Problematic Problems for TDT

A key goal of Less Wrong's "advanced" decision theories (like TDT, UDT and ADT) is that they should out-perform standard decision theories (such as CDT) in contexts where another agent has access to the decider's code, or can otherwise predict the decider's behaviour. In particular, agents who run these theories will one-box on Newcomb's problem, and so generally make more money than agents which two-box. Slightly surprisingly, they may well continue to one-box even if the boxes are transparent, and even if the predictor Omega makes occasional errors (a problem due to Gary Drescher, which Eliezer has described as equivalent to "counterfactual mugging"). More generally, these agents behave like a CDT agent will wish it had pre-committed itself to behaving before being faced with the problem.

However, I've recently thought of a class of Omega problems where TDT (and related theories) appears to under-perform compared to CDT. Importantly, these are problems which are "fair" - at least as fair as the original Newcomb problem - because the reward is a function of the agent's actual choices in the problem (namely which box or boxes get picked) and independent of the method that the agent uses to choose, or of its choices on any other problems. This contrasts with clearly "unfair" problems like the following:

Discrimination: Omega presents the usual two boxes. Box A always contains $1000. Box B contains nothing if Omega detects that the agent is running TDT; otherwise it contains $1 million.

 

So what are some fair "problematic problems"?

Problem 1: Omega (who experience has shown is always truthful) presents the usual two boxes A and B and announces the following. "Before you entered the room, I ran a simulation of this problem as presented to an agent running TDT. I won't tell you what the agent decided, but I will tell you that if the agent two-boxed then I put nothing in Box B, whereas if the agent one-boxed then I put $1 million in Box B. Regardless of how the simulated agent decided, I put $1000 in Box A. Now please choose your box or boxes."

Analysis: Any agent who is themselves running TDT will reason as in the standard Newcomb problem. They'll prove that their decision is linked to the simulated agent's, so that if they two-box they'll only win $1000, whereas if they one-box they will win $1 million. So the agent will choose to one-box and win $1 million.

However, any CDT agent can just take both boxes and win $1001000. In fact, any other agent who is not running TDT (e.g. an EDT agent) will be able to re-construct the chain of logic and reason that the simulation one-boxed and so box B contains the $1 million. So any other agent can safely two-box as well. 

Note that we can modify the contents of Box A so that it contains anything up to $1 million; the CDT agent (or EDT agent) can in principle win up to twice as much as the TDT agent.

 

Problem 2: Our ever-reliable Omega now presents ten boxes, numbered from 1 to 10, and announces the following. "Exactly one of these boxes contains $1 million; the others contain nothing. You must take exactly one box to win the money; if you try to take more than one, then you won't be allowed to keep any winnings. Before you entered the room, I ran multiple simulations of this problem as presented to an agent running TDT, and determined the box which the agent was least likely to take. If there were several such boxes tied for equal-lowest probability, then I just selected one of them, the one labelled with the smallest number. I then placed $1 million in the selected box. Please choose your box."

Analysis: A TDT agent will reason that whatever it does, it cannot have more than 10% chance of winning the $1 million. In fact, the TDT agent's best reply is to pick each box with equal probability; after Omega calculates this, it will place the $1 million under box number 1 and the TDT agent has exactly 10% chance of winning it.
 
But any non-TDT agent (e.g. CDT or EDT) can reason this through as well, and just pick box number 1, so winning $1 million. By increasing the number of boxes, we can ensure that TDT has arbitrarily low chance of winning, compared to CDT which always wins.


Some questions:

1. Have these or similar problems already been discovered by TDT (or UDT) theorists, and if so, is there a known solution? I had a search on Less Wrong but couldn't find anything obviously like them.

2. Is the analysis correct, or is there some subtle reason why a TDT (or UDT) agent would choose differently from described?

3. If a TDT agent believed (or had reason to believe) that Omega was going to present it with such problems, then wouldn't it want to self-modify to CDT? But this seems paradoxical, since the whole idea of a TDT agent is that it doesn't have to self-modify.

4. Might such problems show that there cannot be a single TDT algorithm (or family of provably-linked TDT algorithms) so that when Omega says it is simulating a TDT agent, it is quite ambiguous what it is doing? (This objection would go away if Omega revealed the source-code of its simulated agent, and the source-code of the choosing agent; each particular version of TDT would then be out-performed on a specific matching problem.)

5. Are these really "fair" problems? Is there some intelligible sense in which they are not fair, but Newcomb's problem is fair? It certainly looks like Omega may be "rewarding irrationality" (i.e. giving greater gains to someone who runs an inferior decision theory), but that's exactly the argument that CDT theorists use about Newcomb.

6. Finally, is it more likely that Omegas - or things like them - will present agents with Newcomb and Prisoner's Dilemma problems (on which TDT succeeds) rather than problematic problems (on which it fails)?

 

Edit: I tweaked the explanation of Box A's contents in Problem 1, since this was causing some confusion. The idea is that, as in the usual Newcomb problem, Box A always contains $1000. Note that Box B depends on what the simulated agent chooses; it doesn't depend on Omega predicting what the actual deciding agent chooses (so Omega doesn't put less money in any box just because it sees that the actual decider is running TDT).

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You can construct a "counterexample" to any decision theory by writing a scenario in which it (or the decision theory you want to have win) is named explicitly. For example, consider Alphabetic Decision Theory, which writes a description of each of the options, then chooses whichever is first alphabetically. ADT is bad, but not so bad that you can't make it win: you could postulate an Omega which checks to see whether you're ADT, gives you $1000 if you are, and tortures you for a year if you aren't.

That's what's happening in Problem 1, except that it's a little bit hidden. There, you have an Omega which says: if you are TDT, I will make the content of these boxes depend on your choice in such a way that you can't have both; if you aren't TDT, I filled both boxes.

You can see that something funny has hapened by postulating TDT-prime, which is identical to TDT except that Omega doesn't recognize it as a duplicate (eg, it differs in some way that should be irrelevant). TDT-prime would two-box, and win.

Right, but this is exactly the insight of this post put another way. The possibility of an Omega that rewards eg ADT is discussed in Eliezer's TDT paper. He sets out an idea of a "fair" test, which evaluates only what you do and what you are predicted to do, not what you are. What's interesting about this is that this is a "fair" test by that definition, yet it acts like an unfair test.

Because it's a fair test, it doesn't matter whether Omega thinks TDT and TDT-prime are the same - what matters is whether TDT-prime thinks so.

He sets out an idea of a "fair" test, which evaluates only what you do and what you are predicted to do, not what you are.

Two questions: First, how does is this distinction justified? What a decision theory is is a strategy for responding to decision tasks and simulating agents performing the right decision tasks tells you what kind of decision theory they're using. Why does it matter if it's done implicitly (as in Newcomb's discrimination against CDT) or explicitly. And second why should we care about it? Why is it important for a decision theory to pass fair tests but not unfair tests?

Why is it important for a decision theory to pass fair tests but not unfair tests?

Well, on unfair tests a decision theory still needs to do as well as possible. If we had a version of the original Newcomb's problem, with the one difference that a CDT agent gets $1billion just for showing up, it's still incumbent upon a TDT agent to walk away with $1000000 rather than $1000. The "unfair" class of problems is that class where "winning as much as possible" is distinct from "winning the most out of all possible agents".

Real-world unfair tests could matter, though it's not clear if there are any. However, hypothetical unfair tests aren't very informative about what is a good decision theory, because it's trivial to cook one up that favours one theory and disfavours another. I think the hope was to invent a decision theory that does well on all fair tests; the example above seems to show that may not be possible.

Because it's a fair test

No, not even by Eliezer's standard, because TDT is not given the same problem than other decision theories.

As stated in comments below, everyone but TDT have the information "I'm not in the simulation" (or more precisely, in one of the simulations of the infinite regress that is implied by Omega's formulation). The reason TDT does not have this extra piece of information comes from the fact that it is TDT, not from any decision it may make.

Right, and this is an unfairness that Eliezer's definition fails to capture.

At this point, I need the text of that definition.

The definition is in Eliezer's TDT paper although a quick grep for "fair" didn't immediately find the definition.

This variation of the problem was invented in the follow-up post (I think it was called "Sneaky strategies for TDT" or something like that:

Omega tells you that earlier he flipped a coin. If the coin came down heads, it simulated a CDT agent facing this problem. If the coin came down tails, it simulated a TDT agent facing this problem. In either case, if the simulated agent one-boxed, there is $1000000 in Box-B; if it two-boxed Box-B is empty. In this case TDT still one-boxes (50% chance of $1000000 dominates a 100% chance of $1000), and CDT still two-boxes (because that's what CDT does). In this case, even though both agents have an equal chance of being simulated, CDT out-performs TDT (average payoffs of 500500 vs. 500000) - CDT takes advantage of TDT's prudence and TDT suffers for CDT's lack of it. Notice also that TDT cannot do better by behaving like CDT (both would get payoffs of 1000). This shows that the class of problems we're concerned with is not so much "fair" vs. "unfair", but more like "those problem on which the best I can do is not necessarily the best anyone can do". We can call it "fairness" if we want, but it's not like Omega is discriminating against TDT in this case.

This is not a zero-sum game. CDT does not outperform TDT here. It just makes a stupid mistake, and happens to pay it less dearly than TDT

Let's say Omega submit the same problem to 2 arbitrary decision theories. Each will either 1-box or 2-box. Here is the average payoff matrix:

  • Both a and b 1-box -> They both get the million
  • Both a and b 2-box -> They both get 1000 only.
  • One 1-boxes, the other 2-boxes -> the 1-boxer gets half a million, the other gets 5000 more.

Clearly, 1 boxing still dominates 2-boxing. Whatever the other does, you personally get about half a million more by 1-boxing. TDT may have less utility than CDT for 1-boxing, but CDT is still stupid here, while TDT is not.

Not exactly. Because the problem statement says that it simulates "TDT", if you were to expand the problem statement out into code it would have to contain source code to a complete instantiation of TDT. When the problem statement is run, TDT or TDT-prime can look at that instantiation and compare it to its own source code. TDT will see that they're the same, but TDT-prime will notice that they are different, and thereby infer that it is not the simulated copy. (Any difference whatsoever is proof of this.)

Consider an alternative problem. Omega flips a coin, and asks you to guess what it was, with a prize if you guess correctly. If the coin was heads, he shows you a piece of paper with TDT's source code. If the coin was tails, he shows you a piece of paper with your source code, whatever that is.

I'm not sure the part about comparing source code is correct. TDT isn't supposed to search for exact copies of itself, it's supposed to search for parts of the world that are logically equivalent to itself.

The key thing is the question as to whether it could have been you that has been simulated. If all you know is that you're a TDT agent and what Omega simulated is a TDT agent, then it could have been you. Therefore you have to act as if your decision now may either real or simulated. If you know you are not what Omega simulated (for any reason), then you know that you only have to worry about the 'real' decision.

Suppose that Omega doesn't reveal the full source code of the simulated TDT agent, but just reveals enough logical facts about the simulated TDT agent to imply that it uses TDT. Then the "real" TDT Prime agent cannot deduce that it is different.

Yes. I think that as long as there is any chance of you being the simulated agent, then you need to one box. So you one box if Omega tells you 'I simulated some agent', and one box if Omega tells you 'I simulated an agent that uses the same decision procedure as you', but two box if Omega tells you 'I simulated an agent that had a different copywrite comment in its source code to the comment in your source code'.

This is just a variant of the 'detect if I'm in a simulation' function that others mention. i.e. if Omega gives you access to that information in any way, you can two box. Of course, I'm a bit stuck on what Omega has told the simulation in that case. Has Omega done an infinite regress?

That's an interesting way to look at the problem. Thanks!

Indeed. These are all scenarios of the form "Omega looks at the source code for your decision theory, and intentionally creates a scenario that breaks it." Omega could do this with any possible decision theory (or at last, anything that could be implemented with finite resources), so what exactly are we supposed to learn by contemplating specific examples?

It seems to me that the valuable Omega thought experiments are the ones where Omega's omnipotence is simply used to force the player to stick to the rules of the given scenario. When you start postulating that an impossible, acausal superintelligence is actively working agaisnt you it's time to hang up your hat and go home, because no strategy you could possibly come up with is going to do you any good.

The trouble is when another agent wins in this situation and in the situations you usually encounter. For example, an anti-traditional-rationalist, that always makes the opposite choice to a traditional rationalist, will one-box; it just fails spectacularly when asked to choose between different amounts of cake.

You can see that something funny has hapened by postulating TDT-prime, which is identical to TDT except that Omega doesn't recognize it as a duplicate (eg, it differs in some way that should be irrelevant). TDT-prime would two-box, and win.

I don't think so. If TDT-prime two boxes, the TDT simulation two-boxes, so only one box is full, so TDT-prime walks away with $1000. Omega doesn't check what decision theory you're using at all - it just simulates TDT and bases its decision on that. I do think that this ought to fall outside a rigorously defined class of "fair" problems, but it doesn't matter whether Omega can recognise you as a TDT-agent or not.

I don't think so. If TDT-prime two boxes, the TDT simulation two-boxes, so only one box is full, so TDT-prime walks away with $1000.

No, if TDT-prime two boxes, the TDT simulation still one-boxes.

Hmm, so TDT-prime would reason something like, "The TDT simulation will one-box because, not knowing that it's the simulation, but also knowing that the simulation will use exactly the same decision theory as itself, it will conclude that the simulation will do the same thing as itself and so one-boxing is the best option. However, I'm different to the TDT-simulation, and therefore I can safely two-box without affecting its decision." In which case, does it matter how inconsequential the difference is? Yep, I'm confused.

I also had thoughts along these lines - variants of TDT could logically separate themselves, so that T-0 one-boxes when it is simulated, but T-1 has proven that T-0 will one-box, and hence T-1 two-boxes when T-0 is the sim.

But a couple of difficulties arise. The first is that if TDT variants can logically separate from each other (i.e. can prove that their decisions aren't linked) then they won't co-operate with each other in Prisoner's Dilemma. We could end up with a bunch of CliqueBots that only co-operate with their exact clones, which is not ideal.

The second difficulty is that for each specific TDT variant, one with algorithm T' say, there will be a specific problematic problem on which T' will do worse than CDT (and indeed worse than all the other variants of TDT) - this is the problem with T' being the exact algorithm running in the sim. So we still don't get the - desirable - property that there is some sensible decision theory called TDT that is optimal across fair problems.

The best suggestion I've heard so far is that we try to adjust the definition of "fairness", so that these problematic problems also count as "unfair". I'm open to proposals on that one...

But a couple of difficulties arise. The first is that if TDT variants can logically separate from each other (i.e. can prove that their decisions aren't linked) then they won't co-operate with each other in Prisoner's Dilemma. We could end up with a bunch of CliqueBots that only co-operate with their exact clones, which is not ideal.

I think this is avoidable. Let's say that there are two TDT programs called Alice and Bob, which are exactly identical except that Alice's source code contains a comment identifying it as Alice, whereas Bob's source code contains a comment identifying it as Bob. Each of them can read their own source code. Suppose that in problem 1, Omega reveals that the source code it used to run the simulation was Alice. Alice has to one-box. But Bob faces a different situation than Alice does, because he can find a difference between his own source code and the one Omega simulated, whereas Alice could not. So Bob can two-box without effecting what Alice would do.

However, if Alice and Bob play the prisoner's dilemma against each other, the situation is much closer to symmetric. Alice faces a player identical to itself except with the "Alice" comment replaced with "Bob", and Bob faces a player identical to itself except with the "Bob" comment replaced with "Alice". Hopefully, their algorithm would compress this information down to "The other player is identical to me, but has a comment difference in its source code", at which point each player would be in an identical situation.

You might want to look at my follow-up article which discusses a strategy like this (among others). It's worth noting that slight variations of the problem remove the opportunity for such "sneaky" strategies.

In a prisoners dilemma Alice and Bob affect each others outcomes. In the newcomb problem, Alice affects Bobs outcome, but Bob doesn't affect Alices outcome. That's why it's OK for Bob to consider himself different in the second case as long as he knows he is definitely not Alice (because otherwise he might actually be in a simulation) but not OK for him to consider himself different in the prisoners dilemma.

However, if Alice and Bob play the prisoner's dilemma against each other, the situation is much closer to symmetric. Alice faces a player identical to itself except with the "Alice" comment replaced with "Bob", and Bob faces a player identical to itself except with the "Bob" comment replaced with "Alice". Hopefully, their algorithm would compress this information down to "The other player is identical to me, but has a comment difference in its source code", at which point each player would be in an identical situation.

Why doesn't that happen when dealing with Omega?

Because if Omega uses Alice's source code, then Alice sees that the source code of the simulation is exactly the same as hers, whereas Bob sees that there is a comment difference, so the situation is not symmetric.

So why doesn't that happen in the prisoner's dilemma?

Because Alice sees that Bob's source code is the same as hers except for a comment difference, and Bob sees that Alice's source code is the same as his except for a comment difference, so the situation is symmetric.

Newcomb:

Bob sees that there is a comment difference, so the situation is not symmetric.

Prisoner's Dilemma:

Bob sees that Alice's source code is the same as his except for a comment difference, so the situation is symmetric.

Do you see the contradiction here?

Newcomb, Alice: The simulation's source code and available information is literally exactly the same as Alice's, so if Alice 2-boxes, the simulation will too. There's no way around it. So Alice one-boxes.

Newcomb, Bob: The simulation was in the situation described above. Bob thus predicts that it will one-box. Bob himself is in an entirely different situation, since he can see a source code difference, so if he two-boxes, it does not logically imply that the simulation will two-box. So Bob two-boxes and the simulation one-boxes.

Prisoner's Dilemma: Alice sees Bob's source code, and summarizes it as "identical to me except for a different comment". Bob sees Alice's source code, and summarizes it as "identical to me except for a different comment". Both Alice and Bob run the same algorithm, and they now have the same input, so they must produce the same result. They figure this out, and cooperate.

Ignore Alice's perspective for a second. Why is Bob acting differently? He's seeing the same code both times.

Don't ignore Alice's perspective. Bob knows what Alice's perspective is, so since there is a difference in Alice's perspective, there is by extension a difference in Bob's perspective.

Bob looks at the same code both times. In the PD, he treats it as identical to his own. In NP, he treats it as different. Why?