From: Emlyn O'regan (oregan.emlyn@healthsolve.com.au)
Date: Tue Oct 29 2002 - 16:36:36 MST
>
> > Thanks, Lee. This is just the procedure I wanted to know
> if there was a
> > simpler version of before I did all that math :) but it
> counts as an
> > answer to my question. I was hoping someone would say,
> "Oh, the answer
> > must be" followed by a compact factorial expression.
> >
> > Emlyn, did your procedure eliminate cases where there were
> four heads or
> > fewer in the sample, thus causing the biased sampler to
> report at least
> > one tail? The original problem specified that the biased
> sampler reported
> > five heads.
"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"
I purposely included cases where there were less than 5 heads, based on the
above statement. It actually made things a bit more complex.
When I omit that case, I get 0.6642 .
>
> I thought Emlyn's estimate was rather ingenious, but no, it didn't
> take that into account nor were his numbers exact, but it was an
> interesting shortcut approach that gets damn close to the right
> answer with a lot less work. I guess those of us who are spoiled
> on spreadsheets and Monte-Carlo programs get lazy and fail to think
> of creative shortcuts like that.
LOL! I thought that your method was ingenious, and mine was just a dumb
brute force approach. I wouldn't have called it creative; just the most
obvious method to a programmer.
Emlyn
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