shokwave comments on Bayes Slays Goodman's Grue - Less Wrong Discussion
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Let's do actual Bayesian math on this problem. Let Green_n be "the green premises 1 through n", and so on.
Pr( An emerald is grue | Emerald is green, it is before time T ) = ~1.
Pr( An emerald is grue | Emerald is green, it is after time T ) = ~0.
Pr( An emerald is grue | Emerald is blue, it is before time T ) = ~0.
Pr( An emerald is grue | Emerald is blue, it is after time T ) = ~1.
These are our grue axioms - the probabilistic representation of "grue iff green before time T or blue after time T".
Pr( New Emerald is green | Green_n ) = 0.99 (this is our first sentence axiom)
Pr( New Emerald is grue | Emerald is green ) = undefined. We need to know if we are pre-T or post-T. Without the prior probability for being pre-T (from which we can derive its complement, post-T, or vice versa).
But that is wussing out; Bayesian agents should always be able to assign some level of credence. Assume maximum ignorance about T: it is equally likely to be pre-T or post-T.
We can find Pr( New Emerald is grue | Emerald is green ) by finding Pr( New Emerald is grue | Emerald is green, pre-T ) and adding it to Pr( New Emerald is grue | Emerald is green, post-T ) then normalising.
The pre-T case: Pr( New Emerald is grue | Green_n, pre-T ) = Pr( emerald is grue | emerald is green, pre-T ) * Pr( emerald is green). We know that Pr( emerald is green ) is 0.99. We have Pr( emerald is grue | emerald is green, pre-T ) as an axiom of ~1 above. So 0.99 * ~1 = ~0.99.
The post-T case: Pr( New Emerald is grue | Green_n, post-T ) = Pr( emerald is grue | emerald is green, post-T ) * Pr( emerald is green). We know that Pr( emerald is green ) is 0.99. We have Pr( emerald is grue | emerald is green, post-T ) as an axiom of ~0 above. So 0.99 * ~0 = ~0.
Normalising gives us: Pr( New Emerald is grue | Emerald is green ) = ~.495.
This is for the case where we don't know if pre-T or post-T. When you say
we can ask a better question than "Pr( New Emerald is grue | Emerald is green ) ?". We can ask Pr( An emerald is grue | Emerald is green, it is after time T ). This is an axiom from before! We now know it's ~0, which resolves the problem: the probabilities sum to 0.99 + epsilon, which is below 1, which conserves the Kolmogorov axioms.
Whence the problem? The error creeps in when you use the pre-T case to get one .99, and then you use the complement of the post-T case to get another .99, and then add them together. If you specify pre-T or post-T, but then swap T in calculating some of the posterior probabilities, of course you can violate probability theory! You're already violating it by varying the state of T within the calculation!
If I am not mistaken, this is my independent formulation of the formal Bayesian resolution of the Goodman Grue paradox.