Let’s say that the green premises brings the probability of “A new unobserved emerald is green.” to 99%. In the skeptic’s hypothesis, by symmetry it should also bring the probability of “A new unobserved emerald is grue.” to 99%. But of course after time T, this would mean that the probability of observing a green emerald is 99%, and the probability of not observing a green emerald is at least 99%, since these sentences have no intersection, i.e., they cannot happen together, to find the probability of their disjunction we just add their individual probabilities. This must give us a number at least as big as 198%...
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
But of course after time T
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.
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.