wk) = 1/6 / 3/8 = 4/9.
>
> I can prove at least the first statement...
>
> I imagine a more complex proof could show that the probability of white
> on both of any two rounds is constant, and presumably this could be
> generalized to any n rounds. But I don't see how to prove all this
> in a simple way.
Evil Hint #3:
The classical version of the problem has a bowl where we sample without
replacement - take the chip, and throw it away. Suppose there are M red
chips and N chips. If you don't know what you got on the first round, the
probability of getting a red chip on the second round is still M/N. The
probability of getting a red chip on the second round, given that you got
a red chip on the fourth round, is just (M-1)/(N-1), since there's one
less red chip to go around. We can think of sprinkling the chips over
holes, instead of drawing them in successive rounds, and it doesn't matter
if you swap around the labels on the holes.
The evil version of the bowl, instead of running out of the chips already
sprinkled, has more and more of them, like a P2P file-sharing network.
But that doesn't change anything important.
-- Eliezer S. Yudkowsky http://singinst.org/ Research Fellow, Singularity Institute for Artificial Intelligence
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