# Re: Church Turing be dammed. (Probability Question)

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 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 it later, but I had a strange thought while reading about 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 see clearly that the probability of remaining on earth after the first teleportation is 50%, but as the teleportations continue, does it remain 50%? 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 the probability 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 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 in 32 (as I have caused 5 splittings), but what if the experiment is: measure the spin states of up to 5 electrons, but stop once you find one in the up state. In this case it seems there are 6 copies of me, with the following records:

1. D
2. DU
3. DDU
4. DDDU
5. DDDDU
6. DDDDD

However, 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 transporter experiment, it seems the end result (having 1 copy on earth, and 5 copies on mars) is no different from the case where the transporter creates all 5 copies on Mars at once. In that case, it is clear that the chance of remaining on Earth should be (1/6th) but if the beginning and end states of the experiment are the same, why should it matter if the replication is done iteratively 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.

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/

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