[I assume the reader is familiar with Newcomb’s problem and causal decision theory. Some familiarity with basic theoretical computer science ideas also helps.]

In a recently (Open-Access-)published paper, I (together with my PhD advisor, Vince Conitzer) proposed the following Newcomb-like scenario as an argument against causal decision theory:

> Adversarial Offer: Two boxes, B1 and B2, are on offer. A (risk-neutral) buyer may purchase one or none of the boxes but not both. Each of the two boxes costs \$1. Yesterday, the seller put \$3 in each box that she predicted the buyer would not acquire. Both the seller and the buyer believe the seller’s prediction to be accurate with probability 0.75.

If the buyer buys one of the boxes, then the seller makes a profit in expectation of \$1 – 0.25 * \$3 = \$0.25. Nonetheless, causal decision theory recommends buying a box. This is because at least one of the two boxes must contain \$3, so that the average box contains at least \$1.50. It follows that the causal decision theorist must assign an expected causal utility of at least \$1.50 to (at least) one of the boxes. Since \$1.50 exceeds the cost of \$1, causal decision theory recommends buying one of the boxes. This seems undesirable. So we should reject causal decision theory.

The randomization response

One of the obvious responses to the Adversarial Offer is that the agent might randomize. In the paper, we discuss this topic at length in Section IV.1 and in the subsection on ratificationism in Section IV.4. If you haven’t thought much about randomization in Newcomb-like problems before, it probably makes sense to first check out the paper and only then continue reading here, since the paper makes more straightforward points.

The information-theoretic variant

I now give a new variant of the Adversarial Offer, which deals with the randomization objection in a novel and very interesting way. Specifically, the unique feature of this variant is that CDT correctly assesses randomizing to be a bad idea. Unfortunately, it is quite a bit more complicated than the Adversarial Offer from the paper.

Imagine that the buyer is some computer program that has access to a true random number generator (TRNG). Imagine also that the buyer’s source code plus all its data (memories) has a size of, say, 1GB and that the seller knows that it has (at most) this size. If the buyer wants to buy a box, then she will have to pay \$1 as usual, but instead of submitting a single choice between buying box 1 and buying box 2, she has to submit 1TB worth of choices. That is, she has to submit a sequence of 2^43 (=8796093022208) bits, each encoding a choice between the boxes.

If there is at least one 1GB program that outputs w deterministically, then the seller forgets w again. She then picks an index i of w at random. She predicts w[i] based on what she knows about the buyer and based on w[1…i-1], i.e., on all the bits of w preceding i. Call the prediction w’[i]. She fills the box based on her prediction w’[i] and the buyer receives (in return for the \$1) the box specified by w[i].

Why the information-theoretic variant is interesting

The scenario is interesting because of the following three facts (which I will later argue to hold):

1. The seller makes a profit off agents who try to buy boxes, regardless of whether they do so using randomization or not.
2. CDT and related theories (such as ratificationism) assess randomizing to be worse than not buying any box.
3. CDT will recommend buying a box (mostly deterministically).

I’m not aware of any other scenarios with these properties. Specifically, the novelty is item 2. (Our paper offers a scenario that has the other two properties.) The complications of this scenario – letting the agent submit a TB worth of choices to then determine whether they are random – are all introduced to achieve item 2 (while preserving the other items).

In the following, I want to argue in more detail for these three points and for the claim that the scenario can be set up at all (which I will do under claim 1).

1. For this part, we need to show two things:

A) Intuitively, if the agent submits a string of bits that uses substantially more than 1GB worth of randomness, then he is extremely likely to receive no box at all.

B) Intuitively, if the agent uses only about 1GB or less worth randomness, then the seller – using the buyer’s source code will likely be able to predict w[i] with high accuracy based on w[1…i-1].

I don’t want to argue too rigorously for either of these, but below I’ll give intuitions and some sketches of the information-theoretic arguments that one would need to give to make them more rigorous.

A) The very simple point here is that if you create, say, a 2GB bitstring w where each bit is determined by a fair coin flip, then it is very unlikely that there exists a program that deterministically outputs w. After all, there are many more ways to fill 2GB of bits than there are 1GB programs (about 2^(2^33) as many). From this one may be tempted to conclude that if the agent determines, say, 2GB of the TB of choices by flipping coin, he is likely to receive no box. But this argument is incomplete, because there are other ways to use coin flips. For example, the buyer might use the following policy: Flip 2GB worth of coins. If they all come up heads, always take box B. Otherwise follow some given deterministic procedure.

To make the argument rigorous, I think we need to state the claim information-theoretically. But even this is a bit tricky. For example, it is not a problem per se for w to have high entropy if most of the entropy comes from a small part of the distribution. (For example, if with probability 0.01 the seller randomizes all choices, and with the remaining probability always chooses Box 1, then the entropy of w is 0.01*1TB = 10GB (??), but the buyer is still likely to receive a box.) So I think we’d need to make a more complicated claim, of the sort: if there is no substantial part (say, >p) of the distribution over w that has less than, say, 2GB of entropy, then with high probability (>1-p), the agent will receive no box.

B) Again, we can make a simple but incomplete argument: If of the 1TB of choices, only, say, 2GB are determined by random coin flips, then a randomly sampled bit is likely to be predictable from the agent’s source code. But again, the problem is that the random coin flips can be used in other ways. For example, the buyer might use a deterministic procedure to determine w (say, w=01010101…), but then randomly generate a number n (with any number n chosen with probability 2^-n, for instance), then randomly sample n indices j of w and flip w[j] for each of them. This may have relatively low entropy. But now the seller cannot perfectly predict w[i] given w[1…i-1] for any i.

Again, I think a rigorous argument requires information theory. In particular, we can use the fact that H(w) = H(w[0])+H(w[1]|w[0])+H(w[2]|w[0,1])+…, where H denotes entropy. If H(w) is less than, say, 2GB, then the average of H(w[i]|w[1…i-1]) must be at most 2GB/1TB = 1/500. From this, it follows immediately that with high probability, w[i] can be predicted with high accuracy given w[1…i-1].

2. This is essential, but straightforward: Generating w at random causes the seller to determine that w is determined at random. Therefore, CDT (accurately) assesses randomly generating w to have an expected utility near \$-1.

3. Finally, I want to argue that CDT will recommend buying a box. For this, we only need to argue that CDT prefers some method of submitting w over not submitting a box. So consider the following procedure: First, assign beliefs over the seller’s prediction w’[0] of the first bit. Since there are only two possible boxes, for at least one of the boxes j, it is the case that P(w’[0]=j)<=½, where P refers to the probability assigned by the buyer. Let w[0] = j. We now repeat this inductively. That is, for each i given the w[1…i-1] that we have already constructed, the buyer sets w[i]=k s.t. P(w’[i]=k||w[1…i-1])<=½.

What’s the causal expected utility of submitting w thus constructed? Well, for one, because the procedure is deterministic (if ties are broken deterministically), the buyer can expect that she will receive a box at all. Now, for all i, the buyer thinks that if i is sampled by the seller for the purpose of determining which box to give to the buyer, then the buyer will in causal expectation receive \$1.50, because the seller will predict the wrong box, i.e. w’[i] ≠ w[i], with probability at least ½.