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

From: Wei Dai <>
Date: Mon, 15 Jul 2002 18:30:56 -0700

On Thu, Jun 27, 2002 at 12:50:31PM -0700, Wei Dai wrote:
> P.S. I retract my claim that the self-sampling assumption is incorrect. I
> think I was just using it incorrectly. More on this in another post.

So to elaborate, the reason applying the SSA seemed to lead to a bad
result is that I forgot to apply game theory, which is required here
because we have more than one person making decisions that interact with
each other (even though they're copies of one original person and have the
same preferences).

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.

According to my incorrect analysis, SSA would imply that you choose option
2, because that gives you .5*U(TV2) + .5*U(TV) > .5*U(TV) + .5*U(stereo)
since U(TV2) > U(stereo). I argued that you should consider yourself A and
B simultaneously so you could rationally choose option 2, because
U({TV,stereo}) > U({TV2, TV}). However taking both SSA and game theory
into account implies that option 2 is rational. Furthermore, my earlier
suggestion leads to unintuitive results in general, when the two players
do not share the same utility function.

The game theoretic analysis goes like this. There are two possible
outcomes with pure strategies (I'll ignore mixed strategies for now).
Either A and B both choose 1, or they both choose 2. The first one is a
Nash equilibrium, the second may or may not be. To understand what this
means, suppose you are one of the players in this game (either A or B but
you don't know which) and you expect the other player to choose option 1.
Then your expected utility if you choose option 1 is .5*U({TV,stereo}) +
.5*U({TV,stereo}). If you choose option 2, the expected utility is
.5*U({TV2,stereo}) + .5*({TV,TV}) which is strictly less. So you have no
reason not to choose option 1 if you expect the other player to choose
option 1. Whether or not the second possible outcome is also a Nash
equilibrium depends on whether U({TV2,TV}) > .5*U({TV2,stereo}) +
.5*({TV,TV}). But even if it is, the players can just coordinate ahead of
time (or implicitly) to choose option 1 and obtain the better equilibrium.

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.
Received on Mon Jul 15 2002 - 18:31:36 PDT

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