X-Message-Number: 16099
Date: Sun, 22 Apr 2001 11:17:24 -0700 (PDT)
From: Scott Badger <>
Subject: Re: Identity Issue ... Again

On Sat, 21 Apr 2001, Lee Corbin wrote

> Suppose that you are taken into the next room where
> a
> frozen duplicate of you, made five minutes ago, lies
> encased in ice.  There is a briefcase on top the ice
> cask containing ten million dollars, and you may
> either
> (a) choose to have your duplicate and the money be
> disintegrated, or (b) choose to be disintegrated 
> yourself.  If you select the latter, then the
> duplicate
> is defrosted and gets to deposit the money.
> The question is, given that you are to make the most
> self-interested decision you can, for the benefit of
> the person you consider yourself to be, would you
> choose
> (a) or (b)?

Great thought experiment. Now I'm unresolved again
darn it. Thanks a lot!

> But in our discussion during the fall, it was clear
> that
> only a small minority would go so far.  How about
> you?
> Lee

First of all, my drive to survive in the current body
would likely preclude my choosing the disintegration
option. Even if I accepted the abstract notion that
the copy "is" me and "I" continue when the original is
gone, I can't imagine allowing myself to be killed on
the basis of that belief. Why take the chance that I'm
wrong? But OK, I'll suspend all that for now and play
the game.

Wouldn't making the "most self-interested decision"
involve assessing outcome probabilities and outcome
valuations. Here's the process:

1. What is the probability that there really is no
difference between the copy and you and that either
one could exist and "you" would continue?  Let's say
you determine that it's a 75% chance while the odds of
"you" continuing if you kill the copy are very close
to 100%.

2. What value do you place on "you" continuing with $1
million? Let's say you determine that value to be 75
with 100 being the highest value. (e.g. 85 might
represent "you" continuing with $5 million)

3. What value do you place on "you" continuing without
$1 million additional dollars? Let's say you determine
that value to be 65.

So, option (a)  X = 1.00 x 65 = 65.00

    option (b)  X =  .75 x 75 = 56.25

In this case, you would destroy the copy. Your numbers
may vary. Of course, there are other variables that
are likely to enter the equation as well.

Since I can't figure out how we'll ever resolve this
issue, it seems to boil down to this kind of personal

Of course my math is rusty. Actually, I like Kennita's
worldview about as well as anything.

Best regards,

Scott Badger

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