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SilasBarta comments on The Absent-Minded Driver - Less Wrong
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I'm not confused; I probably should have stopped you at your original derivation for the partial-knowledge case but didn't want to check your algebra. And setting up problems like these is important and tricky, so this discussion belongs here.
So, I think the problem with your setup is that you don't make the outcome space fully symmetric because you don't have an equal chance of drawing Y at X and X at Y (compared to your chance of drawing X at X and Y at Y).
To formalize it for the general case of partial knowledge, plus probabilistic knowledge given action, we need to look at four possibilities: Drawing XY, XX, YY, and YX, only the first of which is correct. If, as I defined it before, the probability of being right at any given exit is r, the corresponding probabilities are: r^2, r(1-r), r(1-r), and (1-r)(1-r).
So then I have the expected payoff as a function of p, q, and r as:
(r^2)(p*q + 4p*(1 - q)) + r(1 - r)(p^2 + 4p(1 - p)) +r(1 - r)(q^2 + 4q(1 - q)) + (1 - r)(1 - r)(p*q + 4q(1 - p))
This nicely explains the previous results:
The original problem is the case of complete ignorance, r=1/2, which has a maximum 4/3 where p and q are such that they average out to choosing "continue" at one's current intersection 2/3 of the time. (And this, I think, shows you how to correctly answer while explicitly and correctly representing your probability of being at a given intersection.)
The case of (always) continuing on (guessing) X and not continuing on (guessing) Y corresponds to p=1 and q=0, which reduces to r*(3r+1), the equation I originally had.
Furthermore, it shows how to beat the payoff of 4/3 when your r is under 52%. For 51% (the original case you looked at), the max payoff is 1.361 {p=1, q=0.306388} (don't know how to show it on Alpha, since you have to constrain p and q to 0 thru 1).
Also, it shows I was in error about the 52% threshold, and the mixed strategy actually dominates all the way up to about r = 61%, at which point of course p=1 and q has shrunk to zero (continue when you see X, exit when you see Y). This corresponds to 32 millibits, much higher than my earlier estimate of 1.3 millibits.
Interesting!