From: Eliezer S. Yudkowsky (sentience@pobox.com)
Date: Mon Oct 28 2002 - 16:55:36 MST
mlorrey@datamann.com wrote:
>
>[quote from: Eliezer on 2002-10-28 at 16:08:43]
>> 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.)
> The probability of what you ask is 50%. The prob that any given flip is one or
> the other is 50% in any situation. The odds that every other flip IN SEQUENCE
> is tails is dependent upon how long your sequence is. In the case of getting
> five tails, the odds are 3.125%.
Thanks, Mike, but this answer is incorrect. See the description of the
problem. If a *random* sampler reports five heads, the other coinflips
have a 50% chance of being heads. If a *biased* sampler reports five
heads, that's an entirely different issue.
-- Eliezer S. Yudkowsky http://singinst.org/ Research Fellow, Singularity Institute for Artificial Intelligence
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