From: Eliezer S. Yudkowsky (sentience@pobox.com)
Date: Tue Oct 29 2002 - 04:27:51 MST
Lee Daniel Crocker wrote:
>>(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.
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.
-- Eliezer S. Yudkowsky http://singinst.org/ Research Fellow, Singularity Institute for Artificial Intelligence
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