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by Daniel_Eth
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Imagine a variation of Newcomb's problem where the Oracle is correct 99% of the time (say, half of people 1-box, half 2-box, and the Oracle is correct 99% of the time on each). What would FDT imply is the correct course of action then?

Functional Decision TheoryNewcomb's ProblemWorld Modeling
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What does Functional Decision Theory say to do in imperfect Newcomb situations?
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Stephen Bennett

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The expected value under FDT would be:

  • 1-boxing: 0.99 * big prize + 0.01 * 0

  • 2-boxing: 0.99 * small prize + 0.01 * (small prize + big prize)

Making a decision based on that will depend on the specifics of the problem (how much bigger is the big prize than the small prize?) and your circumstances (what is your utility function with respect to the big and small prizes?)

[-]Daniel_Eth20

Hmm, does this not depend on how the Oracle is making its decision? I feel like there might be versions of this that look more like the smoking lesion problem – for instance, what if the Oracle is simply using a (highly predictive) proxy to determine whether you'll 1-box or 2-box? (Say, imagine if people from cities 1-box 99% of the time, and people from the country 2-box 99% of the time, and the Oracle is just looking at where you're from).

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2Stephen Bennett
It seems like this might become a discussion of Aleatory vs Epistemic Uncertainty. I like this way of describing the distinction between the two (from here - pdf): I believe that the differences between classical decision theory and FDT's only occur in the context of aleatory uncertainty (although in some formulations of newcomb's paradox there's no actual uncertainty). That is, if you are in an epistemically uncertain environment, then FDT and classical decision theory will agree on all problems (hopefully by saying this I can cause someone to come up with a juicy counterexample). In your example, it is unclear to me what sort of uncertainty the problem possesses because I don't know enough about the oracle. In the simple example where a quantum coin with a 99% chance of coming up heads is flipped to determine whether the oracle gives the right answer or the wrong answer, then the answer I gave above is right. Use expected value under the assumptions of FDT; classical decision theory will lead you to 2-box and that would lower your expected gains. In your example relying on demographic information, it will depend a bit on what sorts of information count as "demographic" in nature. If you are, in this moment by reading this comment on lesswrong, forming the sort of self that will result in you 1-boxing or 2-boxing and that information is also an input to this sort of oracle, then I encourage you to 1-box on the oracle you had in mind.

JBlack

10

The exact values of the payoff depend upon the unknown details of how the Oracle achieves its 99% accuracy. Here is an inconvenient scenario that is consistent with the given description:

Suppose that 99% of the population don't even think about the decision, they just follow inherent preferences that are nearly equally prevalent. The Oracle gets almost all of these correct, but fails on 1% of inherent one-boxers giving them nothing instead of $1,000,000. In the remaining cases where people do actually think about it, the Oracle is always wrong, and everyone who thinks about the situation knows this.

Since you are actually thinking about it, then you're one of the people for whom the Oracle's prediction is always wrong. If you end up taking one box, then it will be empty. If you take both boxes, you will find $1,001,000. In this scenario CDT, EDT, and FDT all agree that you should take both boxes.

In many other scenarios, possibly even "most" in some ill-defined sense, FDT says you should take one box.

1 Related Questions

7JBlack
Just to follow up on that third point a little more: FDT depends upon counterfactual responses, how you would have responded to inputs that you didn't in fact observe. If you go into a scenario where you can observe even as little as 6 bits of information, then there are 2^6 = 64 possible inputs to your decision function. FDT requires that you adopt the function with the greatest expected value over the weighted probabilities of every input, not just the one you actually observed. In the simplest possible case, each output is just one of two deterministic actions and there are only 2^(2^6) = 18446744073709551616 such functions to evaluate. If your space of possible actions is any larger than a deterministic binary action, then there will be vastly more such functions to evaluate. Unlike both CDT and EDT, FDT is massively super-exponential in computational complexity. Pruning the space down to something that is feasibly solvable by actual computers is not something that we know how to do for any but the most ridiculously simple scenarios.
2Shmi
Hmm, if FDT is so complex, how can humans evaluate it? Where does the pruning happen and how?
3JBlack
Mostly they can't, which is why there are a lot more questions posted about it than people who answer correctly. I can't think of any FDT problem that has been answered correctly where there were more than 3 binary inputs to the decision function, and even some with 2 bits have been controversial. For the few cases where they can, it's the same way that humans solve any mathematical problem: via an ill defined bunch of heuristics, symmetry arguments, experience with similar problems, and some sort of intuition or insight.
2Shmi
Hmm, that limits its usefulness quite a bit. For math, one can at least write an unambguous expression and use CAS like mathematica or maple and click "solve for ..." Would be nice to have something like that for various DTs.
2Pattern
Before making this comment, the comment section said:   Edit: as soon as I posted this comment, it changed to 
2Answer by Pattern
FDT operates on a causal graph. (Maybe the other requirements are also hard to satisfy?* I think this is the major obstacle.) You would probably have to create the graph, for the problem, then pass it to the program (if you write one). One could argue that the trick in real world situations is figuring out the causal graph anyway.   When people talk about Bayesian methods here, they don't seem like they're using code for the Monte-Carlo stuff, or stuff like that.   Edit: *JBlack's answer indicates this the other requirements, are an issue.
9Answer by jessicata
There's a Haskell implementation of modal UDT. Previous discussion of modal UDT: 1, 2. FDT is a family of decision theories, modal UDT is a specific algorithm. As stated it requires a hypercomputer, however has bounded variants.
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