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antigonus 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: 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: potato 17 November 2011 10:59:11AM 0 points [-]

Correct, but my solution rests around there being a semantic method for testing greenness. This is what breaks the symmetry which the skeptic was abusing. Because the test stays the same the meaning of green stays the same.

Comment author: antigonus 17 November 2011 11:09:51AM 0 points [-]

I don't think I really understand what this means. Could you give more detail?

Comment author: potato 17 November 2011 11:30:24AM 0 points [-]

Read my conclusion over, I made some edits, if you still don't understand comment and i'll explain.

Comment author: antigonus 17 November 2011 07:44:12PM 3 points [-]

I'm not sure I've understood that very well, either. From what I can gather, it seems like you're arguing that 1. the meaning and physical tests for grue change over time, and consequently 2. grue is a more complicated property than green is, so we're justified in privileging the green hypothesis. If that's so, then I no longer see what role the reft/light example plays in your argument. You could've just started and finished with that.

Comment author: potato 17 November 2011 09:19:23PM 0 points [-]

yea, the reft light argument is just what made it obvious to me, i though it might help my readers to.