Le 06-juil.-05, à 02:44, Lee Corbin a écrit :
Bruno wrote about whether or not we are all the same person.
Sent: Tuesday, July 05, 2005 1:59 AM
Subject: Re: What does "ought" mean? (was RE: Duplicates Are Selves)
I have changed the subject line once again, because this is
no longer about what "ought" ought to mean.
Le 04-juil.-05, à 22:18, Lee Corbin a écrit :
Yes, but I contend that while there are two organisms present,
there is only one person. It's much as though some space
aliens kidnapped you and tried to say that Pete at spacetime
coordinates (X1,T1) could not possibly be the same person as
Pete at coordinates (X1,T2) because the times weren't the same.
You'd have to get them to wrap their heads around the idea that
one person could be at two different times in the same place.
They might find this bizarre.
I'm trying to tell you a possibility that you think equally
bizarre: namely that Pete(X1,T1) is the same person as Pete(X2,T1),
namely that the same person may be at two different locations at
the same time. That's all.
I like that idea, but if they are the *same* person then we are
all the same person.
Or, perhaps you were just meaning that they are very close/similar;
That is so: but moreover, being very very close/similar is what
should be meant by "the same person".
in which case you can say Pete(X2,T1) is much closer to
Pete(X1,T1) than Bruno(x, now) is close to Lee(y, now).
But then, strictly speaking Pete(X1,T1) is not the same
person as Pete(X2,T1).
Well, that's up for discussion! That's what we are trying
I say that it gets pretty silly to formulate our ideas so
that we turn out not to be the same person from second to
second. Now, yes, in order to evade the notion that one is
the same person as one's duplicate across the room, people
will try anything, even denying that they have any identity
whatsoever. They are, apparently, more comfortable with the
notion that they are not the same person from second to
second that the shocking idea that they and their duplicates
are the same person.
In any case I am not sure that those distinctions have any
bearing on the existence of first person indeterminacy and
the problem to quantify that indeterminacy.
(Yes, maybe it is detached from the question you are trying
Imagine you are duplicated iteratively. At the start you
are in room R. You are scanned and destroyed, painlessly,
given some of the discussions we have, :-) this is rather
pleasant to entertain
and we tell you that you will be reconstituted in room 0
and in room 1. Then Lee0 and Lee1 are invited in room R
again and the experience is repeated. Rooms 0 and 1 are
identical and quite separate. The only difference is
that in room 0 there is a big 0 drawn on the wall and
in room 1 there is a big 1 drawn on the wall.
So as this is repeated, there are 2, then 4, then 8, etc.,
copies, and each of them remembers a different sequence of
0's and 1's.
You are asked to bet on your immediate and less immediate
future feeling. Precisely: we ask you to choose among the
A. I will see 0 on the wall.
B. I will see 1 on the wall.
C. I will see 0 on the wall and I will see 1 on the wall.
D. I will see 0 on the wall or I will see 1 on the wall.
A'. I will always see 0 on the wall.
B'. I will always see 1 on the wall
C'. I will see as many 0 and 1 on the wall
D'. I will see an incompressible sequence of 0 and 1 on the wall
And there are three versions of the experiences. In a first version
are always reconstituted in the two rooms.
Okay, let's handle just that for now.
We suppose obviously that you want maximize "your" benefit(s).
Well, since you are asking *me*, then naturally I'll want a
global maximum for me, and so a maximal sum for each instance.
Good idea. And quite coherent with your idea that we are our duplicates.
Each Lee-i is offered 5$ each time his bet is confirmed, but
loses 5$ if he makes a wrong bet.
And yes, it would be possible to emphasize to each instance that
he is to attempt to maximize "his own instance's" earnings.
What will be your strategy in each version? Will your strategy differ?
Now if the Lees know all these facts, then they'll anticipate being
in both rooms upon each iteration. Therefore, they'll anticipate
losing $5 in one room and gaining $5 in the other. They'll also
realize that all bit sequences are being carried out. Therefore,
it doesn't make any difference whatsoever. The expectation of
each sequence is exactly the same number of dollars: zero.
I don't get the significance of this.
I don't understand your answer, and actually you did not answer. It
looks like you are forgetting I give you the choice between A, B, C, D.
I guess you did choose C, without saying.
In that case you are correct the expectation will be zero. Are you sure
there is not a better strategy among A, B, C, D?
And what about the long run (A', B', C', D') ?
(I agree with you about forgetting the second and third version).