Decision Theory and the Context Set

Consider the following decision problem. You have two choices, which we’ll call 1 and 2. If you choose option 1, you’ll get $1,000,000. If you choose option 2, you’ll get $1,000. There are no other consequences of your actions, and you prefer more money to less. What should you do?

It sounds easy enough right? You should take option 1. I think that’s the right answer, but getting clear as to why it is the right ansewr, and what question it is the right answer to, is a little tricky.

Here’s something that’s consistent with the initial description of the case. You’re in a version of Newcomb’s problem. Option 1 is taking one box, Option 2 is taking two boxes. You have a crystal ball, and it perfectly reliably detects (via cues from the future) whether the demon predicted your choice correctly. And she did; so you know (with certainty) that if you pick option one, you’ll get the million, and if you pick option two, you’ll get the thousand. Still, I think you should pick option one, since you should pick option one in any problem consistent with the description in the first paragraph. (And I think that’s consistent with causal decision theory, properly understood, though the reasons why that is so are a little beyond the scope of this post.)

Here’s something else that’s consistent with a flat-footed interpretation of the case, though not I think with the intended interpretation. Option 2 is a box with $1000 in it. Option 1 is a box with a randomly selected lottery ticket in it, and the ticket has an expected value of $1. Now as a matter of fact, it will be the winning ticket, so you will get $1,000,000 if you take option 1. Still, if everything you know is that option 2 is $1,000, and option 1 is a $1 lottery ticket, you should take option 2.

Now I don’t think that undermines what I said above. And I don’t think it undermines it because when we properly interpret descriptions of games/decision problems, we’ll see that this situation isn’t among the class of decision problems described in the first paragraph. When we describe the outcomes of certain actions in a decision problem, those aren’t merely the actual outcomes, they are things that are properly taken as fixed points in the agent’s deliberation. They are, in Stalnakerian terms, the limits of the context set. In the lottery ticket example, it is not determined by the context set that you’ll get $1,000,000 if you take option 1, even though it is in fact true.

I think “things the agent properly takes as fixed points” are all and only the things the agent knows, but that’s a highly controversial theory of knowledge. (In fact, it’s my version of interest-relative invariantism.) So rather than wade into that debate, I’ll simply talk about proper fixed points.

Saying that something is a fixed point is a very strong claim. It means the agent doesn’t even, shouldn’t even, consider possibilities where they fail. So in Newcomb’s problem, the agent shouldn’t be at all worrying about possibilities where the demon miscounts the money she puts into box 1 or 2. Or possibilities where there is really a piranha in box 2 who’ll bite your hand, rather than $1000. And when I say that she shouldn’t be worrying about them, I mean they shouldn’t be in the algebra of possibilities over which her credences are defined.

There’s a big difference formally between something being true at all points over which a probability function is defined, and something (merely) having probability 1 according to that function. And that difference is something that I’m relying on heavily here. In particular, I think the following two things are true.

First, when we state something in the set up of a problem, then we say that the agent can take it as given for the purposes of a problem.

Second, when we are considering the possible outcomes of a situation, the only situations we need to consider are ones that are not fixed points. So in my version of Newcomb’s Problem, the right thing to do is to take one box, because there is no outcome where you do better than taking one box. On the other hand, some things that we now know to be false might (in some sense of might) become relevant, even though we now assign them probability 0. That’s what goes on in cases where backwards induction fails; the context set shifts over the course of the game, and so we have to take into account new things.

Having said all that, there is one hard question that I don’t know the answer to. It’s related to some things that Adam Elga, John Collins and Andy Egan were discussing at a reading group on the weekend. In the kind of puzzle cases that we usually consider in textbooks, the context set consists of the Cartesian product of some hypotheses about the world, and some choices. That’s to say, the context set satisfies this principle: If S is a possible state of the world (excluding my choice), and C is a possible choice, then there is a possibility in the context set where S and C both obtain. I wonder if that’s something we should always accept. I’ll leave the pros and cons of accepting that proposal for another post though.

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Defending Causal Decision Theory

Late last year I wrote up a small note replying to Andy Egan’s paper Some Counterexamples to Causal Decision Theory (PDF). I was hoping to revise it a little and then post it, but I haven’t had the chance to do the revisions (and fill in the references), but it looks interesting enough that it might be worth having up here. Without further ado…

Defending Causal Decision Theory

The core argument is that if Andy is right about what modifications are needed to causal decision theory, then we end up saying bizarre things about cases where there are three choices available. I think it’s less costly to simply keep causal decision theory than to say those bizarre things, though this is a bit of a judgment call.

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Lewis on Google Scholar

I’m (slowly) writing the entry on David Lewis for the Stanford Encyclopaedia. Here’s the tentative table of contents.

1. Convention and Linguistic Meaning
2. Counterfactuals
3. Philosophy of Mind
4. Modal Metaphysics
5. Everything Else

The last section could do with a snappier title. But the idea is that I start with the two early books, and the papers that build directly on those books (esp “Languages and Language” and “Time’s Arrow”). Then I look at what I think of as the three big themes of Lewis’s career. These are (a) his theory of mind, (b) his reductionism about the nomic (and related topics), (c) modal realism and its consequences for metaphysics, especially modal metaphysics.

The problem is that this leaves out quite a lot. For instance, it leaves out practically everything from “Papers in Philosophical Logic” and “Papers in Ethics and Social Philosophy”. But I do think that trying to find another theme on a par with those three would amount to shoehorning material into a category in which it doesn’t quite fit. (Not that the three themes are entirely distinct.)

But that doesn’t mean I shouldn’t say anything about the rest of Lewis’s career. So I was wondering what I should focus on outside those five sections. To that end, I made a crude search on Google Scholar of which were Lewis’s most cited papers. The full results are below the fold, but the top 15 is a little surprising.

Note that the “II” in “Probabilities of Conditionals and Conditional Probabilities” is misleading. Google Scholar thinks that the two papers with roughly this title are just one paper, and it has merged their citations together.

That the books are up top is no surprise. Books generally do much better than papers on Google Scholar. And it isn’t a surprise to some extent that older papers lead the way. They have more time to collect citations. But the showing of the language papers, and in particular the formal semantics papers, is quite stunning. I think I follow Lewis, and Lewisiania, quite a bit, and I can’t recall the last time I saw someone cite “General Semantics”, for instance. So maybe this isn’t the best measure of the importance and influence of the various works.

Full table, and methodology, below the fold. Read the rest of this entry »

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Against Ratificationism

In The Logic of Decision, second edition, Richard Jeffrey endorsed a version of evidential decision theory supplemented with a principle of ratification. A decision is ratifiable iff, conditional on making that decision, the expected utility of carrying out the decision is at least as high as than the expected utility of doing something else. In Newcomb’s Problem, although one-boxing has highest expected utility, only two-boxing is ratifiable. In zero sum games with each agent knowing the other agent’s strategy and no Nash equilibrium, only mixed strategies are ratifiable. So there might be some cases where the principle looks attractive. But it isn’t defensible in general. Here I’ll argue against a relatively weak version of the ratification principle.

Weak Ratificationism: In a decision problem with only one ratifiable strategy, that strategy is uniquely rational.

In our puzzle, the Agent has to name some non-negative integer. Call this number a. After the Agent names the integer, another number will be displayed on a scoreboard. The other number, call it s, will be generated by a process that is causally independent of the Agent’s choice. The Agent’s payout is as follows.

• If a = s + 1, and s > 0, then the Agent gets 2 utils
• If either a = 0 or s = 0, the Agent gets 0 utils
• Otherwise, the Agent gets 1 util

The number on the scoreboard will be the number named by a psychological Twin of the agent (called Twin) who is playing the same game. Agent believes, with probability 1, that Twin will follow the same strategy as Agent in producing the number. This doesn’t mean Twin will produce the same number – both players may be following a mixed strategy of some kind. If Agent is following the mixed strategy that leads to them saying 1 with probability 0.4, and saying 2 with probability 0.6, then Twin is following the same mixed strategy. Since the probability of Agent saying 2 and Twin saying 1 is 0.24, Agent’s expected return is 1.24.

The only ratifiable strategy for the Agent is to say 0. This is ratifiable, because given the Agent says 0, then with probability 1 the Twin says 0, so the expected return of any strategy is 0, so saying 0 is as good as any other strategy.

Clearly this is the only ratifiable pure strategy for the Agent. If they say any number greater than 0, say x, then saying x+1 has an expected return of 2 (conditional, that is, on having said x), while saying x has an expected return of 1.

But there are no other ratifiable mixed strategies in the game. Or at least there aren’t if we assume, as I do, countable additivity for probabilities. Given countable additivity, any mixed strategy must be representable by a function f from non-negative integers to [0, 1] such that the various values of f(x) sum to 1 as x ranges over the non-negative integers. When no ambiguity results, we’ll let f name both a function and the strategy it represents. Let F be the sum, as x ranges over the positive integers, of f(x)f(x+1). Then the expected return of f is (1 – f(0))² + F. (The first term is the probability, given that both players are playing f, of Agent’s return being at least 1. The second term is the probability, given that same assumption, of Agent’s return being 2.)

If Agent decides to play f, the Twin will play f with probability 1. We want to prove that no f, other than the f that makes f(0) = 1, is ratifiable. To this end, we aim to construct, for any other f, a rival strategy g such that the expected return of g is greater than the expected return of f, conditional on Twin playing f.

Note that if Twin plays f, and Agent plays g, then Agent’s expected return is (1 – f(0))(1 – g(0)) plus the sum, as x ranges over positive integers, of f(x)g(x+1). The first term is the probability of Agent’s return being at least 1, and the second is the probability of it being 2.

Given that, it is easy to construct g. Let k be the smallest number such that f(k) is positive. Let g agree with f over every value except k and k+1. In those cases, g(k) = 0 and g(k+1) = f(k) + f(k+1). It’s easy to see that making this change doesn’t lower the value of any term in the sum that is the expected return of g, and raises at least one term. (Either the first term, if k = 0, or the value for x=k otherwise.) So f is not ratifiable; given it is being played, g has a higher expected return.

So the only ratifiable strategy is to simply play 0. But this strategy is guarantees getting the worst possible result in the game, and is weakly dominated by every other strategy. (That is, there are some possibilities where the alternative strategy does better, and none where it does worse.) Perhaps, if you know that things will go maximally wrong for you come what may, a strategy that guarantees the worst possible outcome might be rationally permissible, as it doesn’t make things worse. But it is impossible to see how a strategy that guarantees the worst outcome could be uniquely rational. So Weak Ratificationism fails.

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Is there a rational strategy in Finite Iterated Prisoners Dilemma?

(UPDATE: I think there’s a mistake in the argument here – see Bob Stalnaker’s comment 11 below.)

Row and Column are going to play 100 rounds of Prisoners Dilemma. At each round they can either play Co-op or Defect, with standard rules. (So the payoffs are symmetric, and on each round Defect dominats Co-op for each player, but each playing Co-op is Pareto superior to each playing Defect.) The following is true at the start of the game.

1. Each player is rational.
2. No player loses a belief that they have if they receive evidence that is consistent with that belief.
3. For any r, if it is true that if a player were to play Co-op on round r, the other player would play Defect on every subsequent round, then it is also true that if the first player were to play Defect on round r, then the other player would still play Defect on every subsequent round.
4. The first three premises are matters of common belief.

Call a strategy that a player can play consistent with those four assumptions an approved strategy. (Note that one of the assumptions is that the player is rational, so these will all be rational strategies.) Assume for reductio that there are approved strategies S1 and S2 such that if Column plays S2, then Row can play S1, and this will involve sometimes playing Co-op. I will try to derive a contradiction from that assumption.

Let r be the largest number such that there are approved strategies S1, S2 and if Column is playing S2, and Row plays S1, then Row plays Co-op on round r. I will now argue that it is irrational for Row to play Co-op on round r, contradicting the assumption that S1 is an approved strategy.

Since both players are playing approved strategies, they are both acting consistently with the initial assumptions. So by premise 2, the initial assumptions still hold, and this is a matter of common belief. So it is still a matter of common belief that each player is playing an approved strategy.

If Row plays Co-op on round r, that is still, by hypothesis, an approved strategy, so Column would react by sticking to her approved strategy, by another application of premise 2. Since r is the last round under which any playing an approved strategy against an approved strategy co-operates, and Column is playing an approved strategy, Row believes that if she were to play Co-op, Column would play Defect on every subsequent round. By premise 3 (or, more precisely, by her belief that premise 3 still holds), Row can infer that Column will also play Defect on every subsequent round if she plays Defect on this round.

Putting these two facts together, Row believes prior to playing this round that whatever she were to to, Column would react by playing Defect on every subsequent round. If that’s the case, then she would get a higher return by playing Defect this round, since the only reason to ever play Co-op is that it has an effect on play in later rounds. But it will have no such effect. So it is uniquely rational for Row to play Defect at this round. But this contradicts our assumption that S1 is a rational strategy, and according to it Row plays Co-op on round r.

If our assumption is true, then there can be no approved strategy that ever co-operates before observing the other player co-operate. If there were such a strategy, call it S3, then we can imagine a game where both players play S3. By hypothesis there is a round r where the player playing S3 co-operates before the other player co-operates. So if both players play S3, which is approved, then they will both play Defect up to round r, then play Co-op on that round. But that’s to say that they will play Co-op while (a) playing an approved strategy and (b) believing that the other playing will play an approved strategy. And this contradicts our earlier result.

This does not mean that a rational player can never co-operate, but it does mean that they can never co-operate while the initial assumptions are in place. A rational player might, for instance, co-operate on seeing that her co-player is playing tit-for-tat, and hence that the initial assumptions are not operational.

Nor does it mean, as I think some theorists have been too quick to conclude, that playing Defect all the time is an approved, or even a rational, strategy. Assume that there are approved strategies, and that (as we’ve shown so far) they all involve playing Defect on the first round. Now the familiar objections to backward induction reasoning, tracing back at least to Philip Pettit and Robert Sugden’s “The Backward Induction Paradox”, become salient objections.

If Row holds all the initial assumptions, she may also believe that if she were to play Co-op on the first round, then Column would infer that she is an irrational agent, and that as such she’ll play Tit-for-Tat. (This isn’t built into the original assumptions, but it is consistent with them.) And if Row believes that is how Column would react, then Row is rational to play Co-op, or at least more rational on this occasion than playing Defect. Indeed, even if Row thinks there is a small chance that if she plays Co-op, Column will conclude that she is irrationally playing Tit-for-Tat, then the expected return of playing Co-op will be higher, and hence it will be rational. I conclude that, given any kind of plausible assumptions Row might have about Column’s beliefs, playing Co-op on the first round is rational.

In their paper, Pettit and Sugden try to make two arguments. The first I’ve very quickly excerpted here – namely that the assumption that always Defect is uniquely rational leads to contradiction given minimal assumption about Row’s beliefs about how Column would react. The second, if I’m reading them correctly, is that rational players may play some strategy other than always Defect. The argument for the second conclusion involves rejecting premise 2 of my model. They rely on cases where players react to rational strategies by inferring the other player is irrational, or believes they are irrational, or believes they believe that they are irrational etc. Such cases are not altogether implausible, but it is interesting to think about what happens without making such a possibility.

And I conclude that given my initial assumption, there is no approved strategy. And I’m tempted to think that’s because there is no rational strategy to follow. Just like in Death in Damascus, any strategy a player might follow, they have reason to believe is an irrational strategy when they are playing it. This is a somewhat depressing conclusion, I think, but causal decision theory sometimes doesn’t give us straightforward advice, and I suspect finite iterated Prisoners Dilemma, at least given assumptions like my premise 2, is a case where causal decision theory doesn’t give us any advice at all.

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Asymmetric Death in Damascus

Here is the ‘Death in Damascus’ case from Allan Gibbard and William Harper’s classic paper on causal decision theory.

Consider the story of the man who met Death in Damascus. Death looked surprised, but then recovered his ghastly composure and said, ‘I am coming for you tomorrow’. The terrified man that night bought a camel and rode to Aleppo. The next day, Death knocked on the door of the room where he was hiding, and said I have come for you’.

‘But I thought you would be looking for me in Damascus’, said the man.

‘Not at all’, said Death ‘that is why I was surprised to see you yesterday. I knew that today I was to find you in Aleppo’.

Now suppose the man knows the following. Death works from an appointment book which states time and place; a person dies if and only if the book correctly states in what city he will be at the stated time. The book is made up weeks in advance on the basis of highly reliable predictions. An appointment on the next day has been inscribed for him. Suppose, on this basis, the man would take his being in Damascus the next day as strong evidence that his appointment with Death is in Damascus, and would take his being in Aleppo the next day as strong evidence that his appointment is in Aleppo…

If… he decides to go to Aleppo, he then has strong grounds for expecting that Aleppo is where Death already expects him to be, and hence it is rational for him to prefer staying in Damascus. Similarly, deciding to stay in Damascus would give him strong grounds for thinking that he ought to go to Aleppo.

Causal decision theorists often say that in these cases, there is no rational thing to do. Whatever the man does, he will (when he does it) have really good evidence that he would have been better off if he did something else. Evidential decision theorists often say that this is a terrible consequence of causal decsion theory, but it seems plausible enough to me. It’s bad to make choices that bring about your untimely death, or that you have reason to believe will bring about your untimely death, and that’s what the man here does. So far I’m a happy causal decision theorist.

But let’s change the original case a little. (The changes are similar to the changes in Andy Egan’s various counterexamples to causal decision theory.) The man wants to avoid death, and he believes that Death will predict where he will go tomorrow, and go there tomorrow, and that he’ll die iff he is where Death is. But he has other preferences too. Let’s say that his live options are to spend the next 24 hours somewhat enjoyably in Las Vegas, or exceedingly unpleasanty in Death Valley. Then you might think he’s got a reason to go to Vegas; he’ll die either way, but it will be a better end in Vegas than Death Valley.

Let’s make this a little more precise with some demons and boxes. There is a demon who is, as usual, very good at predicting what you’ll do. The demon has put two boxes, A and B, on the table in front of you, and has put money in them by the following rules.

• If the demon has predicted that you’ll choose and take A, then the demon put $1400 in B, and $100 in A.
• If the demon has predicted that you’ll choose and take B, then then demon put $800 in A, and $700 in B.
If the demon has predicted that you’ll play some kind of mixed strategy, then the demon has put no money in either box, because the demon doesn’t stand for that kind of thing.

What should you do? Three possible answers come to mind.

Answer 1: If you take box A, you’ll probably get $100. If you take box B, you’ll probably get $700. You prefer $700 to $100, so you should take box A.

Verdict: WRONG!. This is exactly the reasoning that leads to taking one box in Newcomb’s problem, and one boxing is wrong. (If you don’t agree, then you’re not going to be in the target audience for this post I’m afraid.)

Answer 2: There’s nothing you can rationally do. If you choose A, you would have been better off choosing B, and you’ll know this. If you choose B, you would have been better off choosing A, and you’ll know this. If you walk away, or mentally flip a coin, you’ll get nothing, which seems terrible.

Verdict: I think correct, but three worries.

First, the argument that the mixed strategy is irrational goes by a little quickly. If you are sure you are going to play a mixed strategy, then you couldn’t do any better than by playing it, so it isn’t obviously irrational. So perhaps what’s really true is that if you know that you aren’t going to play a mixed strategy, then playing a mixed strategy would have a lower payoff than playing some pure strategy. For instance, if you are playing B, then if you had have played the mixed strategy (Choose B with probability 0.5, Choose A with probability 0.5), your expected return would have been $750, which is less than the $800 that you would have got if you’d chosen A. And this generalises to any pure strategy that you choose, and any mixed strategy that you could have chosen as an alternative; whatever two strategies you pick, there is a pure strategy that you could have chosen that would have been better. So for anyone who’s not playing a mixed strategy, it would be irrational to play a mixed strategy. And I suspect that condition covers all readers.

Second, this case seems like a pretty strong argument against Richard Jeffrey’s preferred view of using evidential decision theory, but restricting attention to the ratifiable strategies. Only mixed strategies are ratifiable in this puzzle, but mixed strategies seem absolutely crazy here. So don’t restrict yourself to ratifiable strategies.

Third, it seems odd to give up on the puzzle like this. Here’s one way to express our dissatisfaction with answer two. The puzzle is quite asymmetric; box B is quite different to box A in terms of its outcome profile. But our answer is symmetric; either pure strategy is irrational from the perspective of someone who is planning to play it. Perhaps we can put that dissatisfaction to work.

Answer 3: If you choose A, you could have done much much better choosing B. If you choose B, you could have done a little better choosing A. So B doesn’t look as bad as A by this measure. So you should choose B.

Verdict: Tempting, but ultimately I think inconsistent.

I think the intuitions that Andy pumps with his examples are really driven by something like this reasoning. But I don’t think the reasoning really works. Here is a less charitable, but I think more revealing, way of putting the reasoning.

Choosing A is really irrational. Choosing B is only a bit irrational. Since as rational agents we want to minimise irrationality, we should choose B, since that is minimally irrational.

But it should be clear why that can’t work. If choosing B is what rational agents do, i.e. is rational, then one of the premises of our reasoning is mistaken. B is not a little bit irrational, rather, it is not irrational at all. If choosing B is irrational, as the premises state, then we can’t conclude that it is rational.

The only alternative is to deny that B is even a little irrational. But that seems quite odd, since choosing B involves doing something that you know, when you do it, is less rewarding than something else you could just as easily have done.

So I conclude Answer 2 is correct. Either choice is less than fully rational. There isn’t anything that we can, simply and without qualification, say that you should do. This is a problem for those who think decision theory should aim for completeness, but cases like this suggest that this was an implausible aim.

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Decision Theory Notes

I’ve been continuing to write my decision theory notes for the decision theory course I’ve been doing this term. The course hasn’t ended up quite the way I expected. Because I wanted to go into more details on some of the fundamentals, and because I wanted to avoid heavy lifting in the mathematics, I’ve skipped anything to do with infinities. So that meant cutting countable additivity, the two-envelope problem, the St Petersburg Paradox, etc. But the flip side of that is that there’s more than originally intended on the nature of utility, and the most recent additions have been going quite slowly through the foundations of game theory.

One thing that I suspect is not news to many readers of this site, but which was very striking to me, was how much orthodox game theory resembles evidential decision theory. (Many people will be familiar with this because it is a point that Robert Stalnaker has made well. But writing the notes really drove home for me how true it is.)

It is really hard to offer a motivation for exclusively playing equilibrium strategies in one-shot zero-sum games that doesn’t look like a motivation for taking one-box in Newcomb’s problem. Since us causal decision theorists think taking one-box is irrational, this makes me very suspicious of the normative significance of equilibrium strategies in one-shot games. Of course, most real world games are not one-shot, and I can understand playing a mixed strategy in a repeated zero-sum game. But it is much harder to see why we should play a mixed strategy in a one-shot game.

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Newcomb’s Centipede

The following puzzle is a cross between the Newcomb puzzle and the centipede game.

You have to pick a number between 1 and 50, call that number u. A demon, who is exceptionally good at predictions, will try to predict what you pick and will pick a number d, between 1 and 50, that is 1 less than u. If she predicts u is 1, then she can’t do this, so she’ll pick 1 as well. The demon’s choice is made before your choice is made, but only revealed after your choice is made. (If the demon predicts that you’ll use a mixed strategy to choose u, she’ll set d equal to 1 less than the lowest number that you have some probability of choosing.)

Depending on what numbers the two of you pick, you’ll get a reward by the formula below.

If u is less than or equal to d, your reward will be 2u.
If u is greater than d, your reward will be 2d – 1.

For an evidential decision theorist, it’s clear enough what you should do. Almost certainly, your payout will be 2u – 3, so you should maximise u, so you should pick 50, and get a payout of 97.

For a causal decision theorist, it’s clear enough what you should not do. We know that the demon won’t pick 50. If the demon won’t pick 50, then picking 49 has a return that’s better than picking 50 if d = 49, and as good as picking 50 in all other circumstances. So picking 49 dominates picking 50, so 50 shouldn’t be picked.

Now the interesting question. What should you pick if you’re a causal decision theorist. I know of three arguments that you should pick 1, but none of them sound completely convincing.

Backwards Induction
The demon knows you’re a causal decision theorist. So the demon knows that you won’t pick 50. So the demon won’t pick 49; she’ll pick at most 48. If it is given that the demon will pick at most 48, then picking 48 dominates picking 49. So you should pick at most 48. But the demon knows this, so she’ll pick at most 47, and given that, picking 47 dominates picking 48. Repeating this pattern several times gives us an argument for picking 1.

I’m suspicious of this because it’s similar to the bad backwards induction arguments that have been criticised effectively by Stalnaker, and by Pettit & Sugden. But it’s not quite the same as the arguments that they criticised, and perhaps it is successful.

Two Kinds of Conditionals
In his very interesting The Ethics of Morphing, Caspar Hare appears to suggest that causal decision theorists should be sympathetic to something like the following principle. (Caspar stays neutral between evidential and causal decision theory, so it isn’t his principle. And the principle might be slightly stronger than even what he attributes to the causal decision theorist, since I’m not sure the translation from his lingo to mine is entirely accurate. Be that as it may, this idea was inspired by what he said, so I wanted to note the credit.)

Say an option is unhappy if, supposing you’ll take it, there is another option that would have been better to take, and an option is happy if, supposing you take it, it would have been worse to have taken other options. Then if one option is happy, and the others all unhappy, you should take the happy option.

Every option but picking 1 is unhappy. Supposing you pick n, greater than 1, the demon will pick n-1, and given that you would have been better off picking n-1. But picking 1 is happy. Supposing that, the demon will pick 1, and you would have been worse off picking anything else.

There’s something to the pick happy options principle, so this argument is somewhat attractive. But this does seem like a bad consequence of the principle.

Stable Probability
In Lewis’s version of causal decision theory, we have to look at the probability of various counterfactuals of the form If I were to pick n, I would get k dollars. But we aren’t really told where those probabilities come from. In the Newcomb problem that doesn’t matter; whatever probabilities we assign, two boxing comes out best. But the probabilities matter a lot here.

Now it isn’t clear what constrains the probabilities in question, but I think the following sounds like a sensible constraint. If you pick n, the probability the demon picks n-1 (or n if n=1) should be very high. That’s relevant, because the counterfactuals in question (what would I have got had I picked something else) are determined by what the demon picks.

Here’s a constraint that seems plausible. Say an option is Lewis-stable if, conditional on your picking it, it has the highest “causally expected utility”. (“Causally expected utility” is my term for the value that Lewis thinks we should try to maximise.) Then the constraint is that if there’s exactly one Lewis-stable option, you should pick it.

Again, it isn’t too hard to see that only 1 is Lewis-stable. So you should pick it.

Summary
It seems intuitively wrong to me to pick 1. It doesn’t dominate the other options. Indeed, unless the demon picks 1, it is the worst option of all. And I like causal decision theory. So I’d like a good argument that the causal decision theorist should pick something other than 1. But I’m worried (a) that causal decision theory recommends taking 1, and (b) that if that isn’t true, it makes no recommendation at all. I’m not sure either is a particularly happy result.

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Different Ideas About Newcomb Cases

One advantage of going to parties with mathematicians and physicists is that you can describe a problem to them, and sometimes they’ll get stuck thinking about it and come up with an interesting new approach to it, different from most of the standard ones. This happened to me over the past few months with Josh von Korff, a physics grad student here at Berkeley, and versions of Newcomb’s problem. He shared my general intuition that one should choose only one box in the standard version of Newcomb’s problem, but that one should smoke in the smoking lesion example. However, he took this intuition seriously enough that he was able to come up with a decision-theoretic protocol that actually seems to make these recommendations. It ends up making some other really strange predictions, but it seems interesting to consider, and also ends up resembling something Kantian!

The basic idea is that right now, I should plan all my future decisions in such a way that they maximize my expected utility right now, and stick to those decisions. In some sense, this policy obviously has the highest expectation overall, because of how it’s designed.

In the standard Newcomb case, we see that adopting the one-box policy now means that you’ll most likely get a million dollars, while adopting a two-box policy now means that you’ll most likely get only a thousand dollars. Thus, this procedure recommends being a one-boxer.

Now consider a slight variant of the Newcomb problem. In this version, the predictor didn’t set up the boxes, she just found them and looked inside, and then investigated the agent and made her prediction. She asserts the material biconditional “either the box has a million dollars and you will only take that box, or it has nothing and you will take both boxes”. Looking at this prospectively, we see that if you’re a one-boxer, then this situation will only be likely to emerge if there’s already a box with a million dollars there, while if you’re a two-boxer, then it will only be likely to emerge if there’s already an empty box there. However, being a one-boxer or two-boxer has no effect on the likelihood of there being a million dollars or not in the box. Thus, you might as well be a two-boxer, because in either situation (the box already containing a million or not) you get an extra thousand dollars, and you just get the situation described to you differently by the predictor.

Interestingly enough, we see that if the predictor is causally responsible for the contents of the box then we should follow evidential decision theory, while if she only provides evidence for what’s already in the box then we should follow causal decision theory. I don’t know how much people have already discussed this aspect of the causal structure of the situation, since they seem to focus instead on whether the agent is causally responsible, rather than the predictor.

Now I think my intuitive understanding of the smoking lesion case is more like the second of these two problems – if the lesion is actually determining my behavior, then decision theory seems to be irrelevant, so the way I seem to understand the situation has to be something more like a medical discovery of the material biconditional between my having cancer and smoking

Here’s another situation that Josh described that started to make things seem a little more weird. In Ancient Greece, while wandering on the road, every day one either encounters a beggar or a god. If one encounters a beggar, then one can choose to either give the beggar a penny or not. But if one encounters a god, then the god will give one a gold coin iff, had there been a beggar instead, one would have given a penny. On encountering a beggar, it now seems intuitive that (speaking only out of self-interest), one shouldn’t give the penny. But (assuming that gods and beggars are randomly encountered with some middling probability distribution) the decision protocol outlined above recommends giving the penny anyway.

In a sense, what’s happening here is that I’m giving the penny in the actual world, so that my closest counterpart that runs into a god will receive a gold coin. It seems very odd to behave like this, but from the point of view before I know whether or not I’ll encounter a god, this seems to be the best overall plan. But as Josh points out, if this was the only way people got food, then people would see that the generous were doing well, and generosity would spread quickly.

If we now imagine a multi-agent situation, we can get even stronger (and perhaps stranger) results. If two agents are playing in a prisoner’s dilemma, and they have common knowledge that they are both following this decision protocol, then it looks like they should both cooperate. In general, if this decision protocol is somehow constitutive of rationality, then rational agents should always act according to a maxim that they can intend (consistently with their goals) to be followed by all rational agents. To get either of these conclusions, one has to condition one’s expectations on the proposition that other agents following this procedure will arrive at the same choices.

Of course this is all very strange. When I actually find myself in the Newcomb situation, or facing the beggar, I no longer seem to have a reason to behave according to the dictates of this protocol – my actions benefit my counterpart rather than myself. And if I’m supposed to make all my decisions by making this sort of calculation, then it’s unclear how far back in time I should go to evaluate the expected utilities. This matters if we can somehow nest Newcomb cases, say by offering a prize if I predict that you will make the “wrong” decision on a future Newcomb case. It looks like I have to calculate everything all the way back at the beginning, with only my a priori probability distribution – which doesn’t seem to make much sense. Perhaps I should only go back to when I adopted this decision procedure – but then what stops me from “re-adopting” it at some later time, and resetting all the calculations?

At any rate, these strike me as some very interesting ideas.

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FEW 2007

Fourth Annual Formal Epistemology Workshops, FEW 2007

We are in the process of organizing the fourth annual formal epistemology workshop. The purpose of these workshops will be to bring together individuals, both faculty and graduate students, using mathematical methods in epistemology in small focused meetings. Topics treated will include but are not limited to:

• Ampliative inference (including inductive logic);
• Game theory and decision theory;
• Formal learning theory;
• Formal theories of coherence:
• Foundations of probability and statistics;
• Formal approaches to paradoxes of belief and/or action;
• Belief revision;

• Causal discovery.

Besides papers with respondents, each workshop will typically include short introductory tutorials (three or four topically related presentations) on formal methods. These tutorials will be oriented particularly to graduate students.

The fourth workshop is scheduled for 31 May – 3 June 2007 and will be held at Carnegie Mellon University in Pittsburgh.

FEW 2007 SUBMISSION DEADLINES: – Paper submissions (at least a 1/2 page abstract): By Monday, Feb. 26th. – Notification: By Monday, March 30th.

Please send submissions by email to Branden Fitelson <branden@fitelson.org>.

Those interested in participating, either by presenting papers, responding, or providing tutorials, or in helping with organization, should contact one of the organizers listed below. Limited funds are available for graduate student travel. (A workshop web-site will shortly be set up. It will appear at the usual location: http://socrates.berkeley.edu/~fitelson/few/.)

Richard Scheines <scheines@andrew.cmu.edu> CMU

Sahotra Sarkar <sarkar@mail.utexas.edu> UT-Austin

Branden Fitelson <branden@fitelson.org> UC-Berkeley

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