From: Emlyn O'regan (oregan.emlyn@healthsolve.com.au)
Date: Mon Oct 28 2002 - 18:21:36 MST
I just ran a simulation of this; the frequency of tails in the remaining 15
coins over 10,000,000 runs was 66.62 using the biased sampling (50.00 using
unbiased, of course :-) ).
As to the explanation in terms of probabilities, I'm sure there are lots of
factorials and "n choose m" type things and other goodies that my rusty
probability coursework is no longer competent to tackle. :-(
Emlyn
> -----Original Message-----
> From: Eliezer S. Yudkowsky [mailto:sentience@pobox.com]
> Sent: Tuesday, 29 October 2002 9:39
> To: extropians@extropy.org
> Subject: Math question
>
>
> Suppose that we flip a fair coin 20 times, and that a biased sampling
> procedure then randomly selects 5 coinflips from the set of
> coinflips that
> came up heads and reports on those coinflips; if there aren't
> 5 coinflips
> that came up heads, the biased sampling procedure reports all
> available
> heads and enough randomly selected tails to make up the gap.
>
> Suppose the biased sampling procedure reports that the first, sixth,
> eleventh, fifteenth, and eighteenth flips came up heads. Is there a
> simple general formula for calculating the probability that any given
> other coinflip came up tails? Clearly the probability is
> greater than
> 50%, but by how much?
>
> (I don't know the answer to this one; it's posed as a genuine math
> question, not a math problem.)
>
> --
> Eliezer S. Yudkowsky http://singinst.org/
> Research Fellow, Singularity Institute for Artificial Intelligence
>
>
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