Re: Church Turing be dammed. (Probability Question)

[meekerdb]

On 6/2/2012 1:42 AM, Bruno Marchal wrote:


On 01 Jun 2012, at 20:18, meekerdb wrote:


On 6/1/2012 10:23 AM, Bruno Marchal wrote:
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.


Suppose the original goes to Mars and the copy stays behind.  Then the probability 
the original went to Mars is 1.


The question is asked before the guy enter in the box. This is a "step 5" case. The 
probability to feel to stay the original is 1/2.


Everybody feels they are the original.


"original" refer to the third person body. By definition it is the one being 
copied.


It doesn't really solve the identity problem to assume it is physical continuity.  The 
"copy" also  has physical continuity; and in any even slightly realistic case the 
'original' will be destroyed in the process of extracting information, so there will 
really be two copies and no 'original'.





The question before he enters the box is, "Will you find yourself on Mars?"  To which 
he could reply, "What does 'you' refer to?"


The question is about your future subjective feeling as seen from your future first 
person perspective.If you assume comp, you know in advance that you will feel entire and 
unique,


No, I expect that two someones will feel entire and unique.

either on Earth or on Mars, and you know that you cannot that in advance (or give me the 
algorithm).


But all that assumes that 'you' and 'your' have meaningful referents.  According to comp 
they are no more meaningful than referring to this number 2 and that number 2 and asking 
which number 2 counts the moons of Mars.


Brent



Bruno


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





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Re: Church Turing be dammed. (Probability Question)

[Bruno Marchal]

On 01 Jun 2012, at 20:18, meekerdb wrote:


On 6/1/2012 10:23 AM, Bruno Marchal wrote:
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.


Suppose the original goes to Mars and the copy stays behind.  Then  
the probability the original went to Mars is 1.


The question is asked before the guy enter in the box. This is a  
"step 5" case. The probability to feel to stay the original is 1/2.


Everybody feels they are the original.


"original" refer to the third person body. By definition it is the one  
being copied.



The question before he enters the box is, "Will you find yourself on  
Mars?"  To which he could reply, "What does 'you' refer to?"


The question is about your future subjective feeling as seen from your  
future first person perspective. If you assume comp, you know in  
advance that you will feel entire and unique, either on Earth or on  
Mars, and you know that you cannot that in advance (or give me the  
algorithm).


Bruno


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



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Re: Church Turing be dammed. (Probability Question)

[meekerdb]

On 6/1/2012 10:23 AM, Bruno Marchal wrote:
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.


Suppose the original goes to Mars and the copy stays behind.  Then the probability the 
original went to Mars is 1.


The question is asked before the guy enter in the box. This is a "step 5" case. The 
probability to feel to stay the original is 1/2.


Everybody feels they are the original.  The question before he enters the box is, "Will 
you find yourself on Mars?"  To which he could reply, "What does 'you' refer to?"


Brent



Bruno



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Re: Church Turing be dammed. (Probability Question)

[Bruno Marchal]

On 01 Jun 2012, at 19:09, meekerdb wrote:


On 6/1/2012 7:50 AM, Bruno Marchal wrote:



On 31 May 2012, at 21:38, Jason Resch wrote:




On Thu, May 31, 2012 at 2:09 PM, Bruno Marchal   
wrote:


On 31 May 2012, at 18:29, Jason Resch wrote:




On Wed, May 30, 2012 at 3:27 AM, Bruno Marchal  
 wrote:


On 29 May 2012, at 22:26, Jason Resch wrote:




On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal  
 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.


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".





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%?


Yes.



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?


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





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),


OK.


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.


That is a different protocol. The one above is the one  
corresponding to the earth/mars experience.




In this case it seems there are 6 copies of me, with the  
following records:


1. D
2. DU
3. DDU
4. DDDU
5. U
6. D

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. U (1/32)
6. D (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.


This is ambiguous.


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.


Like the diagram on step 4 of UDA:
http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL_fichiers/image012.gif

Except there is no annihilation on Earth, and there are 4 copies  
created with delay on Mars (instead of one with delay).


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?


Yes.
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.





Is the difference with the iterated example receiving the  
knowledge that the other copy made it to Mars before stepping  
into the Teleporter again?


I don't understand the sentence. It looks like what is the  
difference between 24.



I apologize for not being clear.  There are two different  
experiments I am contrasting:


1. A person steps into a teleporter, and 5 copies (with varying  
delays) are reproduced on Mars.


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.


For experiment 1, you and I seem to agree that subjectively, that  
person person has a 1 i

Re: Church Turing be dammed. (Probability Question)

[meekerdb]

On 6/1/2012 7:50 AM, Bruno Marchal wrote:


On 31 May 2012, at 21:38, Jason Resch wrote:




On Thu, May 31, 2012 at 2:09 PM, Bruno Marchal > wrote:



On 31 May 2012, at 18:29, Jason Resch wrote:




On Wed, May 30, 2012 at 3:27 AM, Bruno Marchal mailto:marc...@ulb.ac.be>> wrote:


On 29 May 2012, at 22:26, Jason Resch wrote:




On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal 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 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. 


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".





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%? 


Yes.




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?


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





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), 


OK.



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. 


That is a different protocol. The one above is the one corresponding to 
the
earth/mars experience.




In this case it seems there are 6 copies of me, with the following 
records:

1. D
2. DU
3. DDU
4. DDDU
5. U
6. D

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. U (1/32)
6. D (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. 


This is ambiguous.



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.

Like the diagram on step 4 of UDA:

http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL_fichiers/image012.gif 


Except there is no annihilation on Earth, and there are 4 copies created 
with
delay on Mars (instead of one with delay).

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?


Yes.
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.




Is the difference with the iterated example receiving the knowledge that 
the other
copy made it to Mars before 

Re: Church Turing be dammed. (Probability Question)

[Bruno Marchal]

On 31 May 2012, at 21:38, Jason Resch wrote:




On Thu, May 31, 2012 at 2:09 PM, Bruno Marchal   
wrote:


On 31 May 2012, at 18:29, Jason Resch wrote:




On Wed, May 30, 2012 at 3:27 AM, Bruno Marchal   
wrote:


On 29 May 2012, at 22:26, Jason Resch wrote:




On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal  
 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.


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".





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%?


Yes.



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?


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





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),


OK.


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.


That is a different protocol. The one above is the one  
corresponding to the earth/mars experience.




In this case it seems there are 6 copies of me, with the following  
records:


1. D
2. DU
3. DDU
4. DDDU
5. U
6. D

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. U (1/32)
6. D (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.


This is ambiguous.


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.


Like the diagram on step 4 of UDA:
http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL_fichiers/image012.gif

Except there is no annihilation on Earth, and there are 4 copies  
created with delay on Mars (instead of one with delay).


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?


Yes.
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.





Is the difference with the iterated example receiving the knowledge  
that the other copy made it to Mars before stepping into the  
Teleporter again?


I don't understand the sentence. It looks like what is the  
difference between 24.



I apologize for not being clear.  There are two different  
experiments I am contrasting:


1. A person steps into a teleporter, and 5 copies (with varying  
delays) are reproduced on Mars.


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.


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 himse

Re: Church Turing be dammed. (Probability Question)

[meekerdb]

On 5/31/2012 12:38 PM, Jason Resch wrote:



On Thu, May 31, 2012 at 2:09 PM, Bruno Marchal > wrote:



On 31 May 2012, at 18:29, Jason Resch wrote:




On Wed, May 30, 2012 at 3:27 AM, Bruno Marchal mailto:marc...@ulb.ac.be>> wrote:


On 29 May 2012, at 22:26, Jason Resch wrote:




On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal 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 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. 


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".





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%? 


Yes.




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?


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





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), 


OK.



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. 


That is a different protocol. The one above is the one corresponding to 
the
earth/mars experience.




In this case it seems there are 6 copies of me, with the following 
records:

1. D
2. DU
3. DDU
4. DDDU
5. U
6. D

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. U (1/32)
6. D (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. 


This is ambiguous.



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.

Like the diagram on step 4 of UDA:

http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL_fichiers/image012.gif



Except there is no annihilation on Earth, and there are 4 copies created 
with delay
on Mars (instead of one with delay).

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?


Yes.
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.




Is the difference with the iterated example receiving the knowledge that 
the other
copy made it to Mars before stepping into the Teleporter again?


I

Re: Church Turing be dammed. (Probability Question)

[Jason Resch]

On Thu, May 31, 2012 at 2:09 PM, Bruno Marchal  wrote:

>
> On 31 May 2012, at 18:29, Jason Resch wrote:
>
>
>
> On Wed, May 30, 2012 at 3:27 AM, Bruno Marchal  wrote:
>
>>
>> On 29 May 2012, at 22:26, Jason Resch wrote:
>>
>>
>>
>> On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal 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.
>>
>>
>> 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".
>>
>>
>>
>>
>> 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%?
>>
>>
>> Yes.
>>
>>
>>
>> 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?
>>
>>
>> You cannot use absolute sampling. I don't think it makes any sense.
>>
>>
>>
>>
>> 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),
>>
>>
>> OK.
>>
>>
>> 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.
>>
>>
>> That is a different protocol. The one above is the one corresponding to
>> the earth/mars experience.
>>
>>
>>
>> In this case it seems there are 6 copies of me, with the following
>> records:
>>
>> 1. D
>> 2. DU
>> 3. DDU
>> 4. DDDU
>> 5. U
>> 6. D
>>
>> 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. U (1/32)
>> 6. D (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.
>>
>>
>> This is ambiguous.
>>
>
>
> 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.
>
> Like the diagram on step 4 of UDA:
>
> http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL_fichiers/image012.gif
>
> Except there is no annihilation on Earth, and there are 4 copies created
> with delay on Mars (instead of one with delay).
>
> 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?
>
>
> Yes.
> 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.
>
>
>
> Is the difference with the iterated example receiving the knowledge that
> the other copy made it to Mars before stepping into the Teleporter again?
>
>
> I don't understand the sentence. It looks like what is the difference
> between 24.
>


I apologize for not being clear.  There are two different experiments I am
contrasting:

1. A person steps into a teleporter, and 5 copies (with varying delays) are
reproduced on Mars.

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.

For expe

Re: Church Turing be dammed. (Probability Question)

[Bruno Marchal]

On 31 May 2012, at 18:29, Jason Resch wrote:




On Wed, May 30, 2012 at 3:27 AM, Bruno Marchal   
wrote:


On 29 May 2012, at 22:26, Jason Resch wrote:




On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal   
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.


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".





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%?


Yes.



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?


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





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),


OK.


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.


That is a different protocol. The one above is the one corresponding  
to the earth/mars experience.




In this case it seems there are 6 copies of me, with the following  
records:


1. D
2. DU
3. DDU
4. DDDU
5. U
6. D

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. U (1/32)
6. D (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.


This is ambiguous.


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.


Like the diagram on step 4 of UDA:
http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL_fichiers/image012.gif

Except there is no annihilation on Earth, and there are 4 copies  
created with delay on Mars (instead of one with delay).


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?


Yes.
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.





Is the difference with the iterated example receiving the knowledge  
that the other copy made it to Mars before stepping into the  
Teleporter again?


I don't understand the sentence. It looks like what is the difference  
between 24.


In this thought experience you were supposed to be an external  
observer on earth, not the candidate doing the duplication.
In your diary, you will always write things like, "he try to multiply  
the copy on mars, push on the button and told me "this fails again".


Bruno








In that case, it is clear that the chance of remaining on Earth  
should be (1/6th)


Yes. In that case.



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?


RSSA has to be applied. Your first protocol is faithful, isomorphic,  
to the experience I was describing. Te second is not.

Re: Church Turing be dammed. (Probability Question)

[Jason Resch]

On Wed, May 30, 2012 at 3:27 AM, Bruno Marchal  wrote:

>
> On 29 May 2012, at 22:26, Jason Resch wrote:
>
>
>
> On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal  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.
>
>
> 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".
>
>
>
>
> 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%?
>
>
> Yes.
>
>
>
> 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?
>
>
> You cannot use absolute sampling. I don't think it makes any sense.
>
>
>
>
> 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),
>
>
> OK.
>
>
> 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.
>
>
> That is a different protocol. The one above is the one corresponding to
> the earth/mars experience.
>
>
>
> In this case it seems there are 6 copies of me, with the following records:
>
> 1. D
> 2. DU
> 3. DDU
> 4. DDDU
> 5. U
> 6. D
>
> 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. U (1/32)
> 6. D (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.
>
>
> This is ambiguous.
>


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.

Like the diagram on step 4 of UDA:
http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL_fichiers/image012.gif

Except there is no annihilation on Earth, and there are 4 copies created
with delay on Mars (instead of one with delay).

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?

Is the difference with the iterated example receiving the knowledge that
the other copy made it to Mars before stepping into the Teleporter again?

Thanks,

Jason



>
>
>
> In that case, it is clear that the chance of remaining on Earth should be
> (1/6th)
>
>
> Yes. In that case.
>
>
>
> 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?
>
>
> RSSA has to be applied. Your first protocol is faithful, isomorphic, to
> the experience I was describing. Te second is not.
>
> OK?
>
> 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.
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>

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Re: Church Turing be dammed. (Probability Question)

[Bruno Marchal]

On 30 May 2012, at 18:16, meekerdb wrote:


On 5/30/2012 1:38 AM, Bruno Marchal wrote:



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  
 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. U
6. D

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. U (1/32)
6. D (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.


No more than the guys who went to Mars.  If they compare experiences  
they will find that although they only had probability 1/2 of it  
happening, they all went to Mars.


They almost all went to Mars ... eventually, with one exception.  
Besides this was just used in a protocol where the observer is the one  
looking his friend, that is the exception. It is his 3-view on the 1- 
view of the guy who never succeed to go on Mars. I have a collection  
of strategies that he can try, like  introducing delays, or using  
random coin between "original" and "copy", unfortunately for the guy  
remaining on earth, by "definition", he cannot succeed, and he will  
have hard time to believe things are not conspiring against his will  
to go on Mars, and this proportionally to the ingenuity developed to  
assure to be the one going on Mars.


If you make that experience, the chance to go on mars is always rather  
great, but of course, we, the spectators, will have to live with the  
unlucky (from its first person view) who remains o

Re: Church Turing be dammed. (Probability Question)

[meekerdb]

On 5/30/2012 1:38 AM, Bruno Marchal wrote:


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 > 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. U
6. D

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. U (1/32)
6. D (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.


No more than the guys who went to Mars.  If they compare experiences they will find that 
although they only had probability 1/2 of it happening, they all went to Mars.


Brent



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

Bruno


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



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Re: Church Turing be dammed. (Probability Question)

[Bruno Marchal]

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   
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. U
6. D

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. U (1/32)
6. D (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  
notreally duplicated so that if it went to Mars on the first try  
there would be zero probability of it going on the second.  Then  
theprobability 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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Re: Church Turing be dammed. (Probability Question)

[Bruno Marchal]

On 29 May 2012, at 22:26, Jason Resch wrote:




On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal   
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.


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".





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%?


Yes.



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?


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





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),


OK.


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.


That is a different protocol. The one above is the one corresponding  
to the earth/mars experience.




In this case it seems there are 6 copies of me, with the following  
records:


1. D
2. DU
3. DDU
4. DDDU
5. U
6. D

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. U (1/32)
6. D (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.


This is ambiguous.



In that case, it is clear that the chance of remaining on Earth  
should be (1/6th)


Yes. In that case.



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?


RSSA has to be applied. Your first protocol is faithful, isomorphic,  
to the experience I was describing. Te second is not.


OK?

Bruno


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



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Re: Church Turing be dammed. (Probability Question)

[Stephen P. King]

On 5/29/2012 4:26 PM, Jason Resch wrote:



On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal > 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. U
6. D

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. U (1/32)
6. D (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
--

Hi Jason,

Fascinating! This decrease in probability given an increase in the 
number of copies would also hold if the copies had amnesia and could not 
identify themselves with the "original"?


--
Onward!

Stephen

"Nature, to be commanded, must be obeyed."
~ Francis Bacon

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Re: Church Turing be dammed. (Probability Question)

[meekerdb]

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



On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal > 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. U
6. D

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. U (1/32)
6. D (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.


Brent

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Re: Church Turing be dammed. (Probability Question)

[Jason Resch]

On Tue, May 29, 2012 at 12:55 PM, Bruno Marchal  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. U
6. D

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. U (1/32)
6. D (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

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