shokwave comments on Counterfactual Calculation and Observational Knowledge - Less Wrong

11 Post author: Vladimir_Nesov 31 January 2011 04:28PM

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Comment author: shokwave 02 February 2011 12:53:12AM 0 points [-]

Q is the same both in the counterfactual and the actual world.

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.

Comment author: FAWS 02 February 2011 01:55:36AM 0 points [-]

Q is the same both in the counterfactual and the actual world.

This doesn't square with my interpretation of the premises of the question.

From the original post:

Consider the counterfactual where the calculator displayed "odd" instead of "even", after you've just typed in the (same) formula Q, on the same occasion (i.e. all possible worlds that fit this description). The counterfactual diverges only in the calculator showing a different result (and what follows).

Clarified here and here.

Comment author: shokwave 02 February 2011 02:42:38AM 0 points [-]

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?

The parity of Q is already determined - Fermi and Neumann worked it out long ago and sealed it in a safe. You punch Q into the calculator and receive the answer "even". Omega appears, and asks you to consider the counterfactual where the calculator shows "odd". Omega offers to let factual you determine what is written on the sheet in the counterfactual world. No matter what is written down in either world, Fermi and Neumann's answer in the safe will remain the same.

Your factual-world observation of "even" on the calculator makes you think it very likely the counterfactual world is just the cases of the calculator being wrong. You would desire to have Omega write down "even" in the counterfactual world too.

But consider this situation:

The parity of Q is already determined - Fermi and Neumann worked it out long ago and sealed it in a safe. You punch Q into the calculator and receive the answer "odd". Omega appears, and asks you to consider the counterfactual where the calculator shows "even". Omega offers to let factual you determine what is written on the sheet in the counterfactual world. No matter what is written down in either world, Fermi and Neumann's answer in the safe will remain the same.

Your factual-world observation of "odd" on the calculator makes you think it very likely the counterfactual world is just the cases of the calculator being wrong. You would desire to have Omega write down "odd" in the counterfactual world too.

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?

Comment author: FAWS 02 February 2011 03:08:20AM *  0 points [-]

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.

Comment author: shokwave 02 February 2011 05:24:09AM 0 points [-]

Yes, that would be a counterfactual. But NOT the counterfactual under consideration.

I disagree. Recall that I specified this in each case:

The parity of Q is already determined - Fermi and Neumann worked it out long ago and sealed it in a safe.

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.

Comment author: FAWS 02 February 2011 05:48:49AM *  0 points [-]

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.

Comment author: Vladimir_Nesov 02 February 2011 11:33:32AM 0 points [-]

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.

Comment author: shokwave 02 February 2011 07:31:09AM 0 points [-]

Therefore you can reduce everything to the two cases I used,

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.

Comment author: FAWS 02 February 2011 04:58:47PM *  0 points [-]

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).

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