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orthonormal comments on The Absent-Minded Driver - Less Wrong
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Jaynes' principle is that injecting randomness shouldn't help you figure out what's true; implementing a mixed strategy of action is a different matter.
Although the distinction is a bit too fuzzy (in my head) for comfort.
I'm concerned about more than just Jaynes's principle. Injecting noise between your decision, and what you turn out to do, shouldn't be able to help you either. Choosing "continue" 2/3 of the time is the same as "Always continue, but add a random disturbance that reverses this 1/3 of the time."
How can that addition of randomness help you?
The randomness helps in this case because the strategy of determining your actions by which intersection you are at is not available.
Yes, the problem deletes the knowledge of what intersection I'm at. How does it help to further delete knowledge of what my decision is?
There are 3 possible sequences of actions:
The payoffs are such that sequence 2 is the best available, but, having complete knowledge of your decision and no knowledge of which intersection you are at makes sequence 2 impossible. By sacrificing that knowledge, you make available the best sequence, though you can no longer be sure you will take it. In this case, the possibility of performing the best sequence is more valuable than full knowledge of which sequence you actually perform.
By the way, I went ahead and calculated what effect probabilistic knowledge of one's current intersection has on the payoff, so as to know the value of this knowledge. So I calculated the expected return, given that you have a probability r (with 0.5<r<=1) of correctly guessing what intersection you're in, and choose the optimal path based on whichever is most likely.
In the original problem the max payoff (under p=2/3) is 4/3. I found that to beat that, you only need r to be greater than 52.05%, barely better than chance. Alternately, that's only 0.0012 bits of the 1 bit of information contained by the knowledge of which intersection you're at! (Remember that if you have less than .0012 bits, you can just revert to the p=2/3 method from the original problem, which is better than trying to use your knowledge.)
Proof: At X, you have probability r of continuing, then at Y you have a probability r of exiting and (1-r) of continuing.
Thus, EU(r) = r(4r + (1-r) ) = r(3r+1).
Then solve for when EU(r)=4/3, the optimum in the fully ignorant case, which is at r = 0.5205.
You made a mistake here, which is assuming that when you guess you are at X, you should choose CONTINUE with probability 1, and when you guess you are at Y, you should choose EXIT with probability 1. In fact you can improve your expected payoff using a mixed strategy, in which case you can always do better when you have more information.
Here's the math. Suppose when you are at an intersection, you get a clue that reads either 'X' or 'Y'. This clue is determined by a dice roll at START. With probability .49, you get 'X' at both intersections. With probability .49, you get 'Y' at both intersections. With probability .02, you get 'X' at the X intersection, and 'Y' at the Y intersection.
Now, at START, your decision consists of a pair <p,q> of probabilities, where p is your probability to CONTINUE after seeing 'X', and q is your probability to CONTINUE after seeing 'Y'. Your expected payoff is:
.02 * (p*q + 4*(p*(1-q))) + .49 * (p*p + 4*(p*(1-p))) + .49 * (q*q + 4*(q*(1-q)))
which is maximized at p=0.680556, q=0.652778. And your expected payoff is 1.33389 which is > 4/3.
Wow, good catch! (In any case, I had realized that if you have probability less than 52.05%, you shouldn't go with the most likely, but rather, revert the original p=2/3 method at the very least.)
The formula you gave for the mixed strategy (with coefficients .02, .49, .49) corresponds to a 51% probability of guessing right at any given light. (If the probability of guessing right is r, the coefficients should be 2r-1,1-r,1-r.) It actually raises the threshold for which choosing based on the most probable, with no other randomness, becomes the better strategy, but not by much -- just to about 52.1%, by my calculations.
So that means the threshold is 0.0013 bits instead of 0.0012 :-P
(Yeah, I did guess and check because I couldn't think of a better way on this computer.)
I think you might still be confused, but the nature of your confusion isn't quite clear to me. Are you saying that if r>52.1%, the best strategy is a pure one again? That's not true. See this calculation with coefficients .2, .4, .4.
ETA: Also, I think talking about this r, which is supposed to be a guess of "being at X", is unnecessarily confusing, because how that probability should be computed from a given problem statement, and whether it's meaningful at all, are under dispute. I suggest thinking in terms of what you should plan to do when you are at START.
That yields a payoff of 1.42, which is less than what the pure strategy gives in the equivalent case corresponding to .2/.4/.4, which is a 60% chance of guessing right. Since the payoff is r*(3r+1), the situation you described has a payoff of 1.68 under a pure strategy of choosing based on your best guess.
I specifically avoided defining r as the probability of "being at X"; r is the probability of guessing correctly (and therefore of picking the best option as if it were true), whichever signal you're at, and it's equivalent to choosing 2r-1,r-1,r-1 as the coefficients in your phrasing. The only thing possibly counterintuitive is that your ignorance maximizes at r=0.5 rather than zero. Less than 50%, and you just flip your prediction.
Okay, that's a more specific, helpful answer.