Comment author: sebmathguy 28 June 2013 03:45:15AM *  0 points [-]

but due to hedonistic adaptation, you will come out no less unhappy.

Did you mean "no more unhappy."?

Edit: Formatting of quote.

Comment author: Coscott 17 June 2013 10:11:08PM 0 points [-]

I think that either I have communicated badly, or you are making a big math mistake. (or both)

Say we believe A with probability p and B with probability 1-p. (We therefore believe not A with probability 1-p and not B with probability p.

You claim that if we learn A and B are equivalent then we should assign probability 1 to A. However, a symmetric argument says that we should also assign probability 1 to not A. (Since not A and not B are equivalent and we assigned probabilities adding up to 1.)

This is a contradiction.

Is that clear?

Comment author: sebmathguy 18 June 2013 12:40:44AM 0 points [-]

Yes. Woops.

Comment author: Coscott 17 June 2013 10:15:05PM 0 points [-]

Oh, I just noticed the problem. When you say p(A or B)=1, that assumes that A and B are disjoint, or equivalently that p(A and B)=0.

The theorem you are trying to use when you say p(A or B)=1 is actually:

p(A or B)=p(A)+p(B)-p(A and B)

Comment author: sebmathguy 18 June 2013 12:12:58AM *  0 points [-]

Ok, this is a definition discrepancy. The or that I'm using is (A or B) <-> not( (not A) and (not B)).

Edit: I was wrong for a different reason.

Comment author: Coscott 14 June 2013 07:09:07PM 2 points [-]

I think that when p+q=1, the answer is clearly 1/2 due to symmetry. How did you get p+q?

Comment author: sebmathguy 17 June 2013 08:47:55PM 0 points [-]

If p + q = 1, then p(A or B) = 1. The equivalence statement about A and B that we're updating can be stated as (A or B) iff (A and B). Since probability mass is conserved, it has to go somewhere, and everything but A and B have probability 0, it has to go to the only remaining proposition, which is g(p, q), resulting in g(p, q) = 1. Stating this as p+q was an attempt to find something from which to further generalize.

Comment author: sebmathguy 14 June 2013 07:01:38PM *  -1 points [-]

My first reaction to the second question is to consider the case in which p + q = 1. Then, the answer is clearly that g(p, q) = p + q. I suspect that this is incomplete, and that further relevant information needs to be specified for the answer to be well-defined.

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