A simple-minded theory of "observer fluid"

post by cousin_it · 2011-02-16T16:40:04.315Z · LW · GW · Legacy · 20 comments

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Eliezer has proposed a puzzle:

So here's a simple algorithm for winning the lottery:

Buy a ticket.  Suspend your computer program just before the lottery drawing - which should of course be a quantum lottery, so that every ticket wins somewhere.  Program your computational environment to, if you win, make a trillion copies of yourself, and wake them up for ten seconds, long enough to experience winning the lottery.  Then suspend the programs, merge them again, and start the result.  If you don't win the lottery, then just wake up automatically.

In response, rwallace has proposed a reductio of subjective probability:

So you decide to create your full allowance of 99 copies, and a customer service representative explains how the procedure works: the first copy is made, and informed he is copy number one; then the second copy is made, and informed he is copy number two, etc. That sounds fine until you start thinking about it, whereupon the native hue of resolution is sicklied o'er with the pale cast of thought. The problem lies in your anticipated subjective experience.

After step one, you have a 50% chance of finding yourself the original; there is nothing controversial about this much. If you are the original, you have a 50% chance of finding yourself still so after step two, and so on. That means after step 99, your subjective probability of still being the original is 0.5^99, in other words as close to zero as makes no difference.

I think they have both missed the simple-minded and "normal" approach to such problems. Let's hypothesize this:

1) If a single universe contains N identical copies of your current information state (possibly existing at different times and in different places), and some event will happen to K of them, then you should assign probability K/N to that event.

2) If there are multiple universes existing with different "measure" (whatever that is), then your prior probability of being in a certain universe is proportional to its "measure", regardless of the number of your copies in that universe, as long as it's >0.

In Eliezer's puzzle, my assumptions imply that you cannot win the lottery by using anthropic superpowers, because making many copies of yourself in the winning branch only splits the "observer fluid" in that branch, not creates more of it overall.

In rwallace's puzzle, my assumptions imply that your probability of still being the original after 99 copyings is 1/100, if you didn't receive any indexical information in the meantime. The reason: spacetime contains 100 copies of you about to be told who they are, all of them informationally equivalent. The physical fact of which copy was made from which is irrelevant, only information matters. For example, if A is copied into B and then B is copied into C without anyone getting indexical info, the second act of copying also pulls some of A's "observer fluid" into C, so they end up with 1/3 each, instead of 1/2 1/4 1/4.

Now the disclaimers:

I know that speaking of things like "reality fluid" is confused and that we know next to nothing. I don't know if my idea carries over to other puzzles. I don't know how well it matches our reality and how it might follow from physics. I don't know what happens when observers get deleted; maybe killing someone without giving them indexical information just redistributes their fluid among surviving identical branches ("merging"), but maybe it just gets lost forever. I don't know what counts as a copy of you. I don't know how to count copies and whether thickness of computers matters. I don't know how to determine if one observer-moment is a continuation of another observer-moment; maybe it's about correct stepping of algorithms, maybe something else. These are all open questions.

(Thanks to Wei Dai and Manfred for discussions)

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