- Contemporary messages sorted: [ by date ] [ by thread ] [ by subject ] [ by author ] [ by messages with attachments ]

From: Hal Finney <hal.domain.name.hidden>

Date: Fri, 14 Jun 2002 19:42:27 -0700

Wei writes:

*> Earlier (at http://www.lucifer.com/exi-lists/extropians/2612.html) I
*

*> argued that preference between the two choices is subjective (i.e. depends
*

*> on your utility function). I now realize this implies that the
*

*> self-sampling assumption (or SSA, the idea that you should reason as if
*

*> you were a random sample from the set of all observers, see
*

*> http://www.anthropic-principle.com/index.html for more details) cannot be
*

*> applied universally, because it implies that only choosing the two
*

*> identical experiences is rational.
*

*> ...
*

What about this variant on the experiment (the full experiment is below).

Instead of B1 and B2 both getting E1, let B1 get E1 and B2 get E1'.

E1' is another experience than E1 that is just about as good.

U(E1) > U(E2) and U(E1') > U(E2). The idea is that this eliminates

possible issues regarding whether two people (B1 & B2) who get exactly

the same experience should count twice.

*> Now which strategy I should choose
*

*> depends on whether U({E1,E1}) > U({E1,E2}), which can be independent of
*

*> whether U(E1) > U(E2).
*

We can change this to whether U({E1,E1'}) > U({E1,E2}) in the modified

form.

It does seem that the SSA pretty much implies that if U(E1') > U(E2) then

U({E1,E1'}) > U({E1,E2}). Is it really rational for this to be otherwise?

We know that rationality puts some constraints on the utility function.

We can't have cyclicity in the utility preference graph, for example.

But in the case above, where U({X,Y}) means the utility of having two

different independent experiences X and Y, maybe it does follow that

U({X,Y}) and U({X,Z}) must compare the same as U(Y) and U(Z). You don't

have any choice but to accept the equivalence. As Lewis Carroll wrote,

"Then Logic would take you by the throat, and FORCE you to do it!"

(http://www.mathacademy.com/pr/prime/articles/carroll/index.asp)

Hal

*> Here's a demonstration of this. Suppose you've agreed to participate in
*

*> the following experiment. First you're copied. The original will observe
*

*> while the copy (named A1) is told the following. A1 will be copied into
*

*> A2, B1 and B2. All four will be run on seperate and identical computers.
*

*> A1 and A2 will be shown a number equal to the millionth bit in the binary
*

*> expansion of PI. B1 and B2 will both be shown a number equal to 1 minus
*

*> that bit. All four will be asked to guess the millionth bit of PI. (Assume
*

*> you have no idea what the millionth bit is.) If A1 guesses correctly, it
*

*> will experience a very pleasant experience (call this experience E1). Same
*

*> applies for B1 and B2, each of whom will also have E1 if he guesses
*

*> correctly. If A2 guesses correctly however, he will experience a slightly
*

*> less pleasant experience E2. If anyone guesses incorrectly, he's halted
*

*> immediately. In any event all four copies are halted at the end of the
*

*> experiment. (The setup can be changed so that the four runs are done
*

*> sequentially instead of in parallel. I don't think that affects my
*

*> argument at all.)
*

*>
*

*> Now put yourself in the position of A1 before he's been further copied,
*

*> trying to devise a strategy for guessing the millionth bit of PI. Let's
*

*> call that bit X and the number you'll be shown Y, and consider the two
*

*> strategies A) guess Y, and B) guess 1-Y. It should be obvious at this
*

*> point that if you prefer to have two identical very pleasant experiences
*

*> you'll select strategy B, and if you prefer to have one very pleasant
*

*> experience and one slightly less pleasant experience you'll select
*

*> strategy A. However according to the SSA only strategy B is rational.
*

*>
*

*> Here's how I would analyze the situation given the SSA. After being shown
*

*> Y, there's 1/4 probability that I'm A1, 1/4 probability that I'm A2, 1/4
*

*> probability that I'm B1, and 1/4 probability that I'm B2. So if I guess Y,
*

*> there's 1/4 probability that I cause a copy of me to experience E1 and 1/4
*

*> probability that I cause a copy of me to experience E2, therefore my
*

*> expected utility is U(A) = 1/4*U(E1) + 1/4*U(E2). If I guess 1-Y instead,
*

*> there's 1/4 probability that I cause a copy of me to experience E1 and
*

*> another 1/4 probability that I cause a copy of me to experience E1, so my
*

*> expected utility is U(B) = 1/2*U(E1). Since U(E1) > U(E2), U(B) > U(A).
*

*>
*

*> Here's my proposed non-SSA way of analyzing the situation. After being
*

*> shown Y, I consider myself to be A1, A2, B1, and B2 "simultaneously". If I
*

*> guess 1-Y, there's probability of 1 that I cause two copies of me to
*

*> experience E1 (call this {E1,E1}). If I guess Y, there's probability of 1
*

*> that I cause one copy of me to experience E1 and one copy of me to
*

*> experience E2 (call this {E1,E2}). Now which strategy I should choose
*

*> depends on whether U({E1,E1}) > U({E1,E2}), which can be independent of
*

*> whether U(E1) > U(E2).
*

*>
*

*> So my position is that rather than being a principle of correct reasoning,
*

*> the status of the SSA should be reduced to that of an approximation useful
*

*> when one's utility function is close to satisfying certain constraints
*

*> (for example U({E1,E2})=U(E1)+U(E2) for all E1, E2). More general
*

*> principle(s) need to be worked out that subsumes the SSA as a special
*

*> case.
*

*>
*

Received on Fri Jun 14 2002 - 19:55:49 PDT

Date: Fri, 14 Jun 2002 19:42:27 -0700

Wei writes:

What about this variant on the experiment (the full experiment is below).

Instead of B1 and B2 both getting E1, let B1 get E1 and B2 get E1'.

E1' is another experience than E1 that is just about as good.

U(E1) > U(E2) and U(E1') > U(E2). The idea is that this eliminates

possible issues regarding whether two people (B1 & B2) who get exactly

the same experience should count twice.

We can change this to whether U({E1,E1'}) > U({E1,E2}) in the modified

form.

It does seem that the SSA pretty much implies that if U(E1') > U(E2) then

U({E1,E1'}) > U({E1,E2}). Is it really rational for this to be otherwise?

We know that rationality puts some constraints on the utility function.

We can't have cyclicity in the utility preference graph, for example.

But in the case above, where U({X,Y}) means the utility of having two

different independent experiences X and Y, maybe it does follow that

U({X,Y}) and U({X,Z}) must compare the same as U(Y) and U(Z). You don't

have any choice but to accept the equivalence. As Lewis Carroll wrote,

"Then Logic would take you by the throat, and FORCE you to do it!"

(http://www.mathacademy.com/pr/prime/articles/carroll/index.asp)

Hal

Received on Fri Jun 14 2002 - 19:55:49 PDT

*
This archive was generated by hypermail 2.3.0
: Fri Feb 16 2018 - 13:20:07 PST
*