From: Lee Daniel Crocker (lee@piclab.com)
Date: Mon Oct 28 2002 - 17:58:55 MST
> (Eliezer S. Yudkowsky <sentience@pobox.com>):
> 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 if this qualifies as an "easy way" to figure it out,
but here's the answer:
Probability 1/2 makes the calculating the binomial distribution a
little simpler, so the number of ways the 20 flips can result in
exactly N heads is 20! / (N! * (20 - N!), so that's 1 way to get
no heads, 20 ways to get 1, 190 ways to get 2, 1140 ways to get 3,
4845 ways to get 4, etc. Adding those 5 outcomes, which are the
ones where your sample produces at least one tail and the probability
of a remaining coing being a tail is therefore certainty, you get
6196/1048576 (the denominator here is 2^20), or a little less than
0.6 % as the cases that aren't interesting. That leaves 99.4% of
the cases where your biased selector has chosen 5 heads.
So, for each of these cases (N=5...N=20), calculate the probability
that a single chosen remaining coin is a tail; this will be just
(20 - N) / 15. For N=5, P=1 (because there were exactly five heads,
so the probability of a reminaining coing being a tail is 1), for
N=6, it's 14/15, etc., to N=20, where P=0 (all heads).
Now write that probability (1, 14/15, 13/15....) next to the previous
numbers you calculated in step 1 (15504, 38760, ...) Multiply them
across, and divide by (1048576-6196). Then add the whole column.
The number I get get is 0.66407..., so you should take the "head"
bet if you're offered 2-to-1.
Note: the hardest part of this for me wasn't figuring out how to do
it or calculating the binomials, but having to keep scratching out
"52" all the time...
-- Lee Daniel Crocker <lee@piclab.com> <http://www.piclab.com/lee/> "All inventions or works of authorship original to me, herein and past, are placed irrevocably in the public domain, and may be used or modified for any purpose, without permission, attribution, or notification."--LDC
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