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.]

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 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.

Causal probability and counterfactuals

The causal probability of an event B on an event A is cP_A(B)=∑_KP(K)P(B|AK), where the Ks are a partition based on the relevant dependency hypotheses compatible with A. (Compare to P(B|A)=∑_KP(K|A)P(B|AK).) A standard proposal in the literature is that

1. the degree of the assertibility of an indicative "If A, then B" is equal to the conditional probability P(B|A) of B on A.

Consider the parallel thesis that

2. the degree of assertibility of a subjunctive conditional or counterfactual A→B is equal to the causal probability cP_A(B) of B on A.

This thesis would unify the Stalnaker and Lewis (and Skyrms) approaches to causal decision theory as closely as possible. For according to the Stalnaker version, the causal expected value of an option A is:

3. EV(A) = ∑_BU(BA)P(A→B),

where the sum is over a partition based on outcomes. On the Lewis/Skyrms approach, it will be:

4. EV(A) = ∑_BU(BA)cP_A(B).

Now, if A→B has truth value, then the degree of asssertibility of A→B is equal to P(A→B), and hence by (2) we have P(A→B)=cP_A(B). And so the two formulae are equivalent. If, on the other hand, A→B has no truth value, then P(A→B) in (3) makes no sense. But we can replace it with Assertibility(A→B), which is basically the most natural replacement for P(A→B) when A→B has no truth value, and the revised (3) will come to the same thing as (4). So that's nice.

Notice, however, that this approach may not be compatibility with Molinism. For according to Molinism, God knows some conditionals of free will A→B, where B is a free action and A is a maximally specific set of antecedents, for sure. If P is God's probabilities, then in such cases:

5. 1=cP_A(B)=∑_KP(K)P(B|AK).

But because A is maximally specific, it will be compatible with only one relevant dependency hypothesis, say K₀, describing how B depends on A. So 1=P(K₀)P(B|AK₀). It follows that P(B|AK₀)=1 and P(K₀)=1. But now we see that there is a dependency hypothesis K₀ such that, together with A, it probabilistically necessitates B. But that can't be acceptable to a libertarian.