Talk:1132: Frequentists vs. Bayesians

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Revision as of 16:22, 9 November 2012 by Lmpk (Talk | contribs)

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Note: taking that bet would be a mistake. If the Bayesian is right, you're out $50. If he's wrong, everyone is about to die and you'll never get to spend the winnings. Of course, this meta-analysis is itself a type of Bayesian thinking, so Dunning-Kruger Effect would apply. - Frankie (talk) 13:50, 9 November 2012 (UTC)

You don't think you could spend fifty bucks in eight minutes? ;-) (PS: wikipedia is probably a better link than lmgtfy: Dunning-Kruger effect) -- IronyChef (talk) 15:35, 9 November 2012 (UTC)

Randall has referenced the Labyrinth guards before: xkcd 246:Labyrinth puzzle. Plus he has satirized p<0.05 in xkcd 882:Significant--Prooffreader (talk) 15:59, 9 November 2012 (UTC)

A bit of maths. Let event N be the sun going nova and event Y be the detector giving the answer "Yes". The detector has already given a positive answer so we want to compute P(N|Y). Applying the Bayes' theorem:

P(N|Y) = P(Y|N) * P(N) / P(Y)
P(Y|N) = 1
P(N) = 0.0000....
P(Y|N) * P(N) = 0.0000...
P(Y) = p(Y|N)*P(N) + P(Y|-N)*P(-N)
P(Y|-N) = 1/36
P(-N) = 0.999999...
P(Y) = 0 + 1/36 = 1/36
P(N|Y) = 0 / (1/36) = 0

Quite likely it's not entirely correct. Lmpk (talk) 16:22, 9 November 2012 (UTC)

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