The title of this post most probably deserves a cautious question mark at the end, but I'll go out on a limb and start sawing it behind me: I think I've got a framework that consistently solves correlated decision problems. That it is, those situation where different agents (a forgetful you at different times, your duplicates, or Omega’s prediction of you) will come to the same decision.

After my first post on the subject, Wei Dai asked whether my ideas could be formalised enough that it could applied mechanically. There were further challenges: introducing further positional information, and dealing with the difference between simulations and predictions. Since I claimed this sort of approach could apply to the Newcomb’s problem, it is also useful to see it work in cases were the two decisions are only partially correlated - where Omega is good, but he’s not perfect.

The theory

In standard decision making, it is easy to estimate your own contribution to your own utility; the contribution of others to your own utility is then estimated separately. In correlated decision-making, both steps are trickier; estimating your contribution is non-obvious, and the contribution from others is not independent. In fact, the question to ask is not "if I decide this, how much return will I make", but rather "in a world in which I decide this, how much return will I make".

You first estimate the contribution of each decision made to your own utility, using a simplified version of the CDP: if N correlated decisions are needed to gain some utility, then each decision maker is estimated to have contributed 1/N of the effort towards the gain of that utility.

Then the procedure under correlated decision making is:

1) Estimate the contribution of each correlated decision towards your utility, using CDP.

2) Estimate the probability that each decision actually happens (this is an implicit use of the SIA).

3) Use 1) and 2) to estimate the total utility that emerges from the decision.

To illustrate, apply it to the generalised absent minded driver problem, where the return for turning off at the first and second intersection are x and y, respectively, while driving straight through grants a return of z. The expected return for going straight with probability p is R = (1-p)x + p(1-p)y + p²z.

Then the expected return for the driver at the first intersection is (1-p)x ...