On Mon, Oct 21, 2013 at 6:03 PM, meekerdb <meeke...@verizon.net> wrote:

 On 10/21/2013 9:16 AM, John Clark wrote:
>  >>  Let me put it in this way: accepting that P(W) = P(M) =1/2, with W
>> and M describing the first person experiences of the respective copies, do
>> you accept that P(M) = P(W) = 1/2,
>  > No I don't accept that, not if P(W) is the probability that the
> Washington Man will see Washington; the probability of that would be 1 not
> 1/2. And if P(W) means the probability the Helsinki Man will see Washington
> that would be 0 not 1/2 because the Helsinki Man would have to be turned
> into something that is not the Helsinki Man before the Helsinki Man can see
> a different city.
> > Why?  If he flew to Washington he would still be the Helsinki man.

OK, then if he flew to Moscow he would be the Helsinki man too, and if he
used a Star Trek style transporter instead of a airplane he would still be
the Helsinki Man, and if the transporter sent him to both cities at the
same time he would still be "The Helsinki Man". So you tell me, using logic
and your definition how many cities did "The Helsinki Man" see?

>   > He's the Helsinki man because of the continuity of his memories, just
> as you are still John Clark even though you've changed locations since
> yesterday.

Fine. If that's what you mean by "The Helsinki Man" then in Bruno's thought
experiment with the duplication chamber and using the exact same reasoning
the probability The Helsinki Man will see Washington is 100% and the
probability The Helsinki Man will see Moscow is 100%. And yes yes I know,
each copy will see only one city, but if the definition of  "The Helsinki
Man" is the one you give above, "the continuity of his memories", then it
is irrelevant how many cities each individual copy sees.

  John K Clark

You received this message because you are subscribed to the Google Groups 
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to everything-list+unsubscr...@googlegroups.com.
To post to this group, send email to everything-list@googlegroups.com.
Visit this group at http://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/groups/opt_out.

Reply via email to