Re: SSA and game theory (was: self-sampling assumption is incorrect)

From: Hal Finney <hal.domain.name.hidden>
Date: Tue, 16 Jul 2002 18:58:50 -0700

Wei writes:
> Here's a simplified thought experiment that illustrates the issue. Two
> copies of the subject S, A and B, are asked to choose option 1 or option
> 2. If A chooses 1, S wins a TV (TV), otherise S wins a worse TV (TV2). If
> B chooses 1, S wins a stereo, otherwise S wins TV. S prefers TV to TV2 to
> stereo, but would rather have a TV and a stereo than two TVs. The copies
> have to choose without knowing whether they are A or B.

I am confused about the relation of S to A and B. Did S go into a
copying machine and get two copies, A and B made, in addition to S?
And now A and B are deciding what S will win?

Why should they care? If S gets a TV that does not benefit them. Is it
just that they are similar to S, being recent copies of him, so they
have a sort of brotherly fondness for him and would like to see him happy?

I thought the earlier experiments had a more direct connection, where
the people making the decisions were the same ones who were getting
the reward (or at least, future copies of themselves)


> Now I want to show that while my earlier suggestion to consider yourself
> to be both A and B works in this case, it doesn't work in general when A
> and B have different utility functions. Consider the following game, which
> we can call Amnesiac Prisoner's Dillemma. Players A and B can choose
> Cooperate or Defect. If they both Defect they both get sentenced to 5
> years in prison. If they both Cooperate they both get 1 year. If one
> Cooperates and the other Defects they get 3 months and 10 years
> respectively. The twist is that you have no memory of who you are and
> don't know whether you're A or B. If you consider yourself to be both A
> and B, then you would choose Cooperate, which contradicts our intuition. A
> game theoretic analysis with the SSA would go like this. Suppose you
> expect the other player to Cooperate. Then EU(Cooperate) = .5*EU(Cooperate
> | I'm A) + .5*EU(Cooperate | I'm B) = .5*U(I'm A and A gets 1 year and B
> gets 1 year) + .5*U(I'm B and A gets 1 year and B gets 1 year) <
> EU(Defect) = .5*EU(Defect | I'm A) + .5*EU(Defect | I'm B) = .5*U(I'm A
> and A gets three months and B gets 10 years) + .5*EU(I'm B and A gets 10
> years and B gets three months). So Cooperate is not a Nash equilibrium.

I'm not sure I understand this; since the payoff matrix is symmetric it
doesn't matter if you are A or B so I don't see what the point is of
introducing amnesia. Would there be any cases with symmetric payoffs
where an amnesiac would behave differently than someone who knew whether
he was A or B?

Tangentially, it seems that the PD is a case where the "evidential"
vs "causal" principles of decision theory would show a difference.
The evidentialist would argue that by cooperating, it increases the
chance that the other person will cooperate (perhaps to a certainty, in
some versions), hence cooperating can be justified. This is basically
Hofstadter's principle of super-rationality. The causalist would reject
the possible correlation of choices and choose the dominant strategy
of defecting. Does that seem correct?

What about this: you are going to be copied, and your two copies are going
to play a PD game. You know this ahead of time and so you can decide
on whatever strategies you intend to follow during this time before the
copying occurs. Do you think in that case it would be rational to firmly
decide beforehand to cooperate?

And would "amnesia" make a difference? We might arrange for amnesia by
having the duplicates immediately play the game, without any knowledge
of which they are; and remove the amnesia by simply telling them that
one is A and one is B, before they play. Again, with a symmetric game
I don't see how the amnesia or its absence would be relevant. Maybe I
am misunderstanding that aspect.

Hal Finney
Received on Tue Jul 16 2002 - 19:12:23 PDT