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shokwave comments on Counterfactual Calculation and Observational Knowledge - Less Wrong
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This would be correct if Q could be different, but Q is the same both in the counterfactual and the actual word. There is no possibility for the actual world being Even World and the counterfactual Odd World.
The possibilities are:
Actual: Even World, Right Calculator (99% of Even Words); Counterfactual: Even World, Wrong Calculator (1% of Even Worlds).
Actual: Odd World, Wrong Calculator (1% of Odd Words); Counterfactual: Odd World, Right Calculator (99% of Odd Words).
The prior probability of either is 50%. If we assume That Omega randomly picks one you out of 100% of possible words(either 100% of all Even Worlds or 100% of all Odd Words) to decide for all possible words where the calculator result is different (but the correct answer is the same), then there is a 99% chance all worlds are Even and your choice affects 1% of all worlds and a 1% chance all words are Odd and your choice affects 99% of all worlds. The result of the calculator in the counterfactual world doesn't provide any evidence on whether all words are Even or all worlds are Odd since in either case there would be such a world to talk about.
If we assume that Omega randomly visits one world and randomly mentions the calculator result of one other possible world and it just happened to be the case that in that other world the result was different; or if Omega randomly picks a world, then randomly picks a world with the opposite calculator result and tosses a coin as to which world to visit and which to mention then the calculator result in the counterfactual word is equally relevant and hearing Omega talk about it just as good as running the calculator twice. In this case you are equally likely to be in a Odd world and might just as well toss a coin as to which result you fill in yourself.
This doesn't square with my interpretation of the premises of the question. We are unsure of Q's parity. Our prior is 50:50 odd, even. We are also unsure of calculator's trustworthiness. Our prior is 99:1 right, wrong. Therefore - on my understanding of counterfactuality - both options for both uncertainties need to be on the table.
I am unconvinced you can ignore your uncertainty on Q's parity by arguing that it will come out only one way regardless of your uncertainty - this is true for coinflips in deterministic physics, but that doesn't mean we can't consider the counterfactual where the coin comes up tails.
From the original post:
Clarified here and here.
We cannot determine Q's parity, except by fallible calculator. When you say Q is the same, you seem to be including "Q's parity is the same".
Hmm. Maybe this will help?
But consider this situation:
These situations are clearly the counterfactuals of each other - that is, when scenario 1 says "the counterfactual world" it is saying "scenario 2", and vice versa. The interpretations given in the second half of each contradict each other - the first scenario attempts to decide for the second scenario and gets it wrong; the second scenario attempts to decide for the first and gets it wrong. Whence this contradiction?
Yes, that would be a counterfactual. But NOT the counterfactual under consideration. The counterfactual under consideration was the calculator result being different but Q (both the number and the formula, and thus their parity) being the same. Unless Nesov was either deliberately misleading or completely failed his intention to clarify anything the comments linked to. If Q is the same formula is supposed to be clear in any way then everything about Q has to be the same. If the representation of Q in the formula was supposed be the same, but the actual value possibly counterfactually different then only answering that the formula is the same is obscuration, not clarification.
I disagree. Recall that I specified this in each case:
Q (both the number and the formula, and thus the parity) is the same in both scenarios. The actual value is not counterfactually different - it's the same value in the safe, both times.
If you agree that Q's parity is the same I'm not sure what you are disagreeing with. Its not possible for Q to be odd in the counterfactual and even in actuality, so if Q is odd in the counterfactual that implies it is also odd in actuality and vice versa. Thus it's not possible for the calculator to be right in both counterfactual and reality simultaneously, and assuming it to be right in the counter-factual implies that it's wrong in actuality. Therefore you can reduce everything to the two cases I used, Q even/actual calculator right/counterfactual calculator wrong or Q odd/actual calculator wrong/counterfactual calculator right.
Maybe this could be more enlightening. When you control things, one of the necessary requirements is that you have logical uncertainty about some property of the thing you control. You start with having a definition of the control target, but not knowing some of its properties. And then you might be able to infer a dependence of one of its properties on your action. This allows you to personally determine what is that property of a structure whose definition you already know. See my posts on ADT for more detail.
I have been positing that these two cases are counterfactuals of each other. Before one of these two cases occurs, we don't know which one will occur. It is possible to consider being in the other case.
The problem is symmetrical. You can just copy everything, replace odd with even and vice versa and multiply everything with 0.5, then you also have the worlds where you see odd and Omega offers you to replace the result in counterfactuals where it came up even and where Q has the same parity. Doesn't change that Q is the same in the world that decides and the counterfactuals that are effected. Omega also transposing your choice to impossible worlds (or predicting what would happen in impossible worlds and imposing that on what happens in real worlds) would be a different problem (that violates that Q be the same in the counterfactual, but seems to be the problem you solved).