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A comment on Paul Graham is right (using AVC’s data). Parent comment

Re:Paul Graham is right (using AVC’s data) by india on Friday October 29, 2010 @01:12PM (#1848005)
No. The event being bimodal only implies that the probability is centered around two values, not that p =1 or p = 0 for those values. For example, take a biased coin that turns up head once in a million. This is bimodal with p(head) = 10^-6.

PG's argument appears to be that the error in your estimate of p is almost always larger than the value of p and one should invest as long as 2*p is also within the estimation error.

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Re:Paul Graham is right (using AVC’s data) by jessriedel on Saturday October 30, 2010 @04:14PM (#1851703)
> PG's argument appears to be that the error in your estimate of p is almost always larger than the value of p and one should invest as long as 2*p is also within the estimation error.

I don't think this is correct under canonical Bayesian reasoning. All uncertainties (whether because of ignorance or "objective uncertainties") can be bundled into your probability assessment. There is no "error" on your probability.

I'm not expert though. I'm only....80% sure.

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