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LawrenceC comments on Open thread, Dec. 21 - Dec. 27, 2015 - Less Wrong Discussion

2 Post author: MrMind 21 December 2015 07:56AM

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Comment author: FrameBenignly 22 December 2015 05:27:17PM 0 points [-]

So what if p(H) = 1, p(H|A) = .4, p(H|B) = .3, and p(H|C) = .3? The evidence would suggest all are wrong. But I have also determined that A, B, and C are the only possible explanations for H. Clearly there is something wrong with my measurement, but I have no method of correcting for this problem.

Comment author: LawrenceC 22 December 2015 05:31:41PM 0 points [-]

Wait, how would you get P(H) = 1?

Comment author: FrameBenignly 22 December 2015 05:36:30PM 0 points [-]

Fine. p(H) = 0.5, p(H|A) = 0.2, p(H|B) = 0.15, p(H|C) = 0.15 It's not really relevant to the problem.

Comment author: Vaniver 22 December 2015 06:28:44PM *  0 points [-]

It's not really relevant to the problem.

The relevance is that it's a really weird way to set up a problem. If P(H)=1 and P(H|A)=0.4 then it is necessarily the case that P(A)=0. If that's not immediately obvious to you, you may want to come back to this topic after sleeping on it.

Comment author: FrameBenignly 22 December 2015 06:37:09PM 0 points [-]

Fair enough.

Comment author: IlyaShpitser 22 December 2015 05:38:06PM 0 points [-]

\sum_i p(H|i) need not add up to p(H) (or indeed to 1).

Comment author: FrameBenignly 22 December 2015 06:00:03PM *  0 points [-]

No, it doesn't.

Edit - I'm agreeing with you. Sorry if that wasn't clear.

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