= 27d567bc2d48d9f40e9ccc20ea370b15)
GuySrinivasan comments on [Link] The Bayesian argument against induction. - Less Wrong Discussion
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (27)
[Correction: In the original version of this comment, I claimed that the linked post was mistaken on a point of mathematics: specifically, I said that it is not the case that P(A|B) – P(A)= P(A v B|B) – P(A v B) + P(A v ~B|B) – P(A v ~B). However, Guy Srinivasan pointed out that my supposed disproof contained a mistake. Redoing my calculations, I find that the equation in question is in fact an identity. I regret the error.]
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.
Whoops! You're right. I'll edit the comment to reflect this correction.
Just for general convenience: P((A+~B)B) = P(AB).