On 5/29/2012 1:26 PM, Jason Resch wrote:

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On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal <marc...@ulb.ac.be<mailto:marc...@ulb.ac.be>> wrote:To see this the following thought experience can help. Some guy won a price consisting in visiting Mars by teleportation. But his state law forbid annihilation of human. So he made a teleportation to Mars without annihilation. The version of Mars is very happy, and the version of earth complained, and so try again and again, and again ... You are the observer, and from your point of view, you can of course only see the guy who got the feeling to be infinitely unlucky, as if P = 1/2, staying on earth for n experience has probability 1/2^n (that the Harry Potter experience). Assuming the infinite iteration, the guy as a probability near one to go quickly on Mars. Bruno,Thanks for your very detailed reply in the other thread, I intend to get back to itlater, but I had a strange thought while reading about the above experiment that Iwanted to clear up.You mentioned that the probability of remaining on Earth is (1/2)^n, where n is thenumber of teleportations. I can see clearly that the probability of remaining on earthafter the first teleportation is 50%, but as the teleportations continue, does it remain50%? Let's say that N = 5, therefore there are 5 copies on Mars, and 1 copy on earth.Wouldn't the probability of remaining on Earth be equal to 1/6th?While I can see it this way, I can also shift my perspective so that I see theprobability as 1/32 (since each time the teleport button is pressed, I split in two).It is easier for me to see how this works in quantum mechanics under the followingexperiment:I choose 5 different electrons and measure the spin on the y-axis, the probability thatI measure all 5 to be in the up state is 1 in 32 (as I have caused 5 splittings), butwhat if the experiment is: measure the spin states of up to 5 electrons, but stop onceyou find one in the up state. In this case it seems there are 6 copies of me, with thefollowing 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 isthey 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 transporter experiment, it seems theend result (having 1 copy on earth, and 5 copies on mars) is no different from the casewhere the transporter creates 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 the beginning and endstates of the experiment are the same, why should it matter if the replication is doneiteratively or all at once? Do RSSA and ASSA make different predictions in this case?Thanks, Jason

`I 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's not really duplicated so that if it went to Mars`

`on the first try there would be zero probability of it going on the second. Then the`

`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 the probability 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 who have some common history.`

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