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sebmathguy comments on How should Eliezer and Nick's extra $20 be split - Less Wrong
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Comments (66)
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.
I think that when p+q=1, the answer is clearly 1/2 due to symmetry. How did you get p+q?
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.
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)
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.
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?
Yes. Woops.