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damang comments on Bayes Slays Goodman's Grue - Less Wrong Discussion

0 Post author: potato 17 November 2011 10:45AM

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Comment author: damang 16 November 2011 10:45:56PM 1 point [-]

He never said where it was, the problem was to find where the line was on the table.

Comment author: endoself 16 November 2011 11:15:25PM 1 point [-]

A better objection would be to ask whether the line chooved, where 'chooved' means 'stayed in the same place before time T and moved to the opposite location afterward'.

Comment author: antigonus 16 November 2011 10:52:12PM *  0 points [-]

To be honest, I don't have a clear sense of what he's saying. However, from a snippet like this:

As long as the line doesn't move, and we know what symbols we are using for what results of what random variables, observing many green emeralds is evidence that the next one will be grue, as long as it is before time T, after time T, every record of an observation of a green emerald is evidence against the next one being grue in the same way it is evidence against the next emerald being not being green.

it sounds like he's trying to draw some conclusion from an assumption ("the line doesn't/won't move") that ultimately rests on inductive support. Is that not the case? If so, how does that supposed support not fall victim to the new problem of induction?

Comment author: potato 17 November 2011 02:34:08AM 0 points [-]

It doesn't seem to me like the problem is to justify induction, but to justify the induction on green over grue after time T.

Comment author: antigonus 17 November 2011 02:49:51AM *  0 points [-]

The problem is to justify any inductively-obtained statement vulnerable to a grue-like variant. "X will remain in the same place" is one such statement. (Namely: Any evidence that X will remain in a given place is prima facie evidence that it will same in the same place', where place' refers to its current location at T and some other location afterward.) Grue is just an example.

Comment author: potato 17 November 2011 03:00:00AM 2 points [-]

That the line will stay in the same place is not something I induce, it is a premise in the hypothetical. The line, or really the area of right of the line on the table, represents the actual frequency with which an emerald turns out green, out of all the cases where an emerald is observed, this is certainly a non-moving line, since there is one and only one answer to that question.

Comment author: antigonus 17 November 2011 07:24:10AM 2 points [-]

That the line will stay in the same place is not something I induce, it is a premise in the hypothetical.

But that's question-begging. Let me put this another way. Define the function reft-distance(x) = x's distance to the rightmost edge of the table before time T, or the distance to the leftmost edge of the table after time T. (Then "x is reft of y" is definable as reft-distance(x) < reft-distance(y). Similarly for the function light-distance(x).) Assuming the line doesn't move is equivalent to assuming that the line's right-distance remains constant, but that its reft-distance changes after T. But that's not a fair assumption, the skeptic will insist: he prefers to assume the line doesn't "anti-move," which means its reft-distance remains constant but its right-distance changes.

If we're simply stipulating that your assumption (that the line doesn't move) is correct and the skeptic's assumption (that the line doesn't anti-move) is incorrect, that's not very useful. We might as well just stipulate that emeralds remain green for all time or whatever.

Comment author: potato 17 November 2011 07:38:43AM *  0 points [-]

you forgot to adress this part:

The line, or really the area of right of the line on the table, represents the actual frequency with which an emerald turns out green, out of all the cases where an emerald is observed, this is certainly a non-moving line, since there is one and only one answer to that question.

The line is constant because the area to its right represents the frequency with which a certain result is observed out of the number of trials. What the skeptic would have to be assuming is that the first 98 balls just happened to fall on the first 100th of the table by chance.

Comment author: antigonus 17 November 2011 07:45:14AM *  0 points [-]

Assuming that the line is constant is analogous to assuming that emeralds' color won't change after T, correct? The skeptic will refuse to do either of these, preferring instead to assume that the line is anti-constant and that emeralds' anti-color won't change after T.

Comment author: potato 17 November 2011 10:27:49AM *  0 points [-]

Assuming that the line is constant is analogous to assuming that emeralds' color won't change after T, correct?

No, that's a common misunderstanding. No emerald ever has to change color for the grue hypothesis to be true

It is analogous to assuming that there is a definite frequency of green emeralds out of emeralds ever made.

Comment author: antigonus 17 November 2011 10:54:43AM *  0 points [-]

No, that's a common misunderstanding. No emerald ever has to change color for the grue hypothesis to be true

Well, O.K. "The next observed emerald is green if before T and blue otherwise" doesn't entail any change of color. I suppose I should have said, "Analogous to assuming that the emeralds' color (as opposed to anti-color) distribution doesn't vary before and after T."

It is analogous to assuming that there is a definite frequency of green emeralds out of emeralds ever made.

I'm really not seeing that analogy. It seems more analogous to assuming there's a single, time-independent probability of observing a green emerald. (Holding the line fixed means there's a single, time-independent probability of landing right of the line.) Which is again an assumption the skeptic would deny, preferring instead the existence of a single, time-invariant probability of observing a grue emerald.

Comment author: alex_zag_al 17 November 2011 06:20:03AM *  2 points [-]

It seems that the assumption in your hypothetical is of an unchanging process producing the random variable, about which we have partial knowledge. In the case of the ball, we know of the unmoving invisible line, the throws uniformly distributed over the table, and whatever mechanism it is that lets us know whether the ball has fallen to the left or the right of the line. However, we don't know enough to know exactly where the balls will land. In the case of the emeralds, we know enough about the emerald construction sites to know that they are grue-blind, and that they will stay grue-blind no matter how many emeralds they produce. In both cases, we know something of the mechanism behind the random variable, and that it will not change. Is that correct?

You talk of a threat to the whole of science. How does your answer to this hypothetical respond to that threat? Do scientists ever have the knowledge assumed in your hypotheticals? How can scientists gain that knowledge in the first place without getting grued up, if they need it that knowledge to stay gruefree? It reminds me of Bugs Bunny pulling himself out of a magicians hat, by holding his ears.

Comment author: potato 17 November 2011 07:35:54AM *  1 point [-]

It seems that the assumption in your hypothetical is of an unchanging process producing the random variable

It is not, my assumption is of a definite frequency with which some result comes, out of trials.

When you realize that the reason you don't determine the meaning of green using grue and bleen because there is a physical test which has higher authority in defining greenhood, the threat disolves.

Comment author: alex_zag_al 17 November 2011 04:02:38PM *  0 points [-]

By “frequency” I suppose you mean the fraction of balls dropped on the right out of all ball drops, past and future? And with emeralds... I guess you mean the fraction of green emeralds out of all emeralds that hbe been or will be observed?

I suppose the physical test in the ball problem is the ball landing on one side or the other of the line. In the emerald problem, the physical test is, what is it?