Re: Holiday Exercise

[Brent Meeker]

On 7/13/2016 6:50 PM, Bruce Kellett wrote:

On 14/07/2016 11:40 am, Bruce Kellett wrote:

On 14/07/2016 11:31 am, John Clark wrote:

On Wed, Jul 13, 2016  Bruce Kellett wrote:

​ > ​
Of course, as I said some time ago, the easiest resolution of
you logical conundrums is that JC ~ H does not survive, and that
there are three new persons, one in each city, so the
probability that JC in H will see Sydney is exactly zero.


​Well... if that's what survival means then I don't care if I 
survive or not, if we use your easy resolution then I died 
yesterday, but my death doesn't seem to have cramped my style any.


So you were duplicated yesterday? One, quite reasonable, way of 
regarding our continuing existence from day to day, moment to moment, 
is that in every instant we die, to be reborn in the next instant. In 
fact, I am sure that someone else has already made this point (I 
can't, at the moment, remember who).


I found a relevant quote:
"I think maybe we die every day. Maybe we're born new each dawn, a 
little changed, a little further on our own road. When enough days 
stand between you and the person you were, you're strangers. Maybe 
that's what growing up is. Maybe I have grown up."

Mark Lawrence, Prince of Thorns (The Broken Empire #1)


Or looked at another way:

The person I was when I was 3 years old is dead. He died because
too much new information was added to his brain.
 -- Saibal Mitra

Brent

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Re: Holiday Exercise

[Bruce Kellett]

On 14/07/2016 11:40 am, Bruce Kellett wrote:

On 14/07/2016 11:31 am, John Clark wrote:

On Wed, Jul 13, 2016  Bruce Kellett wrote:

​ > ​
Of course, as I said some time ago, the easiest resolution of you
logical conundrums is that JC ~ H does not survive, and that
there are three new persons, one in each city, so the probability
that JC in H will see Sydney is exactly zero.


​Well... if that's what survival means then I don't care if I survive 
or not, if we use your easy resolution then I died yesterday, but my 
death doesn't seem to have cramped my style any.


So you were duplicated yesterday? One, quite reasonable, way of 
regarding our continuing existence from day to day, moment to moment, 
is that in every instant we die, to be reborn in the next instant. In 
fact, I am sure that someone else has already made this point (I 
can't, at the moment, remember who).


I found a relevant quote:
"I think maybe we die every day. Maybe we're born new each dawn, a 
little changed, a little further on our own road. When enough days stand 
between you and the person you were, you're strangers. Maybe that's what 
growing up is. Maybe I have grown up."

Mark Lawrence, Prince of Thorns (The Broken Empire #1)

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Re: Holiday Exercise

[Bruce Kellett]

On 14/07/2016 11:31 am, John Clark wrote:
On Wed, Jul 13, 2016  Bruce Kellett >wrote:


​ > ​
Of course, as I said some time ago, the easiest resolution of you
logical conundrums is that JC ~ H does not survive, and that there
are three new persons, one in each city, so the probability that
JC in H will see Sydney is exactly zero.


​Well... if that's what survival means then I don't care if I survive 
or not, if we use your easy resolution then I died yesterday, but my 
death doesn't seem to have cramped my style any.


So you were duplicated yesterday? One, quite reasonable, way of 
regarding our continuing existence from day to day, moment to moment, is 
that in every instant we die, to be reborn in the next instant. In fact, 
I am sure that someone else has already made this point (I can't, at the 
moment, remember who).


If I know that something tomorrow will remember being John Clark today 
then I'm content, after all that procedure worked pretty well in 
conserving what I want conserved during the transition form yesterday 
to today, so if the same thing happens in the transition from today to 
tomorrow I should be OK. And if tomorrow more than one thing remembers 
being John Clark today then that's even better. The more the merrier

​ .​


If you are happy, then fine. But are your dopplegangers equally happy -- 
they no longer exist, after all.


Bruce

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Re: Holiday Exercise

[John Clark]

On Wed, Jul 13, 2016  Bruce Kellett  wrote:

​> ​
> Of course, as I said some time ago, the easiest resolution of you logical
> conundrums is that JC ~ H does not survive, and that there are three new
> persons, one in each city, so the probability that JC in H will see Sydney
> is exactly zero.


​Well... if that's what survival means then I don't care if I survive or
not, if we use your easy resolution then I died yesterday, but my death
doesn't seem to have cramped my style any.

If I know that something tomorrow will remember being John Clark today then
I'm content, after all that procedure worked pretty well in conserving what
I want conserved during the transition form yesterday to today, so if the
same thing happens in the transition from today to tomorrow I should be OK.
And if tomorrow more than one thing remembers being John Clark today then
that's even better. T
he more the merrier
​.​


 John K Clark

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Re: Holiday Exercise

[Bruce Kellett]

On 13/07/2016 11:36 pm, Bruno Marchal wrote:

On 11 Jul 2016, at 13:49, Bruce Kellett wrote:

On 11/07/2016 9:31 pm, Bruno Marchal wrote:


*Holiday Exercise:*

A guy undergoes the Washington Moscow duplication, starting again 
from Helsinki.
Then in Moscow, but not in Washington,  he (the one in Moscow of 
course) undergoes a similar Sidney-Beijing duplication.


I write P(H->M) the probability in H to get M.

In Helsinki, he tries to evaluate his chance to get Sidney.

With one reasoning, he (the H-guy)  thinks that P(H-M) = 1/2, and 
that P(M-S) = 1/2, and so conclude (multiplication of independent 
probability) that P(H-S) = 1/2 * 1/2 = 1/4.


But with another reasoning, he thinks that the duplications give 
globally a triplication, leading eventually to a copy in W, a copy 
in S and a copy in B, and so, directly conclude P(H-S) = 1/3.


So, is it 1/4 or 1/3 ?


Neither. The probability that the guy starting from Helsinki gets to 
Sydney is unity.


Try to convince the guy who gets to Beijing, or the one who stayed in 
Washington. He knows that the probability evaluated in Helsinki was 
not P(Sidney) = 1.


We start with John Clark in Helsinki, so P(JC ~ H) = 1. By construction, 
after the duplication and so on, P(JC ~ W) = P(JC ~ S) = P(JC ~ B) = 1. 
(I use '~' as a shorthand for 'in' or 'sees'.) JC in Helsinki knows the 
protocol, so he can easily see that these are the correct probabilities. 
So, as I said, the probability that the guy starting from Helsinki gets 
to Sydney is unity. Any other interpretation of this scenario involves 
an implicit appeal to dualism -- there is "one true JC" that goes 
through these duplications, and he can only ever end up in just one place.


As John Clark has correctly pointed out, your intuition and formalism 
simply does not work in the presence of person-duplicating machines. 
There is no single 1p view -- there are three possible 1p views in the 
triplication scenario. So, again, John Clark is right when he says that 
JC ~ H will see three cities (W, S, and B) after the experiment is 
completed. If, as you claim, he will see only one city, you have to have 
some dualist 'nut or core' that survives in only one of your copies.


Of course, as I said some time ago, the easiest resolution of you 
logical conundrums is that JC ~ H does not survive, and that there are 
three new persons, one in each city, so the probability that JC in H 
will see Sydney is exactly zero.


Looking at the more realistic quantum realization of this triplication 
scenario, we can formulate that as follows. We prepare a spin-half atom 
with spin along the x-axis, then pass it through an S-G magnet oriented 
along the y-axis, getting two possibilities, which we can call up and 
down. We then take the up channel and pass that through a further S-G in 
the x-direction, getting two further possibilities of left or right.


Let us perform this experiment many times and count the number of 
particles in each of the three possible final states (down, left, and 
right). If this is a real laboratory experiment, in which detection of a 
particle in any channel leads to irreversible decoherence and the 
formation of a separate world containing just that result, we will find 
approximately half the particles end up in the down state, and 
approximately a quarter in each of the left and right states. This gives 
the most reliable estimate of the real probabilities for the outcome 
from the given initial state.


If you take the MWI view, then you get one down, one left, and one right 
in every run of the experiment, so the probability for each outcome is 
unity. In order to get probability of 1/4 for left, say, you have to 
detect the absence of a particle in the down state (so that the particle 
is certainly in the up state) for which the probability is 1/2.


Actually, your preferred answer -- that the probability P(H->S) = 1/3, 
is possible only in a fully dualist model. You are essentially claiming 
that as the scenario puts John Clark's in all three cities, it is purely 
a random chance that selects one of them to be the "true" John Clark -- 
a dualist "core" is assigned to one of these copies purely by chance.



This is the problem with probabilities in the MWI -- how do you 
interpret probabilities when all possible outcomes occur with 
probability one?


The probabilities concern the relative first person experiences. 
Computationalism guaranties that there will be only one outcome.


You are simply talking nonsense, here. "Relative first person 
experiences"? Relative to what? The scenario, and computationalism 
guarantees that there will be all three outcomes. If there is only one 
"1p" experience, then that can only be chosen dualistically. If there is 
duplication (triplication) then your intuitions break down. I have to 
say it -- John Clark has been right all along.


Bruce

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Re: Holiday Exercise (was: self (was Re: Aristotle the Nitwit

[John Clark]

On Wed, Jul 13, 2016 at 10:35 AM, Bruno Marchal  wrote:

​>> ​
>> ​Which turned out to be the better prediction, Moscow or Washington?​ And
>> was the prediction about John Clark or was it about some mysterious figure
>> named "you"?
>
>
> ​> ​
> The better prediction was "W v M and I have no clue which one".
>

​The better prediction about WHAT? Even after the experiment is over nobody
knows what was the better prediction because nobody knows who exactly the
prediction was supposed to be about. ​

​
>> ​>> ​
>> Then asking the Helsinki Person "what city will *you* see?" or "how many
>> cities will *you *see?" is a nonsense question because this is a world
>> with people duplicating machines. ​
>
>
> ​> ​
> Yes, but it should be obvious to anyone understand the difference between
> the 1p and the 3p views
>

​
Then answer the question!
​ ​
After the experiment was over what ONE city turned out to be the correct
answer, Moscow or Washington? If you can't answer that question then it's
not a experiment or even a thought experiment and so it's not science
​​
and assigning a probability to anything concerning it is just ridiculous.

​
As for being obvious, if modern physics and mathematics has taught us
anything it's that common sense is not always a reliable guide,
​
and
​
a lot of obvious things would not be true in a world with people
duplicating machines.
​
We didn't evolve in a environment
​
where
​ ​
things move close to the speed of light so out intuition in that area is
poor, common sense tells us that Einstein's relativity just can'r be true,
but it is.



> ​> ​
> When the H-guy pushes on the button in Helsinki, he knows with certainty
> (assuming computationalism and the protocol and the default hypotheses)
> that such a guy will find itself in a box, in front of a door, behind which
> only one city will be seen (in the 1p view).
>

​If after it's all over you can't name ​what one city "he" ended up seeing
then "*the*" 1p view does not exist, only "*a*" 1p view does.

​
>> ​>> ​
>> you're interested in "THE 1p view​" but as you just pointed out in a
>> world with people duplicating machine "THE 1p view​" is meaningless, there
>> is only "A 1p view".
>
>
> ​> ​
> Exactly, that is the root of the 1p indeterminacy.
>

​We agree, although not very profound it is certainly true that a
meaningless question (like which ONE will have the THE 1p view)
has no answer, and that is the root cause of "1p indeterminacy". ​



> ​
>> ​> ​
>> You are asking about what one and only one city was seen
>
>
> ​> ​
> The question concerns the future, or the next state.
>

​The next state of what?​

​I assume you mean the next state of something that remembers being in
Helsinki, if so then there is certainly no law of physics that demands only
one state can meet those specifications. If  you means something else then
I repeat my question, the next state of what? And please, no personal
pronouns with no clear referent in the answer.

​>> ​
>> John Clark will see two cities.
>
>
> ​> ​
> That is the 3-1-view.
>

​All I know is that
John Clark
​in his 1-view sees Moscow and John Clark in his 1-view sees Washington and
I have no idea what Mr. 3-1-view sees. ​

​> ​
> As you are John Clark, you need to go out of your body to conceive it. But
> to complete the thought experience, you need to re-integrate your body
> after the duplication.
>

​OK even better, after the re-integration I have vivid memories of BOTH
Washington and Moscow and so I John K Clark from John K Clark's 1p ended up
seeing Washington *and* Moscow. ​


​
>> ​>>​
>> There are 2 "1-views", and Bruno Marchal demands to know which *ONE* and
>> only *ONE* *you* will see, and that demand is pure gibberish.
>
>
> ​> ​
> You seem to be unable to understand that despite there are many 1-views
> obtained, all the 1-views feel to be one individual in a specific city.
>

​So what? ​A
ll the ​1-views that saw all those cities have an equal right to call
themselves John Clark, so the answer
​to the question "what is the probability John Clark will see city X?" is
100%. And if you ask just one John Clark how many cities he saw and he just
says only one that does NOT disprove the statement "John Clark will see 2
cities" because there is still another John Clark out there that you
haven't asked yet.​

​> ​
> By computationalism, you know that you will survive, and that you can only
> feel to survive as a unique individual in
> ​ ​
> only one city. You *know* that in advance. ​
>

Ah, more
​
duplicate people and more duplicate personal pronouns
​
with no clear referent!


> ​> ​
> remembering that the question was about that "future personal memory".
>

​I'm not the one who has forgotten that in the future 2 people not just one
will have memories of being in Helsinki, and 2 people not just one will
remember wondering about what city they will end up in; I think you're the
one who has forgotten about that and that's why you call *both* of these
people 

R: Re: What are among the world's most important problems to solve, why?

['scerir' via Everything List]

  
  

I've been close friends with two mathematicians.  The both say,
  "I'm a Platonist Monday thru Friday.  On the weekend I'm a
  nominalist."
Brent


"I raised just
this objection with the (extreme)ultrafinitist
Yessenin Volpin during a lecture of his. He

asked me to be
more specific. I then proceeded to start

with 2^1 and
asked him whether this is "real" or something to

that effect.
He virtually immediately said yes. Then I

asked about
2^2, and he again said yes, but with a

perceptible
delay. Then 2^3, and yes, but with more delay.

This continued
for a couple of more times, till it was

obvious how he
was handling this objection. Sure, he was

prepared to
always answer yes, but he was going to take 2^100

times as long
to answer yes to 2^100 then he would to

answering 2^1.
There is no way that I could get very far with

this."

-Harvey M.
Friedman



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Re: What are among the world's most important problems to solve, why?

[Brent Meeker]

I've been close friends with two mathematicians.  The both say, "I'm a 
Platonist Monday thru Friday.  On the weekend I'm a nominalist."


Brent


On 7/13/2016 5:18 AM, spudboy100 via Everything List wrote:
Interesting. I wonder how many mathematicians (superstitious?) believe 
in non-material "things"? One would think, platonists, neo-platonists, 
whatever, would lean this way, since numbers can be looked at in 2 
ways. That we made them up, or they pre-existed. I could buy into 
easily the notion of an extra physical dimension interlocuted in what 
we all can sense, where other things can exist, but not in "this" 
space, but that also, is superstition. To study mathematics I would 
need sufficient intellect which I do not possess, sad to admit. I am 
guessing you might be referring to pharmacological experiences, which, 
here also, seems to be avoid-worthy, for me.


To get a grip on non material existence, you can either study
mathematics, or use more direct techniques, but those are badly
seen in our obscurantist superstitious lasting middle-age ...




-Original Message-
From: Bruno Marchal 
To: everything-list 
Sent: Tue, Jul 12, 2016 11:11 am
Subject: Re: What are among the world's most important problems to 
solve, why?



On 09 Jul 2016, at 02:53, spudboy100 via Everything List wrote:

We cannot yet leave the Milky Way as yet, Professori! I feel
humans and other animals need something better to dwell on then,
The Big Sleep. Andromeda must wait a few billion years, eh. I do
not have a good mental grip on non-material existence, but neither
am I fluent in network engineering yet. It's back to casting your
fate to the wind as the old folk song went. Pattern Identity, I
understand I think.


We don't need to leave the Milky Way to avoid falling on a (bad) black 
hole. We need to scrutinize and map the black hole in the two 
galaxies, and make quantum simulation, and then disperse ourselves so 
as to minimize the risk to be sent on one of them. To do that, I guess 
we will need to go out of the solar system, if only because our sun 
will grow, but we have plenty of time. Yet, there is a moment where it 
might be too late, if we get distracted by our human affairs, or by 
some other events.


To get a grip on non material existence, you can either study 
mathematics, or use more direct techniques, but those are badly seen 
in our obscurantist superstitious lasting middle-age ...


Bruno




/Let us be careful so that humanity stays connected when the Milky
Way will meet Andromeda. That might be quite a big challenge. It
will involve huge amounts of computations. Meeting the mini galaxy
Magellan, as we do today, will be a good preparation, perhaps. /

/But we might count on some surprises too, as we know
virtually nothing about 'reality'./
/
/
/With Digital Mechanism, there is an inflation of type of
immortality possible, some of which are accessible here and
now (the progress here is that we have stopped to burn on the
stake those who practice them. We send them in jail or in
asylum)./
/
/
/Bruno/

-Original Message-
From: Bruno Marchal >
To: everything-list >
Sent: Fri, Jul 8, 2016 11:14 am
Subject: Re: What are among the world's most important problems to
solve, why?


On 08 Jul 2016, at 02:05, spudboy100 via Everything List wrote:

Well, for psychological reasons, I agree with John Clark's
initial response.The goal, I am guessing is restoration, in
principle, as Brent Meeker's comment, earlier. I am guessing
that file restoration from the past, if done exact enough
would render full resurrection or file restoration of us,
complete to the point of engaging in these emails, and then
doubting that a replica would really be you? If it is based on
quantum information, one physicist writer, started off her
essay, stating that the information that is us, that is the
universe is preserved in the quantum. The physicist in
question is Sabine Hossenfelder, in here article about natural
evolving quantum computers, aka block holes.
After you die, your body’s atoms will disperse and find new
venues, making their way into oceans, trees and other bodies.
But according to the laws of quantum mechanics, all of the
information about your body’s build and function will prevail.
The relations between the atoms, the uncountable particulars
that made you /you/, will remain forever preserved, albeit in
unrecognisably scrambled form – lost in practice, but immortal
in principle.
Good enough 

Re: Holiday Exercise (was: self (was Re: Aristotle the Nitwit

[Bruno Marchal]

On 12 Jul 2016, at 00:15, John Clark wrote:


On Mon, Jul 11, 2016, Bruno Marchal  wrote:


​​>> ​1) Each​  copy saw only one city.

​> ​Excellent! That is the correct 1-view description. Now, you  
just need to interview each copy about the prediction made in  
Helsinki and written in the diary to evaluate the better one.


​How? Which turned out to be the better prediction, Moscow or  
Washington?​ And was the prediction about John Clark or was it  
about some mysterious figure named "you"?


The better prediction was "W v M and I have no clue which one".

In that case, both copies agrees with each other and all subsequent  
similar experiences.







​2) ​All the copies together saw 2 cities.

​> ​Correct 3p description of the experiences of all copies. That  
is the 3-1 view. We need it to get the correct "1)", but "all the  
copies" is not a person,


​Then asking the Helsinki Person "what city will you see?" or "how  
many cities will you see?" is a nonsense question because this is a  
world with people duplicating machines. ​



Yes, but it should be obvious to anyone understand the difference  
between the 1p and the 3p views, that the 1p views are not duplicated  
from the 1p view (the 1-1-view as opposed to the 3-1-views, and the  
3-1-1 views) etc.


When the H-guy pushes on the button in Helsinki, he knows with  
certainty (assuming computationalism and the protocol and the default  
hypotheses) that such a guy will find itself in a box, in front of a  
door, behind which only one city will be seen (in the 1p view).







​> ​that is why you correctly add "together"​ (which is the 3-1  
view, in which we are not interested).


​I know, you're interested in "THE 1p view​" but as you just  
pointed out in a world with people duplicating machine  "THE 1p  
view​" is meaningless, there is only "A 1p view".



Exactly, that is the root of the 1p indeterminacy. There will be at  
all moments only one 1p view, from the points of you of all copies.






​> ​we are asked about the 1-views.

​You are asking about what one and only one city was seen


The question concerns the future, or the next state. Then, the  
verification is asked to all copies, and those which are verified by  
all copies, when discussing together for example, are the correct one.




by "the 1-views​" and that is a incoherent question with no  
coherent answer.​ Garbage in garbage out.



There is nothing incoherent, and indeed, those writing "W or M, and I  
have no clue which one" all win. and all other prediction fails.
In the finite case, just one fail refutes the prediction. In the  
infinite iteration of duplication, we can dismiss the negligible set  
(in the analytical or computer-science theoretical sense).









​>> ​​4) ​The statement "John Clark will see two cities"  
turned out to be unambiguously true.


​> ​In the 3-1 view, sure.

If they were logical it would be true from ​true from ANYBODIES  
view, Helsinki man Moscow Man Washington man you name it;



Yes, and only "W v M" is true from anybodies views when of course they  
keep in mind we are talking about the 1-views, and not the 3-1-views.  
All the copies agree that they expected and eventually verified to see  
only once city, and not knowing in advance which one.





John Clark will see two cities.


That is the 3-1-view.


As you are John Clark, you need to go out of your body to conceive it.  
But to complete the thought experience, you need to re-integrate your  
body after the duplication. As you have two bodies now, you have to do  
a choice, or more seriously, you need to develop just enough empathy  
toward BOTH copies, and listen to them: and both say that they see  
only one city, and could not have guessed that one city in advance,  
nor could they guess it again if we repeat the experience.








​> ​But we asked about the 1-views.

​There are 2 "1-views", and Bruno Marchal demands to know which ONE  
and only ONE *you* will see, and that demand is pure gibberish.


You seem to be unable to understand that despite there are many 1- 
views obtained, all the 1-views feel to be one individual in a  
specific city.


By computationalism, you know that you will survive, and that you can  
only feel to survive as a unique individual in only one city. You  
*know* that in advance. The gibberish is only apparent to you because  
you stop in the middle of the experience: you get first the correct  
3-1 view, but you don't complete the thought experience, by, notably  
looking at the personal memory of each copy, and remembering that the  
question was about that "future personal memory". If you do that, you  
can see easily that "W v M" win, and all others fail.









​>> ​So which one was right?

​> ​Trivially both when in Helsinki the prediction written in the  
diary was "W v M",


​But what exactly was the prediction about?



Given that the guy knows he will survive, and that he will feel to be  
in one city, the 

Re: Holiday Exercise

[Bruno Marchal]

On 11 Jul 2016, at 18:56, smitra wrote:


On 11-07-2016 13:49, Bruce Kellett wrote:

On 11/07/2016 9:31 pm, Bruno Marchal wrote:

HOLIDAY EXERCISE:
A guy undergoes the Washington Moscow duplication, starting again
from Helsinki.
Then in Moscow, but not in Washington, he (the one in Moscow of
course) undergoes a similar Sidney-Beijing duplication.
I write P(H->M) the probability in H to get M.
In Helsinki, he tries to evaluate his chance to get Sidney.
With one reasoning, he (the H-guy) thinks that P(H-M) = 1/2, and
that P(M-S) = 1/2, and so conclude (multiplication of independent
probability) that P(H-S) = 1/2 * 1/2 = 1/4.
But with another reasoning, he thinks that the duplications give
globally a triplication, leading eventually to a copy in W, a copy
in S and a copy in B, and so, directly conclude P(H-S) = 1/3.
So, is it 1/4 or 1/3 ?

Neither. The probability that the guy starting from Helsinki gets to
Sydney is unity. This is the problem with probabilities in the MWI --
how do you interpret probabilities when all possible outcomes occur
with probability one?
Bruce
In duplication experiments the prior probability to exist at all in  
any of the possible states increases after the duplication, while in  
unitary QM this is conserved (except if one or more of the possible  
outcomes is death).


It is plausibly conserved with Mechanism too, if the arithmetical  
quantum logics (the one extracted from Z1*, or X1*, or S4Grz1)  
justifies the linear rule Y = II. And normally, if QM is empirically  
correct, and computationalism is correct, they should match.





The correct way to analyze Bruno style duplication arguments is to  
start with assigning some measure m to the observer before any  
duplication is carried out, in this case the observer at H.


The probabilities are relative, and conditioned implicitly by P(H) = 1.





Then H gives rises to two copies in states W and M (we can call them  
copies, but they are actually different observers as they have  
different memories stored in their brains, so they are different  
algorithms).


They are the same programs, but with different input. By the SMN  
theorem, you *can* conceive them as different programs. OK.



The measures will be m for each of these observers. Then W is not  
going to be copied, while M gives rise to S and B, so we end up with  
3 observers each with measure m. The probability is thus 1/3.


That is the correct answer  if the guy remains unconscious in  
Moscow. But I doubt it is 1/3 in case he does wake up. In that case I  
would say it is 1/4.  (To be sure, I use Gödel-Löb and self-reference,  
which works only for the case P=1, already non trivial, and quantum- 
like, to avoid such probability question which are premature, but we  
can also speculate a bit).







In an analogue MWI setting, the outcome is different, at each  
duplication the measure for a particular outcome is halved. W thus  
has a measure of m/2, while S and B each have a measure of m/4, the  
probability is thus 1/4.


Hmm... I would like to see the analogue. I think the devil is in the  
details here. The analogue of letting the Moscow guy unconscious is  
the interference when we don't make a measurement at some intermediate  
state of a quantum computation (say).


My opinion is P(H->S) = 1/4, if the guy in Moscow remains awake, and  
P(H->S) = 1/3 if he remains unconscious there.


Bruno





Saibal

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Re: Holiday Exercise

[Bruno Marchal]

On 11 Jul 2016, at 13:49, Bruce Kellett wrote:


On 11/07/2016 9:31 pm, Bruno Marchal wrote:


Holiday Exercise:

A guy undergoes the Washington Moscow duplication, starting again  
from Helsinki.
Then in Moscow, but not in Washington,  he (the one in Moscow of  
course) undergoes a similar Sidney-Beijing duplication.


I write P(H->M) the probability in H to get M.

In Helsinki, he tries to evaluate his chance to get Sidney.

With one reasoning, he (the H-guy)  thinks that P(H-M) = 1/2, and  
that P(M-S) = 1/2, and so conclude (multiplication of independent  
probability) that P(H-S) = 1/2 * 1/2 = 1/4.


But with another reasoning, he thinks that the duplications give  
globally a triplication, leading eventually to a copy in W, a copy  
in S and a copy in B, and so, directly conclude P(H-S) = 1/3.


So, is it 1/4 or 1/3 ?


Neither. The probability that the guy starting from Helsinki gets to  
Sydney is unity.


Try to convince the guy who gets to Beijing, or the one who stayed in  
Washington. He knows that the probability evaluated in Helsinki was  
not P(Sidney) = 1.




This is the problem with probabilities in the MWI -- how do you  
interpret probabilities when all possible outcomes occur with  
probability one?


The probabilities concern the relative first person experiences.  
Computationalism guaranties that there will be only one outcome.


Bruno





Bruce


Can you modify a bit the protocol so that we get any of those  
results?



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Re: What are among the world's most important problems to solve, why?

[spudboy100 via Everything List]

Interesting. I wonder how many mathematicians (superstitious?) believe in 
non-material "things"? One would think, platonists, neo-platonists, whatever, 
would lean this way, since numbers can be looked at in 2 ways. That we made 
them up, or they pre-existed. I could buy into easily the notion of an extra 
physical dimension interlocuted in what we all can sense, where other things 
can exist, but not in "this" space, but that also, is superstition. To study 
mathematics I would need sufficient intellect which I do not possess, sad to 
admit. I am guessing you might be referring to pharmacological experiences, 
which, here also, seems to be avoid-worthy, for me. 

To get a grip on non material existence, you can either study mathematics, or 
use more direct techniques, but those are badly seen in our obscurantist 
superstitious lasting middle-age ...




-Original Message-
From: Bruno Marchal 
To: everything-list 
Sent: Tue, Jul 12, 2016 11:11 am
Subject: Re: What are among the world's most important problems to solve, why?




On 09 Jul 2016, at 02:53, spudboy100 via Everything List wrote:


 
We cannot yet leave the Milky Way as yet, Professori! I feel humans and other 
animals need something better to dwell on then, The Big Sleep. Andromeda must 
wait a few billion years, eh. I do not have a good mental grip on non-material 
existence, but neither am I fluent in network engineering yet. It's back to 
casting your fate to the wind as the old folk song went. Pattern Identity, I 
understand I think. 



We don't need to leave the Milky Way to avoid falling on a (bad) black hole. We 
need to scrutinize and map the black hole in the two galaxies, and make quantum 
simulation, and then disperse ourselves so as to minimize the risk to be sent 
on one of them. To do that, I guess we will need to go out of the solar system, 
if only because our sun will grow, but we have plenty of time. Yet, there is a 
moment where it might be too late, if we get distracted by our human affairs, 
or by some other events. 


To get a grip on non material existence, you can either study mathematics, or 
use more direct techniques, but those are badly seen in our obscurantist 
superstitious lasting middle-age ...


Bruno








 
 
 
Let us be careful so that humanity stays connected when the Milky Way will meet 
Andromeda. That might be quite a big challenge. It will involve huge amounts of 
computations. Meeting the mini galaxy Magellan, as we do today, will be a good 
preparation, perhaps. 
 
 
But we might count on some surprises too, as we know virtually nothing about 
'reality'.
 

 
 
With Digital Mechanism, there is an inflation of type of immortality possible, 
some of which are accessible here and now (the progress here is that we have 
stopped to burn on the stake those who practice them. We send them in jail or 
in asylum).
 

 
 
Bruno
 
 
 
 
 
 
 
 
-Original Message-
 From: Bruno Marchal 
 To: everything-list 
 Sent: Fri, Jul 8, 2016 11:14 am
 Subject: Re: What are among the world's most important problems to solve, why?
 
 
 

 
 
On 08 Jul 2016, at 02:05, spudboy100 via Everything List wrote:
 

Well, for psychological reasons, I agree with John Clark's initial response.The 
goal, I am guessing is restoration, in principle, as Brent Meeker's comment, 
earlier. I am guessing that file restoration from the past, if done exact 
enough would render full resurrection or file restoration of us, complete to 
the point of engaging in these emails, and then doubting that a replica would 
really be you? If it is based on quantum information, one physicist writer, 
started off her essay, stating that the information that is us, that is the 
universe is preserved in the quantum. The physicist in question is Sabine 
Hossenfelder, in here article about natural evolving quantum computers, aka 
block holes. 
After you die, your body’s atoms will disperse and find new venues, making 
their way into oceans, trees and other bodies. But according to the laws of 
quantum mechanics, all of the information about your body’s build and function 
will prevail. The relations between the atoms, the uncountable particulars that 
made you you, will remain forever preserved, albeit in unrecognisably scrambled 
form – lost in practice, but immortal in principle.
 
Good enough for me, on this Now the rest is engineering and data 
restoration. Easy, huh?
 
 

 
 
You are not made of atoms, which mainly change every decades. You are a pattern 
of immaterial information, and you survive in any reasonable sense only if you 
have the environment (aka universal numbers) capable of processing that pattern 
of information. Now that exists in infinitely many occurrence in elementary 
arithmetic. But that is not necessary consoling, as in principle it could mean 
that life is 100 years of bearable life followed by an eternal