lolbifrons

• lolbifronsMar 29, 2018, 6:21 AM

  1 point

  in reply to: Said Achmiz’s comment on: Every Im­ple­men­ta­tion of You is You: An In­tu­ition Ladder

  I have expanded on 1 in an edit. Let me know if it makes sense.

• lolbifronsMar 29, 2018, 5:59 AM

  1 point

  in reply to: Said Achmiz’s comment on: Every Im­ple­men­ta­tion of You is You: An In­tu­ition Ladder

  This needs more elaboration, if you want to use it in the way you do. I know what you mean here (at least, I think I do), but it may not be obvious to many interlocutors—“the same thing” in what way, exactly? (Especially since this is step #1.)

  I see your point. Admittedly, I built this while talking to someone who was past 1, and I consider the position a gross misunderstanding of current technology. I’ll consider how to describe the proposition better, though.
  

  Or is it your intention to show that this view is incoherent, for exactly this reason

  Precisely. I expect any lesswronger to be past 4. Perhaps I am building a strawman?
  

  How does this step on the ladder work for people who are not learned in quantum mechanics? (Which is to say—I have no idea what #6 means.)

  I suppose it’s enough to intuitively consider it “vaguely less likely” that you’ll wake up as one copy or the other based on “some unknown criteria that possibly includes likelihood of future survival”, but I submit that either:

  1. one doesn’t understand fully why he is at this step if he isn’t familiar with QM, or

  2. I don’t understand QM or the subject at hand as well as I think I do.
  

  I confess to being confused about how the concept of “probability” is being used here (and in similar comments I’ve seen). Can you elaborate?

  That… might be a pretty long discussion. Essentially, whatever you, before you undergo a copying procedure, anticipate happening to you (subjectively experiencing ending up in the grown body vs. ending up in the body that walked in, each with some respective probability), you have the same anticipation about staying on Earth vs. instantaneously mind-warping to where-ever the hell a human brain could possibly have formed that not only implements you, but has exactly your memories of growing up on earth up to that moment.
  

  As for your entire approach… I think it’s an interesting and worthy attempt, but flawed, for this reason: that around #4 or #5 (I’m not sure, due to the unclearness mentioned above), it’s hard to know what intuition to have—whether to reject or accept the higher rungs of the ladder

  One thing that perhaps isn’t obvious, is that this ladder isn’t meant to convince anyone of anything upon reading it, other than perhaps that I might have a consistent position and not be spouting utter nonsense. Once someone and I have agreed about where they are, then we can get to work on the task of advancing up the ladder. One level at a time.
  

  do you have a solution to the Hard Problem?

  Could you elaborate? I may know what you’re talking about, but not by name.

Every Im­ple­men­ta­tion of You is You: An In­tu­ition Ladder

• lolbifronsMar 29, 2018, 1:41 AM

  3 points

  in reply to: TurnTrout’s comment on: Open-Cat­e­gory Classification

  To think he could have had twice as many readers.

• lolbifronsMar 29, 2018, 1:15 AM

  3 points

  on: En­tropy, and Short Codes

  The encoding scheme you’re talking about is Huffman Coding, and the ambiguity you’re trying to avoid explicitly occurs when one symbol is the prefix of another. The mechanism to build an optimal prefix(-free) code is called the Huffman Tree, and it uses a greedy algorithm that builds a full binary tree from the bottom up based on the frequencies of the symbols. Leaves are symbols, and the code for a symbol is the sequence of left or right branches you must traverse the reach that symbol.

  To get more specific, you add all the symbols to a heap based on their frequency of appearance, lowest-at-the-front. You extract two symbols, foo and bar. These become children of a common parent node, the pseudosymbol baz. baz.freq = foo.freq + bar.freq. Add baz to the heap. Loop until the heap contains one node at the end of this process. That node is the root of the Huffman Tree.

• lolbifronsAug 18, 2016, 12:09 PM

  1 point

  in reply to: thrawnca’s comment on: Coun­ter­fac­tual Mugging

  Sorry, but I’m not in the habit of taking one for the quantum superteam.

  If you’re not willing to “take one for the team” of superyous, I’m not sure you understand the implications of “every implementation of you is you.”

  And I don’t think that it really helps to solve the problem;

  It does solve the problem, though, because it’s a consistent way to formalize the decision so that on average for things like this you are winning.

  it just means that you don’t necessarily care so much about winning any more. Not exactly the point.

  I think you’re missing the point here. Winning in this case is doing the thing that on average nets you the most success for problems of this class, one single instance of it notwithstanding.

  Plus we are explicitly told that the coin is deterministic and comes down tails in the majority of worlds.

  And this explains why you’re missing the point. We are told no such thing. We are told it’s a fair coin and that can only mean that if you divide up worlds by their probability density, you win in half of them. This is defined.

  What seems to be confusing you is that you’re told “in this particular problem, for the sake of argument, assume you’re in one of the worlds where you lose.” It states nothing about those worlds being over represented.

• lolbifronsAug 18, 2016, 12:23 AM

  4 points

  on: The Allais Paradox

  So here’s why I prefer 1A and 2B after doing the math, and what that math is.

  1A = 24000
  1B = 26206 (rounded)
  2A = 8160
  2B = 8910

  Now, if you take (iB-iA)/​iA, which represents the percent increase in the expected value of iB over iA, you get the same number, as you stated.

  (iB-iA)/​iA = .0919 (rounded)

  This number’s reciprocal represents the number of times greater the expected value of iA is than the marginal expected value of iB

  iA/​(iB-iA) = 10.88 (not rounded)

  Now, take this number and divide it by the quantity p(iA wins)-p(iB wins). This represents how much you have to value the first $24000 you receive over the next $3000 to pick iA over iB. Keep in mind that ²⁴⁄₃ = 8, so if $1 = 1 utilon in all cases, you should pick iA only when this quotient is less than 8.

  1A/​(1B-1A)/​[p(1A wins)-p(1B wins)] = 369.92
  2A/​(2B-2A)/​[p(2A wins)-p(2B wins)] = 1088

  I have liabilities in excess of my assets of around $15000. That first $15000 is very important to me in a very quantized, thresholdy way, but it is not absolute. I can make the money some other way, but not needing to—having it available to me right now because of this game—represents more utility than a linear mapping of dollars to utility suggests, by a large factor.

  The next threshold like this in my life that I can think of is “enough money to buy a house in Los Angeles without taking out a mortgage,” of which $3000 is a negligible portion.

  I’d say that the utility I assign the first $24000 because of this lies between 370 and 1080 times the utility I assign the next $3000. This is why I take 1A and 2B given that this entire thing is performed only once. Once my debts are paid, all bets (on 1A) are off.

  If we’re dealing with utilons rather than dollars, or I have repeated opportunity to play (which is necessary for you to “money pump” me) iB is the obvious choice in both cases.

• lolbifronsAug 17, 2016, 1:30 PM

  1 point

  in reply to: thrawnca’s comment on: Coun­ter­fac­tual Mugging

  It seems to me the answer becomes more obvious when you stop imagining the counterfactual you who would have won the $10000, and start imagining the 50% of superpositions of you who are currently winning the $10000 in their respective worlds.

  Every implementation of you is you, and half of them are winning $10000 as the other half lose $100. Take one for the team.

• lolbifronsAug 17, 2016, 11:53 AM

  2 points

  in reply to: Cyan2’s comment on: Con­ser­va­tion of Ex­pected Evidence

  If you count the amount of “wanting to switch” you expect to have because the cable guy hasn’t arrived yet, it should equal exactly the amount of “wishing you hadn’t been wrong” you expect to have if you pick the second half because the cable guy arrived before your window started.

  I’m not sure how to say this so it’s more easily parseable, but this equality is exactly what conservation of expected evidence describes.