Here's my response to the rest of your post. I think you're right that
with two identical deterministic computations, there is no
need to apply game theory. I think in that case you should consider
yourself to be both of them. It would not work to think there's 50% chance
you're one and 50% chance you're the other.

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In my analysis I did not make the assumption that all of the copies are
deterministic and have no access to independent random numbers, so game
theoretic considerations do apply. In the first scenario where your copies
are trying to win prizes for you, game theory lead to the same
conclusion. In the Amnesiac Prisoner's Dillemma, the assumption is that
the players are different people who have suffered temporary amnesia, not
copies of one original person, so they can't be identical deterministic
computations.
On Wed, Jul 17, 2002 at 06:49:04PM -0700, Hal Finney wrote:
> The point is that in one of those two states, his decision does in
> fact have a causal effect on the outcome. It is the direct effect of
> his decision that lets the experimenter fill the boxes. So from his
> subjective perspective, where he doesn't know if this is the first or
> second run, he can at least figure that there is a 50% chance that his
> decision has a causal effect on the outcome. It seems to me that this
> might be enough to justify choosing one box even from a causal analysis.
Again, I think you need to think of yourself as both runs, so there is
probability 1 that your decision has a causal effect on the amount of
money in box 2. Otherwise, how do you compute the expected payoff of
chosing only one box?