From: Hal Finney (hal@finney.org)
Date: Thu Sep 11 2003 - 12:17:40 MDT
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, that the probability is always
the same on every round. Suppose on round N there are p white chips and
q chips total. Then the probability of drawing white on that round is
p/q.
Now consider round N+1. If we drew white on round N (which happened with
probability p/q), the probability of white on this round is (p+1)/(q+1).
If we drew red on round N (which happened with probability (q-p)/q),
the probability of white on this round is p/(q+1).
So the total probability of white on this round is the sum of these two
values, weighted by their probabilities, or p/q * (p+1)/(q+1)
+ (q-p)/q * p/(q+1).
= (p^2 + p + qp - p^2) / q(q+1)
= p(q+1) / q(q+1)
= p / q
QED.
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.
Hal
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