Fixed points and free will

I think this argument doesn't deserve anywhere near as much thought as you've given it. Caplan is committing a logical error, nothing else.

He probably reasoned as follows:

1. If determinism is true, I am computable.

2. Therefore, a large enough computer can compute what I will say.

3. Since my reaction is just more physics, those should be computable as well, hence it should also be possible to tell me what I will do after hearing the result.

This is wrong because "what Caplan outputs after seeing the prediction of our physics simualtor" is a system larger than the physics simulator and hence not computable by the physics simulator. Caplan's thought experiment works as soon as you make it so the physics simulator is not causally entangled with Caplan.

I don't think fixed points have any place in this analysis. Obviously, Caplan can choose to implement a function without a fixed point, like $f:x\mapsto -x$ (edit: rather $x\mapsto \neg x$), in fact he's saying this in the comment you quoted. The question is why he can do this, since (as by the above) he supposedly can't.

See also my phrasing of this problem and Richard_Kennaway's answer [LW · GW]. I think the real problem with his quote is that it's so badly phrased that the argument isn't even explicit, which paradoxically makes it harder to refute. You first have to reconstruct the argument, and then it gets easier to see why it's wrong. But I don't think there's anything interesting there.

I see a gap in the argument at the point where you "give" the agent a probability distribution, and it "gets" from that a probability distribution to use to make its choice. A probability distribution is an infinite object, in this case an assignment of a real number to each element of $A$. You can only give such a thing to a finite automaton by streaming it as a string of bits that progressively give more and more information about the distribution, approaching it in the limit. The automation must calculate from this a stream of bits that similarly represent its probability distribution, each bit of output depending on only a finite part of the input. However, it cannot use the distribution to calculate its action until it has the exact values, which it never will. (That is generally true even if the automaton is not calculating based on an input distribution. In practice, of course, one just uses the computer's IEEE 754 "reals" and hopes the result will be close enough. But that will not do for Mathematics.)

If it is going to repeatedly choose actions from the same distribution, then I imagine there may be ways for it to sample from increasingly good finite approximations to the distribution in such a way that as it approximates it more and more precisely, it can correct earlier errors by e.g. sampling an action with a little higher probability than the distribution specifies, to correct for too low an approximation used earlier. But for a single choice, a finite automaton cannot sample from a probability distribution specified as a set of real numbers, unless you allow it a tolerance of some sort.

Furthermore, to do these calculations, some representation of the real numbers must be used that makes the operations computable. The operations must be continuous functions, because all computable functions on real numbers are continuous. This generally means using an ambiguous representation of the reals, that gives more than one representation to most real numbers. Each real is represented by an infinite string drawn from some finite alphabet of "digits". However, the topology on the space of digit-strings in this context has to be the one whose open sets are all the strings sharing any given finite prefix. These are also the closed sets, making the space totally disconnected. Brouwer's fixed point theorem therefore does not apply. There are trivial examples of continuous functions without fixed points, e.g. permute the set of digits. Such a function may not correspond to any function on the reals, because multiple representations of the same real number might be mapped to representations of different real numbers.

I feel like a lot of the angst about free will boils down to conflicting intuitions. 

1. It seems like we live in a universe of cause and effect, thus all my actions/choices are caused by past events.
2. It feels like I get to actually make choices, so 1. obviously can't be right.

The way to reconcile these intuitions is to recognize that yes, all the decisions you make are in a sense predetermined, but a lot of what is determining those decisions is who you are and what sort of thing you would do in a particular circumstance. You are making decisions, that experience is not invalidated by a fully deterministic universe. It's just that you are who you are and you'll make the decision that you would make. 

This is the most interesting assessment of free will I've seen in a long time. I've thought about this a great deal before, particularly when trying to work out my feelings on this post [LW · GW]. Obviously we shouldn't expect any predictor to also be able to predict itself, for halting problem reasons. But yeah, why couldn't it shop around to find a way of telling the prediction which preserves the outcome? Eventually I decided that probably there was something like that, but it would look a lot more like coercion than prediction. I, for instance, could probably predict a persons actions fairly well if I was holding them at gunpoint. Omega could do the same by digging through possible sentences until it worked out how to tell you what you'd do without swaying the odds.

I agree with e.g. Rafael that Caplan's argument is just silly and wrong and doesn't deserve this much analysis.

But: I don't see how any version of this could possibly be the thing that makes Bryan Caplan say "oh, silly me, obviously my argument was no good". It doesn't amount to saying that there must be a way of saying "you will raise your hand" or "you will not raise your hand" that guarantees that he will do what you say. It's more like this: if you say "the probability that you will raise your hand is p" and he responds by raising his hand with probability q, and if q is a continuous function of p, then there is a choice of p for which q=p. True, but again: can you imagine this observation being convincing to him? How? Won't he just say "well, duh, you replaced my thought experiment with a different one and it came out differently; so what?"?

$g$ being continuous does not appear to actually help to resolve predictor problems, as a fixed point of 50%/50% left/right^[1] is not excluded, and in this case Omega has no predictive power over the outcome $A$^[2].

If you try to use this to resolve the Newcomb problem, for instance, you'll find that an agent that simply flips a (quantum) coin to decide does not have a fixed point in $f$, and does have a fixed point in $g$, as expected, but said fixed point is 50%/50%... which means the Omega is wrong exactly half the time. You could replace the Omega with an inverted Omega or a fair coin. They all have the same predictive power over the outcome $A$ - i.e. none.

Is there e.g. an additional quantum circuit trick that can resolve this? Or am I missing something?

1. ^{^}

   Or one-box/two-box, etc.

2. ^{^}

   For the same reason that a fair coin has no predictive power over a (different) fair coin.

My view of this is that Caplain (and likely the poster) are likely confused about what it means for physics to "predict" something.  Assuming something that looks vaguely like a multiverse interpretation is true, a full prediction is across the full set of downstream universes, not a single possible downstream universe out of the set.

From my standpoint, the only reason the future appears to be unpredictable is because we have this misguided notion that there is only "one" future, the one we will find ourselves in.  If the reality is that we're simultaneously in all of those possible futures, then a making a comprehensive future prediction has to contain all of them, and by containing all of them the prediction will be exact.

I think this is interesting because our understanding of physics seems to exclude effects that are truly discontinuous.

This is not true. An electron and a positron will, or will not, annihilate. They will not half-react.

For example, real-world transistors have resistance that depends continuously on the gate voltage

This is incorrect. It depends on the # of electrons, which is a discrete value. It's just that most of the time transistors are large enough that it doesn't really matter. That being said, it's absolutely important for things like e.g. flash memory. Modern flash memory cell might have ~400 electrons per cell^[1].

1. ^{^}

   https://electronics.stackexchange.com/questions/505361/how-many-excess-electrons-are-in-a-modern-slc-flash-memory-cell