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FAWS comments on Counterfactual Calculation and Observational Knowledge - Less Wrong
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Comments (183)
Of course not.
No. I'm assuming that either even is correct in all worlds or odd is correct in all worlds (0.5 prior for either). If Omega randomly picks a world, the chance of the calculator being correct is independent of that and 99% everywhere, then there is a 99% chance of the calculator being correct in the particular world Omega arrives in. If odd is correct Omega is 99% likely to arrive in a world where the calculator says odd, and if the calculator says odd in the particular world Omega arrives in there is a 99% chance that's because odd is correct.
EDIT:
If I were
the probability of even being correct would be 50% no matter what, and there would be a 50% chance each for affecting 99% of all worlds or 1% of all worlds.
I seem to agree with all of the above statements. The conditional probabilities are indeed this way. But it's incorrect to use these conditional probabilities (which is to say, probabilities of Odd/Even after updating on observing "even") to compute expected utility for the counterfactual. In a prior comment, you write:
99% is P(Even|Omega,"even"), that is to say it's probability of Even updated by observations (events) that Omega and "even".
No. There is no problem with using conditional probabilities if you use the correct conditional probabilities, that is the probabilities from wherever the decision happens, not from what you personally encounter. And I never claimed that any of the pieces you were quoting were part of an updateless analysis, just that it made no difference.
I would try to write a Wei Dai style world program at this point, but I know no programming at all and am unsure how drawing at random is supposed to be represented. It would be the same as the program for this game, though:
1 black and 99 white balls in an urn. You prefer white balls. You may decide to draw a ball and change all balls of the other color to balls of the color drawn, and must decide before the draw is made. (or to make it slightly more complicated: Someone else secretly flips a coin whether you get points for black or white balls. You get 99 balls of the color you get points for and one ball of the other color).
It would help a lot if you just wrote the formulas you use for computing expected utility (or the probabilities you named) in symbols, as in P(Odd|"odd")=0.99,
P(Odd|"odd")*100+P(Even|"odd")*0 = 0.99*100+0.01*0 = 99.
Do you need more than that? I don't see how this could possibly help, but:
N(worlds)=100
For each world:
P(correct)=0.99
U_world(correct)=1
U_world(~correct) = 0
P(Omega)=0.01
P(correct|Omega)=P(correct|~Omega) = 0.99
If choosing to replace:
correct ∧ Omega ⇒for all worlds: U_world(~correct) = 1
~correct ∧ Omega ⇒for all worlds: U_world(correct) = 0
This is imprecise in that exactly one world ends up with Omega.
I give up, sorry. Read up on standard concepts/notation for expected utility/conditional probability maybe.
I don't think there is a standard notation for what I was trying to express (if there was formalizing the simple equivalent game I gave should be trivial, so why didn't you do that?) if you are happy with just the end result here is another attempt:
P(Odd|"odd")=P(Even|"even")=P("odd"|Odd)=P("even"|Even)=0.99, P(Odd)=P(Even)=0.5, P("odd" n Odd)= P("even" n Even) =0.495
U_not_replace = P("odd" n Odd)*100 + P("even" n Odd)*0 +P("even" n Even)*100 + P("odd" n Even)*0 = 0.495*100 + 0.005*0 + 0.495*100 + 0.005*0 = 99
U_replace= P("odd"|Odd)*( P("odd" n Odd)*100 + P("even" n Odd)*100) + P("even"|Odd)*( P("odd" n Odd)*0 + P("even" n Odd)*0) + P("even"|Even)*( P("even" n Even)*100 + P("odd" n Even)*100) + P("odd"|Even)*( P("even" n Even)*0 + P("odd" n Even)*0) = 0.99*( 0.495*100 + 0.005*100) + 0.01* ( 0.495*0 + 0.005*0) +0.99*( 0.495*100 + 0.005*100) + 0.01* ( 0.495*0 + 0.005*0) =99
Probabilities correct, U_not_replace correct, U_replace I don't see what's going on with (what's the first conceptual step that generates that formula?). Correct U_replace is just this:
U_replace_updateless = P("odd" n Odd)*0 + P("even" n Odd)*0 +P("even" n Even)*100 + P("odd" n Even)*100 = 0.495*0 + 0.005*0 + 0.495*100 + 0.005*100 = 50
That seems obviously incorrect to me because as an updateless decision maker you don't know you are in the branch where you replace odds with evens. Your utility is half way between a correct updateless analysis and a correct analysis with updates. Or it is the correct utility if Omega also replaces the result in worlds where the parity of Q is different (so either Q is different or Omega randomly decides whether it's actually going to visit anyone or just predict what you would decide if the situation was different and applies that to whatever happens), in which case you have done a horrible job of miscommunication.
I have only a vague idea what exactly required more explanation so I'll try to explain everything.
My U_replace is the utility if you act on the general policy of replacing the result in counterfactual branches with the result in the branch Omega visits. It's the average over all imaginable worlds (imaginable worlds where Q is even and those where Q is odd), the probability of a world multiplied with its utility.
P("odd"|Odd)*( P("odd" n Odd)*100 + P("even" n Odd)*100) + P("even"|Odd)*( P("odd" n Odd)*0 + P("even" n Odd)*0) is the utility for the half of imaginable worlds where Q is odd (all possible worlds if Q is odd).
P("odd"|Odd) is the probability that the calculator shows odd in whatever other possible world Omega visits, conditional on Q being odd (which is correct to use because here only imaginable worlds where Q is odd are considered, the even worlds come later). If that happens the utility for worlds where the calculator shows even is replaced with 100.
P("even"|Odd) is the probability that the calculator shows even in the other possible (=odd) world Omega visits. If that happens the utility for possible worlds where the calculator shows odd is replaced with 0.
At this point I'd just say replace odd with even for the other half, but last time I said something like that it didn't seem to work so here's it replaced manually:
P("even"|even)*( P("even" n even)*100 + P("odd" n even)*100) + P("odd"|even)*( P("even" n even)*0 + P("odd" n even)*0) is the utility for the half of imaginable worlds where Q is even (all possible worlds if Q is even).
P("even"|even) is the probability that the calculator shows even in whatever other possible world Omega visits, conditional on Q being even (which is correct to use because here only imaginable worlds where Q is even are considered, the odd worlds came earlier). If that happens the utility for worlds where the calculator shows odd is replaced with 100.
P("odd"|even) is the probability that the calculator shows odd in the other possible (=even) world Omega visits. If that happens the utility for possible worlds where the calculator shows even is replaced with 0.
If you want to say that updateless analysis is not allowed to take dependencies of this kind into account I ask you for an updateless analysis of the game with black and white balls a few comments upthread. Either updateless analysis as you understand it can't deal with that game (and is therefore incomplete) or I can use whatever you use to formalize that game for this problem and you can't brush me aside with saying that I'm not working updatelessly.
EDIT: The third interpretation of your utility function would be the utility of the general policy of always replacing odds with evens regardless of what the calculator in the world Omega visited showed, which would be so ridiculously stupid that it didn't occur to me anyone might possibly be talking about that, even to point out fallacious thinking.
Consider expected utility [P("odd" n Odd)*100 + P("even" n Odd)*100)] from your formula. What event and decision is this the expected utility of? It seems to consider two events, ["odd" n Odd] and ["even" n Odd]. For both of them to get 100 utils, the strategy (decision) you're considering must be, always answer-odd (since you can only answer in response to indication on the calculators, and here we have both indications and the same answer necessary for success in both events).
But U_replace estimates the expected utility of a different strategy, of strategy where you answer-even on your own "even" branch and also answer-even on the "odd" branch with Omega's help. So you're already computing something different.
Then, in the same formula, you have [P("odd" n Odd)*0 + P("even" n Odd)*0]. But to get 0 utils in both cases, you have to answer incorrectly in both cases, and since we're considering Odd, this must be unconditional answer-even. This contradicts the way you did your expected utility calculation in the first terms of the formula (where you were considering the strategy of unconditional answer-odd).
Expected utility is computed for one strategy at a time, and values of expected utility computed separately for each strategy are used to compare the strategies. You seem to be doing something else.
Since you don't know what parity of Q is, you can't refer to the class of worlds where it's "the same" or "different", in particular because it can't be different. So again, I don't know what you describe here.
(It's still correct to talk about the sets of possible worlds that rely on Q being either even or odd, because that's your model of uncertainty, and you are uncertain about whether Q is even or odd. But not of sets of possible worlds that have your parity of Q, just as it doesn't make sense to talk of the actual state of the world (as opposed to the current observational event, which is defined by past observations).)
I don't understand what this refers to. (Which branch is that? What do you mean by "replace"? Does your 'odd' refer to calculator-shows-odd or it's-actually-odd or 'let's-write-"odd"-on-the-test-sheet etc.?)
Also, updateless decision-maker reasons about strategies, which describe responses to all possible observations, and in this sense updateless analysis does take possible observations into account.
(The downside of long replies and asynchronous communication: it's better to be able to interrupt after a few words and make sure we won't talk past each other for another hour.)