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GuySrinivasan comments on [Link] The Bayesian argument against induction. - Less Wrong Discussion

4 Post author: Peterdjones 18 July 2011 09:52PM

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Tags: bayes probability popper induction

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Comment author: GuySrinivasan 19 July 2011 12:31:18AM *  0 points [-]

P(Av~B|B) does not equal P(A). P((Av~B) & B) equals P(A).

Edit: it doesn't, of course. P(Av~B|B) = P(A|B) and the other thing I said is just silly.

Comment author: Zack_M_Davis 19 July 2011 12:56:47AM 1 point [-]

P(Av~B|B) does not equal P(A) [...] P(Av~B|B) = P(A|B)

Whoops! You're right. I'll edit the comment to reflect this correction.

Comment author: Manfred 19 July 2011 07:23:54AM 0 points [-]

Just for general convenience: P((A+~B)B) = P(AB).

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