Hal Finney writes:

Stathis Papaioannou writes:> Here is another way of explaining this situation. When there aremultiple> parallel copies of you, you have no way of knowing which copy you are,> although you definitely are one of the copies during any given moment,with> no telepathic links with the others or anything like that. If aproportion> of the copies are painlessly killed, you notice nothing, because your> successor OM will be provided by one of the copies still going (afterall,> this is what happens in the case of teleportation). Similarly, if thenumber> of copies increases, you notice nothing, because during any given momentyou> are definitely only one of the copies, even if you don't know which one.> However, if your quantum coin flip causes 90% of the copies to have bad > experiences, you *will* notice something: given that it is impossible to> know which particular copy you are at any moment, or which you will bethe> next moment, then there is a 90% chance that you will be one of thosewho> has the bad experience. Similarly, if you multiply the number of copies> tenfold, and give all the "new" copies bad experiences, then even thoughthe> "old" copies are left alone, you will still have a 90% chance of a bad> experience, because it is impossible to know which copy will provideyour> next OM. I'm not sure I fully understand what you are saying, but it sounds like you agree at least to some extent that "copies count". The number of copies, even running in perfect synchrony, will affect the measure of what that observer experiences, or as you would say, his subjective probability. So let me go back to Bruno's thought experiment and see if I understand you. You will walk into a Star Trek transporter and be vaporized and beamed to two places, Washington and Moscow, where you will have two (independent) copies wake up. Actually they are uploads and running on computers, but that doesn't matter (we'll assume). Bruno suggests that you would have a 50-50 expectation of waking up in Washington or Moscow, and I think you agree. But suppose it turns out that the Moscow computer is a parallel processor which, for safety, runs two copies of your program in perfect synchrony, in case one crashes. Two synchronized copies in Moscow, one in Washington. Would you say in this case that you have a 2/3 expectation of waking up in Moscow? And to put it more sharply, suppose instead that in Washington you will have 10 copies waking up, all independent and going on and living their lives (to the extent that uploads can do so), sharing only the memory of the moment you walked into the transporter. And in Moscow you will have only one instance, but it will be run on a super-parallel computer with 100 computing elements, all running that one copy in parallel and synchronized. So you have 10 independent copies in Washingon, and 100 copies that are all kept in synchrony in Moscow. What do you expect then? A 90% chance of waking up in Washington, because 9/10 of the versions of you will be there? Or a 90% chance of waking up in Moscow, because 9/10 of the copies of you will be there? I think, based on what you wrote above, you will expect Moscow, and that "copies count" in this case. If you agree that copies count when it comes to spatial location, I wonder if you might reconsider whether they could count when it comes to temporal location. I still don't have a good understanding of this situation either, it is counter-intuitive, but if you accept that the number of copies, or as I would say, measure, does make a difference, then it seems like it should apply to changes in time as well as space.

`I agree that you will have a 90% chance of waking up in Moscow, given that`

`that is the *relative* measure of your successor OM when you walk into the`

`teleporter. This is the only thing that really matters with the copies, from`

`a selfish viewpoint: the relative measure of the next moment:`

`(a) If you are copied 100 times and 99% of the copies tortured, you will`

`certainly know this, as there is a 99% subjective probability that you will`

`be tortured.`

`(b) If you are copied 100 times and the copies allowed to diverge, then 99%`

`of the copies painlessly killed, that means you have a 99% chance of being`

`killed, because in two steps, (i) there is a 100% chance your next OM will`

`be one of the 100 copies; and (ii) there will be a 99% chance that you will`

`have become one of the copies that will be killed, and if you are, then`

`there will be 0% chance that you will have any "next moment".`

`(c) If you are copied 100 times and all the copies are kept running in`

`parallel, then 99% of the copies painlessly killed, you can't possibly know`

`that anything odd has happened at all, because there is a 100% chance that`

`your next OM will come from the one remaining copy. Similarly if there were`

`10^100 copies *kept running in perfect sync* and all but one terminated.`

`1/1=100/100=10^100/10^100=1.`

--Stathis Papaioannou _________________________________________________________________

`REALESTATE: biggest buy/rent/share listings`

`http://ninemsn.realestate.com.au`