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From: Wei Dai <weidai.domain.name.hidden>

Date: Thu, 18 Jul 2002 11:20:15 -0700

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.

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
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*> subjective perspective, where he doesn't know if this is the first or
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*> second run, he can at least figure that there is a 50% chance that his
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*> decision has a causal effect on the outcome. It seems to me that this
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*> 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?

Received on Thu Jul 18 2002 - 11:21:02 PDT

Date: Thu, 18 Jul 2002 11:20:15 -0700

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.

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:

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?

Received on Thu Jul 18 2002 - 11:21:02 PDT

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