pnrjulius comments on But There's Still A Chance, Right? - Less Wrong

44 Post author: Eliezer_Yudkowsky 06 January 2008 01:56AM

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Comment author: pnrjulius 03 April 2012 03:08:33AM *  0 points [-]

Suzie's suspicion is correct in general, though the two could work out the same in certain cases.

We know the probability P(draw red | choose random) = 1/20 What we need to know is P(choose random) where P(~choose random) is the prior probability of cheating. We also need to know P(draw red | ~choose random), the probability of drawing red if you cheat (presumably 1, but not necessarily---maybe it's an unreliable cheating method). From all those, we can solve the system and compute P(chose random | drew red).

What you're asking is whether P(chose random | drew red) = P(draw red | choose random); and in general this is not the case.

Indeed, we get to be so Bayesian we actually use Bayes's Theorem explicitly: P(A|B) = P(B|A) * P(A)/P(B)

Unless the priors are equal P(A) = P(B) [P(draw red) = P(choose random)] those two conditional probabilities will be distinct.