On 29 May 2012, at 22:41, meekerdb wrote:

On 5/29/2012 1:26 PM, Jason Resch wrote:On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal <marc...@ulb.ac.be>wrote:To see this the following thought experience can help. Some guy wona price consisting in visiting Mars by teleportation. But his statelaw forbid annihilation of human. So he made a teleportation toMars without annihilation. The version of Mars is very happy, andthe version of earth complained, and so try again and again, andagain ... You are the observer, and from your point of view, youcan of course only see the guy who got the feeling to be infinitelyunlucky, as if P = 1/2, staying on earth for n experience hasprobability 1/2^n (that the Harry Potter experience). Assuming theinfinite iteration, the guy as a probability near one to go quicklyon Mars.Bruno,Thanks for your very detailed reply in the other thread, I intendto get back to it later, but I had a strange thought while readingabout the above experiment that I wanted to clear up.You mentioned that the probability of remaining on Earth is(1/2)^n, where n is the number of teleportations. I can seeclearly that the probability of remaining on earth after the firstteleportation is 50%, but as the teleportations continue, does itremain 50%? Let's say that N = 5, therefore there are 5 copies onMars, and 1 copy on earth. Wouldn't the probability of remainingon Earth be equal to 1/6th?While I can see it this way, I can also shift my perspective sothat I see the probability as 1/32 (since each time the teleportbutton is pressed, I split in two). It is easier for me to see howthis works in quantum mechanics under the following experiment:I choose 5 different electrons and measure the spin on the y-axis,the probability that I measure all 5 to be in the up state is 1 in32 (as I have caused 5 splittings), but what if the experiment is:measure the spin states of up to 5 electrons, but stop once youfind one in the up state. In this case it seems there are 6 copiesof me, with the following records:1. D 2. DU 3. DDU 4. DDDU 5. DDDDU 6. DDDDDHowever, not all of these copies should have the same measure.The way I see it is they have the following probabilities:1. D (1/2) 2. DU (1/4) 3. DDU (1/8) 4. DDDU (1/16) 5. DDDDU (1/32) 6. DDDDD (1/32)I suppose what is bothering me is that in the Mars transporterexperiment, it seems the end result (having 1 copy on earth, and 5copies on mars) is no different from the case where the transportercreates all 5 copies on Mars at once. In that case, it is clearthat the chance of remaining on Earth should be (1/6th) but if thebeginning and end states of the experiment are the same, why shouldit matter if the replication is done iteratively or all at once? DoRSSA and ASSA make different predictions in this case?Thanks, JasonI think you are right, Jason. For the probability to be (1/2^n)implies that there is some single "soul" that is "you" and it'snot really duplicated so that if it went to Mars on the first trythere would be zero probability of it going on the second. Thenthe probability of your "soul" being on Mars is(1/2)+(1/4)+(1/8)+...+(1/2^n).Under the alternative, that "you" really are duplicated theprobability that some "you" chosen at random is on Mars is (n-1/n).But in this case there is really no "you", there are n+1 people whohave some common history.

`The probability bears on the first experiences, which are indeed never`

`duplicated from their 1-pov, and we ask for the probability of`

`"staying" on earth. It is equivalent with the probability of always`

`getting head in a throw of a coin. So, from the perspective of the guy`

`who stays on Earth, he is living an Harry-Potter like experience. But`

`the experience is "trivial" for the observer looking at it from outside.`

Bruno http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.