More on wagers for the perfectly rational

Consider a choice between two wagers on a fair coin:

• W₁: on heads, you get $1 if you are perfectly rational and $3 if you are not

• W₂: on tails, you get $2 if you are perfectly rational and $1 if you are not.

Suppose you are perfectly rational, and that it’s a part of perfect rationality that you know for sure you’re perfectly rational. It’s obvious you should go for W₂. But let’s calculate. We immediately run into the zero-probability problem that I’ve lately been thinking about. For if you’re perfectly rational, the probability that you go for W₁ is zero, so E(U|W₁) seems to be undefined. Of course, E(U|W₂) is unproblematically half of $2, or $1, but you can’t say whether that beats “undefined” or not.

Suppose you think: Maybe E(U|W₁) is undefined in classical probability, but maybe I can use some other way of defining it, say using Popper functions.

Well, let’s think about what E(U|W₁) “should be”. So imagine that you actually go for W₁. Now, only an imperfectly rational agent would go for W₁. So, if you were to go for W₁, you would get $3 on heads, so your expected payoff would be $1.50, which beats anybody’s expected payoff for W₂. So, formally, E(U|W₁) is undefined, but if you close your eyes to that and think intuitively, you get E(U|W₁) equally $1.50, which yields the wrong result that as a perfectly rational agent you should go for W₁.

What if we say that a perfectly rational agent need not know for sure that they are perfectly rational? Suppose, say, you are perfectly rational agent who is 0.99 sure you are perfectly rational. Then E(U|W₁) and E(U|W₂) are both well-defined. But what are they? Well, it’s intuitively clear that if you are 0.99 sure that you are perfectly rational, you should go for W₂. But supposing that’s right, then W₁ entails you are not perfectly rational, and since P(W₁) = 0.01, the expectation E(U|W₁) is well-defined, and must be equal to $1.50. Oops!

This line of reasoning assumed evidential decision theory. What if you go for causal decision theory? Well, there are two causal hypotheses: R (you are perfectly rational) and R^c (you are not) with P(R) = 0.99 and P(R^c) = 0.01. So now your causal expected utility on W₁ equals

• CE(U|W₁) = 0.99E(U|W₁∩R) + 0.01E(U|W₂∩R^c).

What is this? Well, W₁ ∩ R is the empty set! But conditionalizing on an empty set is not a merely technical problem in the way that conditionalizing on a specific zero-probability outcome of a continuous spinner is. Rather, it is simply nonsense. So the first summand is undefined, and hence the sum is undefined. Thus you simply cannot make a decision with causal decision theory here.

It’s obvious that if you’re nearly sure you’re perfectly rational you should go for W₂. But neither evidential nor causal decision theory gives a way to that conclusion.

[By the way, the reason I set up W₁ and W₂ as I did, with one having the payoff on heads and the other on tails, was to ensure that we didn’t have domination. For one might reasonably say that a perfectly rational agent will try to decide on grounds of domination first, before resorting to probabilities.]

Good's Theorem, perfect rationality, and conditioning on zero probability events

Recently, I found myself puzzled by the difficulty in applying “classical” evidential decision theory to a perfectly rational agent. The problem was that the rational agent decides whether to do A or B based on a comparison between the conditional expectations E(U|A) and E(U|B) of the utility function U. But supposing that in fact E(U|A) > E(U|B), the perfectly rational agent has no chance of doing B, so P(B) = 0, and hence E(U|B) is undefined.

But then I thought this isn’t a big deal, because we aren’t perfectly rational agents, so we always have a chance of screwing up and hence P(B) > 0 even if E(U|B) is much less than E(U|A).

I am not entirely satisfied with this. After all, you might think: “I may be pretty imperfect, but if I am choosing between a donut D and a year of torture T, I have zero chance of choosing the year of torture. But then E(U|T) is undefined, so how am I being rational in this choice? Maybe that’s a good objection, maybe not.

But here is another reason why the “We’re imperfect” solution isn’t completely ideal. We want to say that Good’s Theorem tells us something important about rationality—namely, that more information makes rational agents make better decisions. Good’s Theorem is usually interpreted as saying that under some independence conditions, the expected value of a perfectly rational choice given more information is no less than that of a perfectly rational choice given less information. Notice that this is obviously false in the case of an imperfectly rational agent. Thus, we have to make sense of “What a perfectly rational agent would choose” to make sense of the standard interpretation of Good’s Theorem. Moreover, in the setting of Good’s Theorem, the perfectly rational agent has to be choosing based on expected utilities—and that’s precisely what generates the zero-probability-conditioning problem.

Now, the Theorem is still true as an abstract bit of mathematics. But the application is difficult if we can’t make sense of a perfectly rational agent who is certain to maximize expected utility.

Likely we can extend Good’s Theorem to talk about the limiting case of imperfect agents getting more and more perfect. But it would be nice if we didn’t have to.

Extending Good's Theorem to experiments and not just observations

Good’s Theorem basically says that a utility-maximizing agent can expect to make decisions that are at least as good if they get more information. (And under some additional conditions, one can expect the decisions to be better.)

Now consider this case:

1. You will be offered a chance to make a bet at certain odds on the result of a coin toss, where as far as you can tell it’s equally likely that the coin is fair and that it is double-headed. Someone offers to tell you how the previous toss of the coin went.

Good’s Theorem says your decision whether to make the bet will be at least as good given the information about the previous three tosses as without that information. Hence, if the information is being announced, you don’t need to cover your ears. This is, of course, very intuitive. But now consider a slightly different case:

2. Things are set up just as in (1), except now instead of information about the previous toss, you are offered a chance to have the following experiment get performed before your decision: the coin will be tossed an extra time and the result will be announced to you.

The difference is that in (2) you are not simply being offered additional information about how things are. For whether you go for the experiment or not, either way, you have full information about the experiment and its results. If you don’t go for the experiment, that full information is that the coin was not tossed an extra time (and hence did not land either heads or tails). If you do go for the experiment, the full information is that the coin was tossed and it landed heads, or else that it was tossed and it landed tails. In (2), you are not just finding out information by going for the deal: you are making something happen—an extra toss—and then finding out something about that.

So you can’t apply Good’s Theorem directly to (2). It would be nice to have a formulation of Good’s Theorem that works in cases where instead of merely finding out information, you perform an experiment.

I initially thought this would be easy. Maybe it is, but I don’t see it. There are, after all, cases where performing a cost-free experiment is not a good idea. Suppose, for instance, that you will be allowed to bet tomorrow that a certain car has more than 10 gallons of gasoline. The experiment is to start up the car and look at the gas gauge. But starting the car reduces the amount of gasoline in it, and one can easily rig the case so that benefits from the information gain are outweighed by the fact that you have made that bet less favorable.

So, we want to rule out cases where there is dependence between whether you perform the experiment and the payoffs of the wagers. If F is the event of performing the experiment, it may seems initially we should assume something like:

3. E(U|W_i∩F) = E(U|W_i∩F^c) for all i,

where W_i is your choosing wager i and U is the utility random variable. In other words, the expected utility of each wager is unaffected by whether the experiment has been performed. But no! Suppose a coin has been tossed, and you are choosing between W₁ where you get a dollar on heads and W₂ where you get a dollar on tails. But let F be the experiment of looking at the coin. (This is a case for the original Good’s Theorem.) Then E(U|W_i∩F^c) = 0.50, while E(U|W_i∩F) is very close to 1.00 for the reason that when you find out what the coin is like, you are close to certain to bet on what you see, and hence you are close to certain to win your bet.

If F₁ is heads and F₂ is tails, we solve the problem by replacing (3) with:

4. E(U|W_i∩F_j∩F) = E(U|W_i∩F_j∩F^c) for i and j.

Namely, the expected utility of wager W_i given information F_j is independent of whether you performed the experiment F. But that only works because it makes sense to ask what the coin is showing if you aren’t looking: it makes sense to conditionalize on F_j ∩ F^c. But in the cases that interest me, there is no fact of the matter as to the result of the experiment when the experiment is not performed, since Molinism is false and we live in an indeterministic world. And in these cases, F_j ∩ F^c is the empty set: the F_j represent the possible results of the experiment but the experiment has no result when it is not performed.

I can get something by supposing a two-step procedure. You perform the experiment, event F, and you learn the result, event L. Then we can assume:

5. E(U|W_i∩F∩L^c) = E(U|W_i∩F^c) for all i

6. E(U|W_i∩F_j∩F∩L) = E(U|W_i∩F_j∩F∩L^c) for all i and j

7. P(F_j|F∩L) = P(F_j|F∩L^c).

Assumption (5) says that it makes no difference to the expected utility of a wager whether (3) the experiment is performed but its result is not learned or (b) the experiment is not performed at all. In other words, the experiment itself doesn’t affect things. Assumption (6) says that given a specific experimental result, learning the result makes no difference to the expected utility of each wager–result pair. Assumption (7) says that the results of the experiment are unaffected by whether you learn the result of the experiment.

Without (6) or (7), we wouldn’t expect to get the result we want. If we don’t have (6), it might be that utilities are wildly affected by whether you learn the result. (The simplest case is that the wagers all have a big negative payoff on L.) If we don’t have (7), then learning the result might have some evidential or retrocausal impact on what the result is, and then again we shouldn’t expect that learning the result is a good thing.

Given (5)–(7), I think we can now reason as follows. You are choosing between:

1. performing the experiment and learning the results

and

2. not performing the experiment and (hence) not learning the results.

By (5), a rational agent will decide the same way in (ii) as in:

3. performing the experiment and not learning the results,

and the expected utilities of (ii) and (iii) will be the same for this rational agent.

We now apply Good’s Theorem to the choice between (i) and (iii) (we will use (6) and (7) here, and assume the case is non-Newcombian and hence allows the use of Evidential Decision Theory) and get the result that (i) is at least as good as (iii). Since we have indifference between (ii) and (iii), it follows that (i) is at least as good as (ii). (We can also analyze the cases of a strict expected utility inequality.)

This is roundabout, but that’s not my main worry.

What I am really worried about is one technicality. To run the above argument, I had to assume that there is a way of performing the experiment without learning the result, namely that F ∩ L^c is non-empty. In general, however, we cannot assume this. Suppose, for instance, that we have a world with a quantum mechanics where observation causes collapse. Then the experiment of collapsing a wavefunction by means of observation cannot be done without observing the result of the experiment. In such scenarios, I cannot simply introduce a third option of performing the experiment and not learning the results, since that third option may not be consistent with the laws of physics. (And, of course, the utilities for breaking the laws of physics could be wild.)

But without introducing that third option, namely F ∩ L^c, I don’t know how to formulate the independence assumptions that are needed. I also don’t know if the problem is “merely technical” or “deep”. If I had to bet at even odds, I would bet on its being merely technical. But it might be deep.

A problem with perfectly rational agents and decision theory

Suppose I am perfectly rational in the decision theoretic sense. A coin is about to be tossed, and I will get five dollars on heads (H) and one dollar on tails (T). I have a choice whether to leave the coin fair (F) or load it (L) in favor of tails so that the probability of tails is 3/4.

It is obvious what I do. I calculate the expected utilities of my options F and L as follows.

• EU(F) = P(H|F) ⋅ $5 + P(T|F) ⋅ $1 = (1/2) ⋅ $5 + (1/2) ⋅ $1 = $3

and

• EU(L) = P(H|L) ⋅ $5 + P(T|L) ⋅ $1 = (1/4) ⋅ $5 + (3/4) ⋅ $1 = $2.

And then I choose F.

Except it’s not so simple. For I am perfectly rational. But since, as we just saw, the perfectly rational agent has to choose F, it follows that P(L) = 0, and so P(H|L) and P(T|L) are undefined. So I can’t decide! So now there is no guarantee how I will act, and P(H|L) and P(T|L) once again make sense. And then again they don’t. Oops!

What can be done? Causal decision theorists will note that I reasoned like an evidential decision theorist above. But this makes no difference in this case. The causalist’s story will be a bit more complicated but will end up with the same problem.

We might want to introduce primitive conditional probabilities like Popper functions that let you conditionalize on events with zero probability, and then have P(H|L) = 1/4 and P(T|L) = 3/4, even though P(L) = 0. But that is introducing a lot of complications. Primitive conditional probabilities are not unproblematic.

What should we do? Maybe we should suppose something like primitive suppositional decision theory, where what we are primitively given are the suppositional probabilities P^F and P^L, without them being defined in terms of conditional and unconditional credences as in evidential and causal decision theories. But this seems problematic. Do we have to suppose that in addition to conditional and unconditional credences, we have suppositional credences? Maybe.

Or perhaps decision theory only applies to agents that have non-zero credences of going for all the options.

Quantifying saving infinitely many lives

Suppose there is an infinite set of people, all of them worth saving, and you can save some subset of them from drowning. Can you assign a utility U(A) to each subset A of the people that represents the utility of saving the people in A subject to the following pair of reasonable conditions:

1. If A is a proper subset of B, then U(A) < U(B)

2. If A is a subset of the people, and x is one of the people not in A while I is an infinite set of people not in A, then U(A∪{x}) ≤ U(A∪I)?

The first condition says that it’s always better to add extra people to the set of people you save. The second condition says it’s always at least as good to add infinitely many people to the set of people you save as to add just one. (It would make sense to say: it’s always better to add infinitely many, but I don’t need that stronger condition.)

Theorem. For any infinite set of people, there is no real-valued utility function satisfying conditions (1) and (2), but there is a hyperreal-valued one.

It’s obvious we can’t do this with real numbers if we think of the value of saving n lives as proportional to n, since then the value of infinitely many lives will be ∞ which is not a real number. What’s mildly interesting in the result is that there is no way to scale the values of lives saved in some unequal way that preserves (1) and (2).

Proof: The hyperreal case follows from Theorem 2 here, where we let Ω = Ω′ be the set of people, G be the group of permutations of the set of people that shuffle around only finitely many people, and let U be the hyperreal probability (!) generated by the theorem. For this group is clearly locally finite, and any utility satisfying condition (1) and invariant under G will satisfy (2) (apply invariance to a permutation π be that swaps x and a member of I and does nothing else to conclude that U(A∪{x}) = U(A∪{πx}) which must be less than U(A∪I) by (1)).

The real case took me a fair amount of thought. Suppose we have a real U satisfying (1) and (2). Without loss of generality, the set of people is countably infinite, and hence can be represented by rational numbers Q. For a real number x, let D(x) be the Dedekind cut {q ∈ Q : q < x}. Fix a real number x. Choose any rational q bigger than x. Then for any real y > x we will have D(y) ∖ D(x) infinite, and by (1) and (2) we will have:

3. U(D(x)) < U(D(x)∪{q}) ≤ U(D(y)).

Let b = inf_y > xU(D(y)). It follows that U(D(x)) < b ≤ U(D(y)) for all y > x. Let f(x) be the open interval (D(x),b). Then f(x) and f(y) are disjoint and non-empty for x < y. But the collection of disjoint non-empty open intervals of the reals is always countable. (The quick argument is that we can choose a different rational in each such interval.) So f is a one-to-one function on the reals with countable range, a contradiction.

Notes: The positive part of the Theorem uses the Axiom of Choice (I think in the form of the Boolean Prime Ideal Theorem). The negative part doesn’t need the Axiom of Choice if the set of people is countable (the final parenthetical argument about intervals and rationals ostensibly uses Choice but doesn’t need it as the rationals are well-ordered); in general, the argument of the negative part uses the weak version of the Countable Axiom of Choice that says that every infinite set has a countably infinite subset.

Another infinite dice game

Suppose infinitely many people independently roll a fair die. Before they get to see the result, they will need to guess whether the die shows a six or a non-six. If they guess right, they get a cookie; if they guess wrong, an electric shock.

But here’s another part of the story. An angel has considered all possible sequences of fair die outcomes for the infinitely many people, and defined the equivalence relation ∼ on the sequences, where α ∼ β if and only if the sequences α and β differ in at most finitely many places. Furthermore, the angel has chosen a set T that contains exactly one sequence from each ∼-equivalence class. Before anybody guesses, the angel is going to look at everyone’s dice and announce the unique member α of T that is ∼-equivalent to the actual die rolls.

Consider two strategies:

1. Ignore what the angel says and say “not six” regardless.

2. Guess in accordance with the unique member α: if α says you have six, you guess “six”, and otherwise you guess “not six”.

When the two strategies disagree for a person, there is a good argument that the person should go with strategy (1). For without the information from the angel, the person should go with strategy (1). But the information received from the angel is irrelevant to each individual x, because which ∼-equivalence class the actual sequence of rolls falls into depends only on rolls other than x’s. And following strategy (1) in repeats of the game results in one getting a cookie five out of six times on average.

However, if everyone follows strategy (2), then it is guaranteed that in each game only finitely many people get a shock and everyone else gets a cookie.

This seems to be an interesting case where self-interest gets everyone to go for strategy (1), but everyone going for strategy (2) is better for the common good. There are, of course, many such games, such as Tragedy of the Commons or the Prisoner’s Dilemma, but what is weird about the present game is that there is no interaction between the players—each one’s payoff is independent of what any of the other players do.

(This is a variant of a game in my infinity book, but the difference is that the game in my infinity book only worked assuming a certain rare event happened, while this game works more generally.)

My official line on games like this is that their paradoxicality is evidence for causal finitism, which thesis rules them out.

Epistemic utilities and decision theories

Warning: I worry there may be something wrong in the reasoning below.

Causal Decision Theory (CDT) and Epistemic Decision Theory (EDT) tend to disagree when the payoff of an option statistically depends on your propensity to go for that option. The most example of this phenomenon is Newcomb’s Problem (where money is literally put into a box or not depending on what your propensities are), and there is a large literature of other clever and mind-twisting examples. From the literature, one might get a feeling that these cases are all somehow weird, and normally there is no such dependence.

But here is a family of cases that happens literally almost all the time to us. Pretty much whenever we act we gain information relevant to facts about ourselves, and specifically to facts about our propensities to act. For instance, when you choose chocolate over vanilla ice cream you raise your credence for the hypothesis that you have a greater propensity to choose chocolate ice cream than to choose vanilla ice cream. But truth about oneself is valuable and falsehood about oneself is disvaluable. If in fact you have a greater propensity to choose chocolate ice cream, then by eating chocolate ice cream you gain credence in a truth, which is a good thing. If in fact your propensity for vanilla ice cream is at least as great as for chocolate ice cream, then by eating chocolate ice cream, you gain credence in a falsehood. The payoffs of your decision as to flavor of ice cream thus statistically depend on what your propensities actually are, and so this is exactly the kind of case where we would expect CDT and EDT to disagree.

Let’s be more precise. You have a choice between eating chocolate ice cream (C), eating vanilla ice cream (V) or not eating ice cream at all (N). Let H be the hypothesis that you have a greater propensity for eating chocolate ice cream than for eating vanilla ice cream. Then if you choose C, you will gain evidence for H. If you choose V, you will gain evidence for not-H. And if you choose N, you will (plausibly) gain no evidence for or against H. Your epistemic utility with respect to H is, let us suppose, measured by a single-proposition accuracy scoring rule, which we can think of as a pair of functions T_H and F_H, where T_H(p) is the value of having credence p in H if in fact H is true and F_H(p) is the value of having credence p in H if in fact H is false.

The expected evidential utilities of your three options are:

• E_e(C) = P(H|C)T_H(P(H|C)) + (1−P(H|C))F_H(P(H|C))

• E_e(V) = P(H|V)T_H(P(H|V)) + (1−P(H|V))F_H(P(H|V))

• E_e(N) = P(H|N)T_H(P(H|N)) + (1−P(H|N))F_H(P(H|N)) = P(H)T_H(P(H)) + (1−P(H))F_H(P(H)).

The expected causal utilities are:

• E_c(C) = P(H)T_H(P(H|C)) + (1−P(H))F_H(P(H|C))

• E_c(V) = P(H)T_H(P(H|V)) + (1−P(H))F_H(P(H|V))

• E_c(N) = P(H)T_H(P(H|N)) + (1−P(H))F_H(P(H|N)) = P(H)T_H(P(H)) + (1−P(H))F_H(P(H)).

We can make some quick observations in the case where the scoring rule is strictly proper, given that P(H|V) < P(H) < P(H|C):

1. E_c(C) < E_c(N)

2. E_c(V) < E_c(N)

3. At least one of E_e(C) > E_e(N) and E_e(V) > E_e(N) is true.

Observations 1 and 2 follow immediately from strict propriety and the formulas for E_c. Observation 3 follows from the fact that the expected accuracy score after Bayesian update on evidence is better (in non-trivial cases where the scoring rule is strictly proper) than before update, and the expected accuracy score after update on what you’ve chosen is:

• P(C)E_e(C) + P(V)E_e(V) + P(N)E_e(N)

while the expected accuracy score before update is equal to E_e(N). Since P(C) + P(V) + P(N) = 1, it follows from the superiority of the post-update expectation that at least one of E_e(C) and E_e(V) must be bigger than E_e(N).

The above results seem to be a black eye for CDT, which recommends that if what you care about is your epistemic utility with regard to your propensities regarding chocolate and vanilla ice cream, then you should always avoid eating ice cream!

(What about ratifiability? Some CDTers say that only ratifiable options should count. Is N ratifiable? Given that you’ve learned nothing about H from choosing N, I think N should be ratifiable. But I may be missing something. I find the epistemic utility case confusing.)

It also seems to me (I haven’t checked details) that on EDT there are cases where eating either flavor is good for you epistemically, but there are also cases where only one specific flavor is good for you.

More decision theory stuff

Suppose there are two opaque boxes, A and B, of which I can choose one. A nearly perfect predictor of my actions put $100 in the box that they thought I would choose. Suppose I find myself with evidence that it’s 75% likely that I will choose box A (maybe in 75% of cases like this, people like me choose A). I then reason: “So, probably, the money is in box A”, and I take box A.

This reasoning is supported by causal decision theory. There are two causal hypotheses: that there is money in box A and that there is money in box B. Evidence that it’s 75% likely that I will choose box A provides me with evidence that it’s close to 75% likely that the predictor put the money in box A. The causal expected value of my choosing box A is thus around $75 and the causal expected value of my choosing box B is around $25.

On evidential decision theory, it’s a near toss-up what to do: the expected news value of my choosing A is close to $100 and so is that of my choosing B.

Thus, on causal decision theory, if I have to pay a $10 fee for choosing box A, while choosing box B is free, I should still go for box A. But on evidential decision theory, since it’s nearly certain that I’ll get a prize no matter what I do, it’s pointless to pay any fee. And that seems to be the right answer to me here. But evidential decision theory gives the clearly wrong answer in some other cases, such as that infamous counterfactual case where an undetected cancer would make you likely to smoke, with no causation in the other direction, and so on evidential decision theory you refrain from smoking to make sure you didn’t get the cancer.

In recent posts, I’ve been groping towards an alternative to both theories. The alternative depends on the idea of imagining looking at the options from the standpoint of causal decision theory after updating on the hypothesis that one has made a specific choice. In current my predictor cases, if you were to learn that you chose A, you would think: Very likely the money is in box A, so choosing box A was a good choice, while if you chose B, you would think: Very likely the money is in box B, so choosing box B was a good choice. As a result, it’s tempting to say that both choices are fine—they both ratify themselves, or something like that. But that misses out the plausible claim that if there is a $10 fee for choosing A, you should choose B. I don’t know how best to get that claim. Evidential decision theory gets it, but evidential decision theory has other problems.

Here’s something gerrymandered that might work for some binary choices. For options X and Y, which may or may not be the same, let e_X(Y) be the causal expected value of Y with respect to the credences for the causal hypotheses updated with respect to your having chosen X. Now, say that the differential restrospective causal expectation d(X) of option X equals e_X(X) − e_X(Y). This measures how much you would think you gained, from the standpoint of causal decision theory, in choosing X rather than Y by the lights of having updated on choosing X. Then you should the option that provides a bigger d(X).

In the case where there is a $10 fee for choosing box A, d(B) is approximately $100 while d(A) is approximately $90, so you should go for box B, as per my intuition. So you end up agreeing with evidential decision theory here.

You avoid the conclusion you should smoke to make sure you don’t have cancer in the hypothetical case where cancer causes smoking but not conversely, because the differential retrospective causal expectation of smoking is positive while the differential retrospective causal expectation of not smoking is negative, assuming smoking is fun (is it?). So here you agree with causal decision theory.

What about Newcomb’s paradox? If the clear box has a thousand dollars and the opaque box has a million or nothing (depending on whether you are predicted to take just the opaque box or to take both), then the differential retrospective causal expectation of two-boxing is a thousand dollars (when you learned you two-box, you learn that the opaque box was likely empty) and the differential retrospective causal expectation of one-boxing is minus a thousand dollars.

So the differential retrospective causal expectation theory agrees with causal decision theory in the clear case (cancer-causes-smoking), the difficult case (Newcomb), but agrees with evidential decision theory in the $10 fee variant of my two-box scenario, and the last seems plausible.

But (a) it’s gerrymandered and (b) I don’t know how to generalize it to cases with more than two options. I feel lost.

Maybe I should stop worrying about this stuff, because maybe there just is no good general way of making rational decisions in cases where there is probabilistic information available to you about how you will make your choice.

An immediate regret principle

Here’s a plausible immediate regret principle:

1. It is irrational to make a decision such that learning that you’ve made this decision immediately makes it rational to regret that you didn’t make a different decision.

The regret principle gives an argument for two-boxing in Newcomb’s Paradox, since if you go for one box, as soon as you have made your decision to do that, you will regret you didn’t make the two-box decision—there is that clear box with money staring at you, but if you go for two boxes, you will have no regrets.

Interestingly, though, one can come up with predictor stories where one has regrets no matter what one chooses. Suppose there are two opaque boxes, A and B, and you can take either box but not both. A predictor put a thousand dollars in the box that they predicted you won’t take. Their prediction need not be very good—all we need for the story is that there is a better than even probability of their having predicted you choosing A conditionally on your choosing A and a better than even probability of their having predicted you choosing B conditionally on your choosing B. But now as soon as you’ve made your decision, and before you opened the chosen box, you will think the other box is more likely to have the money, and so your knowledge of your decision will make it rational to regret that decision. Note that while the original Newcomb problem is science-fictional, there is nothing particularly science-fictional about my story. It would not be surprising, for instance, if someone were able to guess with better than even chance of correctness about what their friends would choose.

Is this a counterexample to the immediate regret principle (1), or is this an argument that there are real rational dilemmas, cases where all options are irrational?

I am not sure, but I am inclined to think that it’s a counterexample to the regret principle.

Can we modify the immediate regret principle to save it? Maybe. How about this?

2. No decision is such that learning that you’ve rationally made this decision immediately makes it rationally required to regret that you didn’t make a different decision.

On this regret principle, regret is compatible with non-irrational decision making but not with (known) rational decision making.

In my box story, it is neither rational nor irrational to choose A, and it is neither rational nor irrational to choose B. Then there is no contradiction to (2), since (2) only applies to decisions that are rationally made. And applying (2) to Newcomb’s Paradox no longer yields an argument for two-boxing, but only an argument that it is not rational to one-box. (For if it were rational to one-box, one could rationally decide to one-box, and one would then regret that.)

The “rationally” in (2) can be understood in a weaker way or a stronger way (the stronger way reads it as “out of rational requirement”). On either reading, (2) has some plausibility.

An odd decision theory

Suppose I am choosing between options A and B. Evidential decision theory tells me to calculate the expected utility E(U|A) given the news that I did A and the expected utility E(U|B) given the news that I did B, and go for the bigger of the two. This is well-known to lead to the following absurd result. Suppose there is a gene G that both causes one day to die a horrible death and makes one very likely to choose A, while absence of the gene makes one very likely to choose B. Then if A and B are different flavors of ice cream, I should always choose B, because E(U|A) ≪ E(U|B), since the horrible death from G trumps any advantage of flavor that A might have over B. This is silly, of course, because one’s choice does not affect whether one has G.

Causal decision theorists proceed as follows. We have a set of “causal hypotheses” about what the relevant parts of the world at the time of the decision are like. For each causal hypothesis H we calculate E(U|H∧A) and E(U|H∧B), and then we take the weighted average over our probabilities, and then decide accordingly. In other words, we have a causal expected utility of D

• E_c(U|D) = ∑_HE(U|H∧D)P(H)

and are to choose A over B provided that E_c(U|A) = E_c(U|B). In the gene case, the “bad news” of the horrible death on G is a constant addition to E_c(U|A) and to E_c(U|B), and so it can be ignored—as is right, since it’s not in our control.

But here is a variant case that worries me. Suppose that you are choosing between flavors A and B of ice cream, and you will only ever ever get to taste one of them, and only once. You can’t figure out which one will taste better for you (maybe one is oyster ice cream and the other is sea urchin ice cream). However, data shows that not only does G make one likely to choose A and its absence makes one likely to choose B, but everyone who has G derives pleasure from A and displeasure from B and everyone who lacks G has the opposite result, and all the pleasures and displeasures are of the same magnitude.

Now, background information says that you have a 3/4 chance of having G. On causal decision theory, this means that you should choose A, because likely you have G, and those who have G all enjoy A. Evidential decision theory, however, tells you that you should choose B, since if you choose B then likely you don’t have the terrible gene G.

In this case, I feel causal decision theory isn’t quite right. Suppose I choose A. Then after I have made my choice, but before I have consumed the ice cream, I will be glad that I chose A: my choice of A will make me think I have G, and hence that A is tastier. But similarly, if I choose B, then after I have made my choice, and again before consumption, I will be glad that I chose B, since my choice B will make me think I don’t have G and hence that B was a good choice. Whatever I choose, I will be glad I chose it. This suggests to me that my there is nothing wrong with either choice!

Here is the beginning of a third decision theory, then—one that is neither causal nor evidential. An option A is permissible provided that causal decision theory with the causal hypothesis credences conditioned on one’s choosing A permits one to do A. An option A is required provided that no alternative is permissible. (There are cases where no option is permissible. That’s weird, I admit.)

In the initial case, where the pleasure of each flavor does not depend on G, this third decision theory gives the same answer as causal decision theory—it says to go for the tastier flavor. In the second case, however, where the pleasure/displeasure depends on G, it permits one to go for either flavor. In a probabilistic-predictor Newcomb’s Paradox, it says to two-box.

And one more post on comparing experiments

In my last couple of posts, starting here, I’ve been thinking about comparing the epistemic quality of experiments for a set of questions. I gave a complete geometric characterization for the case where the experiments are binary—each experiment has only two possible outcomes.

Now I want to finally note that there is a literature for the relevant concepts, and it gives a characterization of the comparison of the epistemic quality of experiments, at least in the case of a finite probability space (and in some infinite cases).

Suppose that Ω is our probability space with a finite number of points, and that F_Q is the algebra of subsets of Ω corresponding to the set of questions Q (a question partitions Ω into subsets and asks which partition we live in; the algebra F_Q is generated by all these partitions). Let X be the space of all probability measures on F_Q. This can be identified with an (n−1)-dimensional subset of Euclidean Rⁿ consisting of the points with non-negative coordinates summing to one, where n is the number of atoms in F_Q. An experiment E also corresponds to a partition of Ω—it answers the question where in that partition we live. The experiment has some finite number of possible outcomes A₁, ..., A_m, and in each outcome A_i our Bayesian agent will have a different posterior P_Ai = P(⋅∣A_i). The posteriors are members of X. The experiment defines an atomic measure μ_E on X where μ_E(ν) is the probability that E will generate an outcome whose posterior matches ν on F_Q. Thus:

• μ_E(ν) = P(⋃{A_i:P_Ai|_FQ=ν}).

Given the correspondence between convex functions and proper scoring rules, we can see that experiment E₂ is at least as good as E₁ for Q just in case for every convex function c on X we have:

• ∫_Xcdμ_E2 ≥ ∫_Xcdμ_E1.

There is an accepted name for this relation: μ_E2 convexly dominates μ_E1. Thus, we have it that experiment E₂ is at least as good as experiment E₁ for Q provided that there is a convex domination relation between the distributions the experiments induce on the possible posteriors for the questions in Q. And it turns out that there is a known mathematical characterization of when this happens, and it includes some infinite cases as well.

In fact, the work on this epistemic comparison of experiments turns out to go back to a 1953 paper by Blackwell. The only difference is that Blackwell (following 1950 work by Bohnenblust, Karlin and Sherman) uses non-epistemic utility while my focus is on scoring rules and epistemic utility. But the mathematics is the same, given that non-epistemic decision problems correspond to proper scoring rules and vice versa.

Does one's vote make a difference?

Suppose that there is a simple majority election, with two candidates, and there is a large odd number of voters. Suppose polling data makes the election too close to call. How likely is it that you can decide which candidate wins?

I could look up this stuff, but it’s more fun to figure it out.

A quick and dirty model is this. We have N people other than you voting, each choosing between candidates A and B with probabilities p and 1 − p respectively. You don’t know what p and 1 − p are, but polling data tells you that p is between 1/2 − a and 1/2 + b for some positive numbers a and b. Your vote decides the election provided that exactly N/2 people vote for candidate A. This requires that N be even (if N is odd, at best you can decide between a candidate winning and the election being undecided, so you can’t decide which candidate wins), which has probability 1/2. Given that N = 2n is even, the probability that the other votes are exactly balanced is (a+b)⁻¹ C(2n,n)∫_1/2−a^1/2+bpⁿ(1−p)^n − 1dp, where C(m,n) is the binomial coefficient. Assuming n is large as compared to a and b, the integral can be approximated by replacing its bounds by 0 and 1 respectively, and some work with Mathematica shows that for large n the probability is approximately 1/(N(a+b)).

So what? Well, suppose you think that candidate A will on average make a person in the jurisdiction be u units of flourishing better off than candidate B will, and there are K persons, where K ≥ N + 1 (there are at least as many persons as candidates). So, the expected amount of difference that your voting for A will make is at least Ku/(2N(a+b)). This is at least u/(a+b). Thus, if the polling data gives you a range between 0.48 and 0.52 for the probability of a person’s preferring candidate A, and half of the people in the jurisdiction vote, the expected amount of difference that your vote makes is 25u. This is quite a lot if you think that which candidate wins makes a significant difference u per governed person.

Interestingly, some numerical work with Mathematica also shows that as number of people increases, then the expected amount of difference your vote makes also increases asymptotically, up to the limit of Ku/(2N(a+b)). So for larger jurisdictions, even though the probability of your vote making a difference is smaller, the expected difference from your vote is a bit bigger.

My quick and dirty model is not quite right. Of course, people don’t come to the polls and randomly choose whom to vote for. A more likely source of randomness has to do with who actually makes it to the polls (who gets sick, who has something come up, who decides it’s pointless to vote, etc.). A better model might be this. We have M people eligible to vote, of whom pM want to vote for A and (1−p)M want to vote for B. Some random subset of the M people then votes. My probabilist intuitions say that this is not that different from my model if the number of actual voters is, say, half of the eligible voters. If I had an election that I was eligible to vote in coming, I might try to figure our the more complex model, but I don’t.

Fun with St. Petersburg

A generous patron makes an offer to you. You are to pick out a positive integer n and you will get 2ⁿ units of value. You have the ability to pick out any positive integer at no cost to yourself (maybe you can engage in a supertask and name long numbers really fast).

You think about naming a million, but then a billion would pay so much better, and a billion and two is four times better! You agonize. And then you have a brilliant idea. You will randomize by choosing positive integer n with probability 2⁻ⁿ (say, by flipping a coin until you get heads and counting how many flips that took). Your expected payoff will be

• (1/2)(2) + (1/4)(4) + (1/8)(8) + ... = ∞.

That beats any specific number you could choose. So you go for it.

And, poof, you get 4. Regrets! You don’t want to stick to what the random choice gave you, as you’ll “only” get 2⁴ = 16 units of value. Disappointing! So you try again. You choose another positive integer. Now it is, mirabile dictu, a billion and two. But you think: 2^1000000002 may be a lot, but infinity is more, and if you randomly choose another number, your expected payoff is ∞. So you randomly choose again. And whatever you get, you are dissatisfied.

Three invariance arguments

Suppose we have two infinite collections of items L_n and R_n indexed by integers n, and suppose we have a total preorder ≤ on all the items. Suppose further the following conditions hold for all n, m and k:

1. L_n > L_n − 1

2. R_n > R_n + 1

3. If L_n ≤ R_m, then L_n + k ≤ R_m + k.

Theorem: It follows that either L_n > R_m for all n and m, or R_n > L_m for all n and m.

(I prove this in a special case here, but the proof works for the general case.)

Here are three interesting applications. First, suppose that an integer X is fairly chosen. Let L_n be the event that X ≤ n and let R_n be the event that X ≥ n. Let our preorder be comparison of the probabilities of events: A ≤ B means that A is no less likely than B. Intuitively, it is less likely that X is less than n − 1 than that it is less than n, so we have (1), and similar reasoning gives (2). Claim (3) says that the relationship between L_n and R_m is the same as that between L_n + k ≤ R_m + k and that seems right, too.

So all the conditions seem satisfied, but the conclusion of the Theorem seems wrong. It just doesn’t seem right to think that all the left-ward events (X being less than or equal to something) are more likely than all the right-ward events (X being bigger than or equal to something), nor that it be the other way around.

I am inclined to conclude that countable infinite fair lotteries are impossible.

Second application. Suppose that for each integer n, a coin is tossed. Let L_n be the event that all the coins ..., n − 2, n − 1, n are heads. Let R_n be the event that all the coins n, n + 1, n + 2, ... are heads. Let ≤ compare probabilities in reverse: bigger is less likely. Again, the conditions (1)–(3) all sound right: it is less likely that ..., n − 2, n − 1, n are heads than that ..., n − 2, n − 1 are heads, and similarly for the right-ward events. But the conclusion of the theorem is clearly wrong here. The rightward all-heads events aren’t all more likely, nor all less likely, than the leftward ones.

I am inclined to conclude that all the L_n and R_n have equal probability (namely zero).

Third application. Supppose that there is an infinite line of people, all morally on par, standing on numbered positions one meter apart, with their lives endangered in the same way. Let L_n be the action of saving the lives of the people at positions ...., n − 2, n − 1, n and let R_n be the action of saving the lives of the people at positions n, n + 1, n + 2, .... Let ≤ measure moral worseness: A ≤ B means that B is at least as bad as A. Then intuitively we have (1) and (2): it is worse to save fewer people. Moreover, (3) is a plausible symmetry condition: if saving one group of people beats saving another group of people, shifting both groups by the same amount doesn’t change that comparison. But again the conclusion of the theorem is clearly wrong.

I am less clear on what to say. I think I want to deny the totality of ≤, allowing for cases of incommensurability of actions. In particular, I suspect that L_n and R_m will always be incommensurable.

Acting for the sake of rationality alone

Alice is confused about the nature of practical rationality and asks wrong philosopher about it. She is given this advice:

1. For each of your options consider all the potential pleasures and pains for you that could result from the option. Quantify them on a single scale, multiply them by their probabilities, and add them up. Go for the option where the resulting number is biggest.

Some time later, Alice goes to a restaurant and follows the advice to the letter. After spending several hours pouring over the menu and performing back-of-the-envelope calculations she orders and eats the kale and salmon salad.

Traditional decision theory will try to explain Alice’s action in terms of ends and means. What is her end? The obvious guess is that it’s pleasure. But that need not be correct. Alice may not care at all about pleasure. She just cares about doing the action that maximizes the sum of pleasure quantities multiplied by their probabilities. She may not even know that this sum is an “expected value”. It’s just a formula, and she is simply relying on an expert’s opinion as to what formula to use. (If we want to, we could suppose the philosopher gives Alice a logically equivalent formula that was so complicated that she can’t tell that she is maximizing expected pleasure.)

I suppose the right end-means analysis of Alice’s action would be something like this:

• End: Act rationally.

• Means: Perform an action that maximizes the sum of products of pleasures and probabilities.

The means is constitutive rather than causal. In this case, there is no causal means that I can see. (Alice may have been misinformed by the same philosopher that there is no such thing as causation.)

The example thus shows that there can be cases of action where one’s aim is simply to act rationally, where one isn’t aiming at any other end. These may be defective cases, but they are nonetheless possible.

More on benefiting infinitely many people

Once again let’s suppose that there are infinitely people on a line infinite in both directions, one meter apart, on positions numbered in meters. Suppose all the people are on par. Fix some benefit (e.g., saving a life or giving a cookie). Let L_n be the action of giving the benefit to all the people to the left of position n. Let R_n be the action of giving the benefit to all the people to the right of position n.

Write A ≤ B to mean that action B is at least as good as action A, and write A < B to mean that A ≤ B but not B ≤ A. If neither A ≤ B nor B ≤ A, then we say that A and B are noncomparable.

Consider these three conditions:

• Transitivity: If A ≤ B and B ≤ C, then A ≤ C for any actions A, B and C from among the {L_k} and the {R_k}.

• Strict monotonicity: L_n < L_n + 1 and R_n > R_n + 1 for all n.

• Weak translation invariance: If L_n ≤ R_m, then L_n + k ≤ R_m + k and if L_n ≥ R_m, then L_n + k ≥ R_m + k, for any n, m and k.

Theorem: If we have transitivity, strict monotonicity and weak translation invariance, then exactly one of the following three statements is true:

1. For all m and n, L_m and R_n are incomparable

2. For all m and n, L_m < R_n

3. For all m and n, L_m > R_n.

In other words, if any of the left-benefit actions is comparable with any of the right-benefit actions, there is an overwhelming moral skew whereby either all the left-benefit actions beat all the right-benefit actions or all the right-benefit actions beat all the left-benefit actions.

Proposition 1 in this paper is a special case of the above theorem, but the proof of the theorem proceeds in basically the same way. For a reductio, assume that (i) is false. Then either L_m ≥ R_n or L_m ≤ R_n for some m and n. First suppose that L_m ≥ R_n. Then the second and third paragraphs of the proof of Proposition 1 show that (iii) holds. Now suppose that L_m ≤ R_n. Let L_k^* = R_−k and R_k^* = L_−k. Say that A≤^*B iff A^* ≤ B^*. Then transitivity, strict monotonicity and weak translation invariance hold for ≤^*. Moreover, we have L_m ≤ R_n, so R_−m≤^*L_−n. Applying the previous case with  − m and  − n in place of n and m respectively we conclude that we always have L_j>^*R_k and hence that we always have L_j < R_k, i.e., (ii).

I suppose the most reasonable conclusion is that there is complete incomparability between the left- and right-benefit actions. But this seems implausible, too.

Again, I think the big conclusion is that human ethics has limits of applicability.

I hasten to add this. One might reasonably think—Ian suggested this in a recent comment—that decisions about benefiting or harming infinitely many people (at once) do not come up for humans. Well, that’s a little quick. To vary the Pascal’s Mugger situation, suppose a strange guy comes up to you on the street, and tells you that there are infinitely many people in a line drowning in a parallel universe, and asks you if you want him to save all the ones to the left of position 123 or all the ones to the right of position  − 11, because he can magically do either one, and nothing else, and he needs help in his moral dilemma. You are, of course, very dubious of what he is saying. Your credence that he is telling the truth is very, very small. But as any good Bayesian will tell you, it shouldn’t be zero. And now the decision you need to make is a real one.

The hyperreals and the von Neumann - Morgenstern representation theorem

This is all largely well-known, but I wanted to write it down explicitly. The von Neumann–Morgenstern utility theorem says that if we have a total preorder (complete transitive relation) ≾ on outcomes in a mixture space (i.e., a space such that given members a and b and any t ∈ [0,1], there is a member (1−t)a + tb satisfying some obvious axioms) and satisfying:

• Independence: For any outcomes a, b and c and any t ∈ (0, 1], we have a ≾ b iff ta + (1−t)c ≾ tb + (1−t)c, and

• Continuity: If a ≾ b ≾ c then there is a t ∈ [0,1] such that b ≈ (1−t)a + tc (where x ≈ y iff x ≾ y and y ≾ x)

the preorder can be represented by a real-valued utility function U that is a mixture space homomorphism (i.e., U((1−t)a+tb) = (1−t)U(a) + tU(b)) and such that U(a) ≤ U(b) if and only a ≾ b.

Clearly continuity is a necessary condition for this to hold. But what if we are interested in hyperreal-valued utility functions and drop continuity?

Quick summary:

• Without continuity, we have a hyperreal-valued representation, and

• We can extend our preferences to recover continuity with respect to the hyperreal field.

More precisely, Hausner in 1971 showed that in a finite dimensional case (essentially the mixture space being generated by the mixing operation from a finite number of outcomes we can call “sure outcomes”) with independence but without continuity we can represent the total preorder by a finite-dimensional lexicographically-ordered vector-valued utility. In other words, the utilities are vectors (u₀,...,u_n − 1) of real numbers where earlier entries trump later ones in comparison. Now, given an infinitesimal ϵ, any such vector can be represented as u₀ + u₁ϵ + ... + u_n − 1ϵ^n − 1. So in the finite dimensional case, we can have a hyperreal-valued utility representation.

What if we drop the finite-dimensionality requirement? Easy. Take an ultrafilter on the space of finitely generated mixture subspaces of our mixture space ordered by inclusion, and take an ultraproduct of the hyperreal-valued representations on each of these, and the result will be a hyperreal-valued utility representing our preorder on the full space.

(All this stuff may have been explicitly proved by Richter, but I don’t have easy access to his paper.)

Now, on to the claim that we can sort of recover continuity. More precisely, if we allow for probabilistic mixtures of our outcomes with weights in the hyperreal field F that U takes values in, then we can embed our mixing space M in an F-mixing space M_F (which satisfies the axioms of a mixing space with respect to members of the larger field F), and extend our preference ordering ≾ to M_F such that we have:

• F-continuity: If a ≾ b ≾ c then there is a t ∈ F with 0 ≤ t ≤ 1 such that b ≈ (1−t)a + tc (where x ≈ y iff x ≾ y and y ≾ x).

In other words, if we allow for sufficiently fine-grained probabilistic mixtures, with hyperreal probabilities, we get back the intuitive content of continuity.

To see this, embed M as a convex subset of a real vector space V using an embedding theorem of Stone from the middle of the last century. Without loss of generality, suppose 0 ∈ M and U(0) = 0. Extend U to the cone C_M = {ta : t ∈ [0, ∞), a ∈ M} generated by M by letting U(ta) = tU(a). Note that this is well-defined since U(0) = 0 and if ta = ub with 0 ≤ t < u, then b = (1−s) ⋅ 0 + s ⋅ a, where s = t/u, and so U(b) = sU(a). It is easy to see that the extension will be additive. Next extend U to the linear subspace V_M generated by C_M (and hence by M) by letting U(a−b) = U(a) − U(b) for a and b in C_M. This is well-defined because if a − b = c − d, then a + d = b + c and so U(a) + U(d) = U(b) + U(c) and hence U(a) − U(b) = U(c) − U(d). Moreover, U is now a linear functional on V_M. If B is a basis of V_M, then let V_M^F be an F-vector space with basis B, and extend U to an F-linear functional from V_M^F to F by letting U(t₁a₁+...+t_na_n) = t₁U(a₁) + ... + t_nU(a_n), where the a_i are in B and the t_i are in F. Now let M^F be the F-convex subset of V_M^F generated by M. This will be an F-mixing space (i.e., it will satisfy the axioms of a mixing space with the field F in place of the reals). Let a ≾ b iff U(a) ≤ U(b) for a and b in M^F. Then if a ≾ b ≾ c, we have U(a) ≤ U(b) ≤ U(c). Let t between 0 and 1 in F be such that (1−t)U(a) + tU(c) = U(b). By F-linearity of U, we will then have U((1−t)a+tc) = U(b).

Binding and small probabilities

Suppose we neglect events with probability less than ϵ for some small ϵ > 0. Let’s suppose two independent random things have happened. First, an independent event E may or may not have happened, and P(E) = 2ϵ. Second, a fair die was rolled. You don’t have any information on whether E happened or the die was rolled. The following complex deal (The Deal) is offered to you.

If you accept The Deal, it will be revealed to you whether E happened. Then the following will happen:

1. If E happened, you get a choice between:

   1. you pay a dollar, or

   2. one following happens:

      1. you get a dollar if the die showed 1, 2, 3 or 4, but

      2. you get a year of torture if the die showed 5 or 6.

2. If E did not happen, you pay ϵ cents.

The event of E happening and the die showing 5 or 6 has probability 2ϵ ⋅ (2/6) = (2/3)ϵ which we have supposed is negligible. So, it seems that 1.b.ii can be completely neglected. On the other hand, the event of E happening and the die showing 1, 2, 3 or 4 has probability 2ϵ ⋅ (4/6) = (4/3)ϵ, which is not negligible.

What should you do? The difficulty here is that a full probabilistic evaluation of what you will do depends on what your choice in case 1.a will be. One way to handle such cases is through binding: you think of your choice as being a choice between strategies and then you stick to your strategy no matter what. This seems to be a good way to handle various paradoxes like Satan’s Apple.

What are the relevant strategies here? Well, in terms of pure strategies (we can consider mixed strategies, but in this case I think they won’t change anything), they are:

1. Reject The Deal.

2. Accept The Deal, and if E happened, pay the dollar.

3. Accept The Deal, and if E happened, don’t pay the dollar.

If you don’t neglect small probabilities, then clearly (A) is the right strategy to choose (and stick to).

Also, clearly, (B) is never the right strategy: whatever happens, you pay.

Now, suppose you do neglect small probabilities, and let’s evaluate (A) and (C). The payoff for (A) is zero. The payoff for (C) is, in dollars:

• (4/3)ϵ − (1−2ϵ)(0.01)ϵ > 0.

For the torture option drops out, as it has the negligible probability (2/3)ϵ.

So, if you neglect small probabilities, and take binding to a strategy to be the right approach to such puzzles, you should accept The Deal and bind yourself to not pay the dollar. But now notice how psychologically impossible the binding is. If in fact E happened—and the probability of E is 2ϵ, which is not negligible—then you have to choose between paying a dollar and a wager that has a 2/3 chance of yielding a dollar and a 1/3 chance of a year of torture. How could you possibly accept a 1/3 chance of a year of torture in exchange for about $1.67? Real brainwashing would be required, not just a mere resolution to stick to a strategy.

So what? Why can’t the proponent of the binding solution simply agree that (C) is the abstractly best strategy, but since we can’t practically bind ourselves to it, we are stuck with (A)? But there is something counterintuitive about thinking that (C) is the abstractly best strategy when it requires brainwashing that is this extreme.

Brownian motion and regret

Let B_t be a one-dimensional Brownian motion, i.e., Wiener process, with B₀ = 0. Let’s say that time 0 you are offered, for free, a game where your payoff at time 1 will be B₁. Since the expected value of a Brownian motion at any future time equals its current value, this game has zero value, so you are indifferent and go for it.

But here is a fun fact. With probability one, at infinitely many times t between 0 and 1 we will have B_t < 0 (this follows from Th. 27.24 here). At any such time, your expectation of your payout at B₁ to be negative. Thus, at infinitely many times you will regret your decision to play the game.

Of course, by symmetry, with probability one, at infinitely many times between 0 and 1 we will have B_t > 0. Thus if you refuse to play, then at infinitely many times you will regret your decision not to play the game.

So we have a case where regret is basically inevitable.

That said, the story only works if causal finitism is false. So if one is convinced (I am not) that regret should always be avoidable, we have some evidence for causal finitism.

Avoiding regrets

I’ve recently been troubled by cases where you are sure to regret your decision, but the decision still seems reasonable. Some of these cases involve reasonable-seeming violations of expected utility maximization, but there is also the Cable Guy paradox, though admittedly I think I can probably exclude the Cable Guy paradox with causal finitism.

I shared Cable Guy with Clare Pruss, and she said that the principle of avoiding future regrets is false, and should be modified to a principle of avoiding final future regrets, because there are ordinary cases where you expect to regret something temporarily. For instance, you volunteer to do something onerous, and you expect that while volunteering, you will be regreting your choice, but you will be glad afterwards.

In all the cases that I’ve been interested in, while you are sure that there will be regret at some point in the future, you are not sure that there will be regret at the end (half the time the Cable Guy comes at the time you bet on him coming, after all).