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

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On 31 May 2012, at 21:38, Jason Resch wrote:```
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On Thu, May 31, 2012 at 2:09 PM, Bruno Marchal <marc...@ulb.ac.be> wrote:
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On 31 May 2012, at 18:29, Jason Resch wrote:

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On Wed, May 30, 2012 at 3:27 AM, Bruno Marchal <marc...@ulb.ac.be> wrote:
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On 29 May 2012, at 22:26, Jason Resch wrote:

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On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal <marc...@ulb.ac.be> wrote:
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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.
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Bruno,

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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.
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You mentioned that the probability of remaining on Earth is (1/2)^n, where n is the number of teleportations.
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Not really. I pretend that this is the relative probability inferred by the person in front of you. But he is wrong of course. Each time the probability is 1/2, but his experience is "harry- Potter-like".
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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%?
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Yes.

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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?
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You cannot use absolute sampling. I don't think it makes any sense.

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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:
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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),
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OK.

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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.
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That is a different protocol. The one above is the one corresponding to the earth/mars experience.
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In this case it seems there are 6 copies of me, with the following records:
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1. D
2. DU
3. DDU
4. DDDU
5. DDDDU
6. DDDDD

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However, not all of these copies should have the same measure. The way I see it is they have the following probabilities:
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1. D (1/2)
2. DU (1/4)
3. DDU (1/8)
4. DDDU (1/16)
5. DDDDU (1/32)
6. DDDDD (1/32)

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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.
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This is ambiguous.

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What I mean is me stepping into the teleporter 5 times, with the net result being 1 copy on Earth and 5 copies on Mars, seems just like stepping into the teleporter once, and the teleporter then creating 5 copies (with delay) on Mars.
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Like the diagram on step 4 of UDA:
http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL_fichiers/image012.gif

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Except there is no annihilation on Earth, and there are 4 copies created with delay on Mars (instead of one with delay).
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When stepping into the teleporter once, and having 5 copies created on Mars (with various delays between each one being produced) is the probability of remaining on Earth 1/6th?
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Yes.
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That would be a good idea to enhance the probability to be the one, or a one, finding himself of mars. But again, the guy on earth will be in front of the "looser", even if you multiply by 20. billions your delayed copies on mars.
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Is the difference with the iterated example receiving the knowledge that the other copy made it to Mars before stepping into the Teleporter again?
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I don't understand the sentence. It looks like what is the difference between 24.
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I apologize for not being clear. There are two different experiments I am contrasting:
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1. A person steps into a teleporter, and 5 copies (with varying delays) are reproduced on Mars.
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2. A person steps into a teleporter, and a duplicate is created on Mars. To increase the chance of subjectively finding himself on Mars, he does it again (when he fails) and the copy on Earth does so 5 times before giving up.
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For experiment 1, you and I seem to agree that subjectively, that person person has a 1 in 6 chance of experiencing a continued presence on earth, and a 5/6 chance of finding himself on mars.
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For experiment 2, I believe you suggested there is a 1 in 32 (subjective chance) of going through this exercise and not having the subjective experience of ending up on Mars. Have I understood this correctly thus far?
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If so, what accounts for these different subjective probabilities? How can it be that there is a 31/32 chance of finding oneself on mars if there are just 5 copies there?
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I hope I have been clear enough.  Thanks again.
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OK. Thanks, this is clearer.

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In the experience 1, the guy steps in the teleporter only once, and his multiplied by 6, including the original. So, it has a probability 1/6 to stay on earth, and 5/6 to find itself on Mars (accepting the P=1/2, etc.).
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In the experience 2, the guy repeats 5 times a duplication experience, each of which has a probability of 1/2.
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That is what account for the difference. The protocol of the two experiments are very different in term of the relative probabilities. OK?
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Experience 1 is equivalent with a throwing of a dice. Experience 2 is equivalent with 6 throwing of a coin.
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You might be disturbed by the fact that in experience 2, the "original" remains the same person, so we don't count him as a new person, each time he steps in the box. This, in my opinion, illustrates again that we have to use RSSA instead of ASSA.
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All right?

Bruno

http://iridia.ulb.ac.be/~marchal/

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