You cannot assume any α, and choose p based on it, for α depends on p. You just introduced a time loop into your example.

There's an old story about a motorist who gets a flat tire next to a mental hospital. He goes over to change the tire, putting the lug nuts into the wheel cap... and sees that one of the patients is staring at him from behind the fence. Rattled, he steps on the wheel cap, and the lug nuts go into the storm sewer.

The motorist is staring, disheartened, at the sewer drain, when the patient speaks up from behind the fence: "Take one lug nut off each of the other wheels, and use those to keep on the spare tire until you get home."

"That's brilliant!" says the motorist. "What are you doing in a mental hospital?"

"I'm here because I'm crazy," says the patient, "not because I'm stupid."

Robert Aumann, Nobel laureate, is an Orthodox Jew. And Isaac Newton wasted a lot of his life on Christian mysticism. I don't think theists, or Nobel laureate physicists who don't accept MWI, are stupid. I think they're crazy. There's a difference.

Remind me to post at some point about how rationalists, in a certain stage in their development, must, to progress further, get over their feeling of nervousness about leaving the pack and just break with the world once and for all.

"one of them won a Nobel Prize in Economics for his work on game theory. I really don't think we want to call these people "crazy".)"

Aumann is a religious Orthodox Jew who has supported Bible Code research. He's brilliant and an expert, but yes, he's crazy according to local mores.

Of course, that doesn't mean we should dismiss his work. Newton spent much of his life on Christian mysticism.

Your payoff for choosing CONTINUE with probability p becomes α[p^2+4(1-p)p] + (1-α)[p+4(1-p)], which doesn't equal p^2+4(1-p)p unless α = 1.

No. This statement of the problem pretends to represent the computation performed by the driver at an intersection - but it really doesn't. The trouble has to do with the semantics of alpha. Alpha is not the actual probability that the driver is at point X; it's the driver's estimate of that probability. The driver knows ahead of time that he's going to make the same calculation again at intersection Y, using the same value of alpha, which will be wrong. Therefore, he can't pretend that the actual payoff is alpha x (payoff if I am at X) + (1-alpha) x (payoff if I am at Y). Half the time, that payoff calculation will be wrong.

Perhaps a clearer way of stating this, is that the driver, being stateless, must believe P(I am at X) to be the same at both intersections. If you allow the driver to use alpha=.7 when at X, and alpha=.3 when at Y, then you've given the driver information, and it isn't the same problem anymore. If you allow the driver to use alpha=.7 when at X, and alpha=.7 again when at Y, then the driver at X is going to make a decision using the information that he's probably at X and should continue ahead, without taking into account that he's going to make a bad decision at Y because he will be computing with faulty information. That's not an alternative logic; it's just wrong.

The "correct" value for alpha is actually 1/(p+1), for those of us outside the problem; the driver is at Y p times for every 1 time he's at X, so he's at X 1 out of p+1 times. But to the driver, who is inside the problem, there is no correct value of alpha to use at both X and Y. This means that if the driver introduces alpha into his calculations, he is knowingly introducing error. The driver will be using bad information at at least one of X and Y. This means that the answer arrived at must be wrong at at least either X or Y. Since the correct answer is the same in both places, the answer arrived at must be wrong in both places. Using alpha simply adds a source of guaranteed error, and prevents finding the optimal solution.

Does the driver at each intersection have the expected payoff alpha x [p^2+4(1-p)p] + (1-alpha)*[p+4(1-p)] at each intersection? No; at X he has the actual expected payoff p^2+4(1-p)p, and at Y he has the expected payoff p+4(1-p). But he only gets to Y p/1 of the time. The other 1-p of the time, he's parked at A. So, if you want him to make a computation at each intersection, he maximizes the expected value averaging together the times he is at X and Y, knowing he is at Y p times for every one time he is at X:

p^2+4(1-p)p + p[p+4(1-p)] = 2p[p+4(1-p)]

And he comes up with p = 2/3.

There's just no semantically meaningful way to inject the alpha into the equation.

then the expected payoff is p^2^+4(1-p)p

For anyone whose eyes glazed over and couldn't see how this was derived:

There are 3 possible outcomes:

1. you miss both turns
   The probability of missing both turns for a p is p*p (2 turns, the same p each time), and the reward is 1. Expected utility is probability*reward, so 2*p*1. Which is just 2*p or p^2
2. you make the second turn.
   The probability of making the second turn is the probability of missing the first turn and making the second. Since p is for a binary choice, there's only one other probability, q, of missing the turn; by definition all probabilities add to 1, so p+q=1, or q=1-p. So, we have our q*p (for the missed and taken turn), and our reward of 4. As above, the expected utility is q*p*4, and substituting for q gives us (1-p)*p*4, or rearranging, 4*(1-p)*p.
3. or, you make the first turn
   The probability of making the first turn is just p-1 as before, and the reward is 0. So the expected utility is (p-1)*0 or just 0.

Our 3 possibilities are exhaustive, so we just add them together:

p^2 + 0 + 4*(1-p)*p

0 drops out, leaving us with the final result given in the article:

p^2 + 4*(1-p)*p

I could add some groundless speculation, but my general advice would be: Ask, don't guess!

You probably won't get answers like "I wanted to publish a paper.". But I'd bet, it would be very enlightening regardless. It'd be surprising if all of them were so extremely busy that you can't approach anybody in the area. But don't settle for PhD students, go for senior professors.

"The authors were trying to figure out what is rational for human beings, and that solution seems too alien for us to accept and/or put into practice."

I'm not sure about that. A lot of people intuitively endorse one-boxing on Newcomb, and probably a comparable fraction would endorse the 2/3 strategy for Absent-Minded Driver.

"Well, the authors don't say (they never bothered to argue against it)"

They do mention and dismiss mystical /psychic causation, the idea that in choosing what we will do we also choose for all identical minds/algorithms

"The authors were trying to solve one particular case of time inconsistency. They didn't have all known instances of time/dynamic/reflective inconsistencies/paradoxes/puzzles laid out in front of them, to be solved in one fell swoop."

Decision theorists have a lot of experience with paradoxical-seeming results of standard causal decision theory where 'rational agents' lose in certain ways. Once that conclusion has been endorsed by the field in some cases, it's easy to dismiss further such results: "we already know rational agents lose on all sorts of seemingly easy problems, such that they would precommit/self-modify to avoid making rational decisions, so how is this further instance a reason to change the very definition of rationality?" There could be substantial path-dependence here.

What I'm more interested in is: doesn't the UDT/planning-optimal solution imply that injecting randomness can improve an algorithm, which is a big no-no? Because you're saying (and you're right AFAICT) that the best strategy is to randomly choose whether to continue, with a bias in favor of continuing.

Also, could someone go through the steps of how UDT generates this solution, specifically, how it brings to your attention the possibility of expressing the payoff as a function of p? (Sorry, but I'm a bit of a straggler on these decision theory posts.)

In #lesswrong, me, Boxo, and Hyphen-ated each wrote a simple simulation/calculation of the absent-minded driver and checked how the various strategies did. Our results:

• 1/3 = 1.3337, 1.3338174; Hyphen-ated: 1.33332; Boxo: 1.3333333333333335
• 1/4 = 1.3127752; Hyphen-ated: 1.31247; Boxo: 1.3125
• 2/3 = 0.9882
• 4/9 = 1.2745, 1.2898, 1.299709, 1.297746, 1.297292, 1.2968815; Hyphen-ated: 1.29634; Boxo: 1.2962962962962963

As expected, 4/9 does distinctly worse than 1/3, which is the best strategy of the ones we tested. I'm a little surprised at Piccione & Rubinstein - didn't they run any quick* little simulation to see whether 4/9 was beaten by another probability or did they realize this and had some reason bad performance was justifiable? (I haven't read P&R's paper.)

Boxo used his UDT in Haskell code ([eu (Strategy $ const $ chance p) absentMinded id | p <- [0, 1/2, 1/3, 1/4, 4/9]]); I used some quick and dirty Haskell pasted below; and I don't know what Hyphen-ated used.

import System.Random
import Control.Monad
main = foo >>= print -- test out the 1/4 strategy. obvious how to adapt
foo = let n = 10000000 in fmap (\x -> fromIntegral (sum x) / fromIntegral n) $ replicateM n run
run = do x <- getStdRandom (randomR (1,4))
         y <- getStdRandom (randomR (1,4))
         return $ trial x y
trial :: Int -> Int -> Int
trial x y | x == 1 = 0 -- exit at first turn with reward 0
          | y == 1 = 4 -- second turn, reward 4
          | otherwise = 1 -- missed both, reward 1

* It's not like this is a complex simulation. I'm not familiar with Haskell's System.Random so it took perhaps 20 or 30 minutes to get something working and then another 20 or 30 minutes to run the simulations with 1 or 10 million iterations because it's not optimized code, but Boxo and Hyphen-ated worked even faster than that.

Wei, the solution that makes immediate sense is the one proposed by Uzi Segal in Economic Letters, Vol 67, 2000 titled "Don't Fool Yourself to Believe You Won't Fool Yourself Again".

You find yourself at an intersection, and you have no idea whether it is X or Y. You believe that you are at intersection X with probability α. Denote by q the probability of continuing to go straight rather than taking a left. What lottery do you face? Well, if you are at Y, then EXITING will yield a payoff of 4, and CONTinuing will yield a payoff of 1. If you are at X, then EXITING will yield a payoff of 0, and CONTinuing on straight will take you to Y. However, when you do in fact reach Y you will not realize you have reached Y and not X. In other words, you realize that going straight if you are at X will mean that you will return to the same intersection dilemma. So, suppose we denote the lottery you currently face at the intersection by Z. Then the description of the lottery is Z = ( (Z, q; 0, (1-q)), α; (1, q; 4, (1-q)), 1 - α). This is a recursive lottery. The recursive structure might seem paradoxical since there are a finite number of intersections, but is simply a result of the absent-mindedness. When you are at an intersection, you realize that any action you take at the moment that will put you at an intersection will not be distinguishable from the present moment in time, and that when you contemplate your choice at this subsequent intersection, you will not have the knowledge that you had already visited an intersection, inducing this "future" you to account for the belief that the second intersection is yet to be reached. The next crucial step is to recognize that beliefs about where you are ought to be consistent with your strategy. You only ever have to make a choice when you are at an intersection, and since all choice nodes are indistinguishable from each other (from your absent-minded perspective), your strategy can only be a mixture over EXIT and CONT. Note that the only way to be at intersection Y is by having chosen CONT at intersection X. Thus, if you believe that you are at intersection X with probability α, then the probability of being at Y is the probability that you ascribed to being at X when you previously made the choice to CONT multiplied by the probability on CONTinuing, as described by your optimal strategy. That is, Prob(Y) = Prob(X) Prob(CONT). So, consistency of beliefs with your strategy requires that (1 - α) = α q. Then, the value of the lottery Z, V(Z) as a function of q is described implicitly by: V(Z) = αqV(Z) + α(1-q)(0) + (1-α)q(1) + (1-α)(1-q)4. Solving for V(Z), and using the consistent beliefs condition yields V(Z) = (1 + 3q)(1-q), and so the optimal q is 2/3, which is the same mixed strategy as you chose when you were at START.

Of course, Aumann et. al. obtain that optimality of the planning-stage choice, but the argument they make is distinct from the one presented here.

It occurs to me that perhaps "professional rationalist" is a bit of an oxymoron. In today's society, a professional rationalist is basically someone who is paid to come up with publishable results in the academic fields related to rationality, such as decision theory and game theory.

I've always hated the idea of being paid for my ideas, and now I think I know why. When you're being paid for you ideas, you better have "good" ideas or you're going to starve (or at least suffer career failure). But good ideas can't be produced on a schedule, so the only way to guarantee academic survival is to learn the art of self-promotion. Your papers better make your mediocre ideas look good, and your good ideas look great. And having doubts about your ideas isn't going to help your career, even if they are rationally justified.

Given this reality, how can any professional rationalist truly be rational?