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From: Marchal <marchal.domain.name.hidden>

Date: Sat Jan 15 10:39:56 2000

This post is my answer to Jacques Mallah. I have send it yesterday

(saturday) to Jacques only, but that was due to an error of manipulation.

(apology to jacques).

But I realise also that my answer (last paragraph) is most probably

false. I think Christopher Maloney is plausibly right concerning the

matter.

My 'real' opinion is that such probability computations (which I do

not use in UDA!) are so

much conceptually difficult that it is necessary to approach the whole

problem through some conceptual tools (like Goedel's arithmetisation

and modal logics, for exemple).

I nevertheless send you my whole answer, but please keep in mind this

proviso.

Note also the link between the SSA (ASSA, or RSSA) and the need to better

defined or represent notion like worlds, observer-moment, memories, etc.

Here is my not quite good answer to Jacques Mallah:

=========================================================

Jacques M. Mallah wrote:

*> Actually that doesn't matter. For me a physical system is just a
*

*>mathematical structure which "exists". So let all computations, for
*

*>example, exist.
*

*> But for me the term "implementation" has a special definition
*

*>which I am still trying to formulate precisely, due to the implementation
*

*>problem, so I'm still not sure how you distinguish between a computation
*

*>and an implementation of a computation.
*

Fair enough.

*> Well, first there is the question of what you mean by a 1st person
*

*>expectation and by "yourself".
*

1st person expectation = expectation by an agent of his own

personal subjective experience.

Exemple: - a child who expect a nightmare.

- a computationalist practioner who expect to be somewhere

after a self-duplication.

- a dog who expect his "master" giving him a bone.

"myself" : well, for a rough third person description, look at my

signature. For a first person description I'm afraid the best

approximation is you (from your first person point of view).

In fine, I modelise the first person by the knower, and I

modelise knowledge by the modal system S4 and S4Grz.

*> There is no 1 of the 10 copies which is
*

*>linked more than the others to the original copy. So depending on the
*

*>definition I would either say that you are only at Brussels, or that you
*

*>are at all three places.
*

*> But for practical purposes (such as, if you knew only German,
*

*>deciding which language to learn before being scanned, English or
*

*>Russian), I would of course say that, after the reconstitution, your
*

*>effective probability of seeing Moscow is 90%, so Russian is more
*

*>"likely" to come in handy.
*

Nice.

*>> If you say that your expectation to find yourself at Maoscow is bigger
*

*>> than the one to find yourself at Washington, then, by definition, I will
*

*>> say that you link the measure with the number of (actual, here) running
*

*>> computations.
*

*>
*

*> OK, not just link it, but say it is directly proportional
*

*>(but see above). But my point is that if you do that you reject the
*

*>RSSA. For example, suppose that a year later (on 1/1/2001), the copies at
*

*>Moscow are destroyed except for one, while at Washington 8 new copies of
*

*>the version of you that has been there are made. During 2000, what is the
*

*>effective probability that you would be at Moscow? It is 90%; next year,
*

*>it is 10%.
*

*> But according to the RSSA that would not be the case. If you
*

*>are 90% likely to be in Moscow in 2000, each type of version of you could
*

*>only "flow into" its own continuation, so the eff. pr. of being at Moscow
*

*>woulkd remain at 90% for 2001. (Chris Maloney would claim that the
*

*>eff. pr. of being at Moscow would be 10% for both years).
*

*> The ASSA, on the other hand, is naturally compatible with the
*

*>simple rule that the effective probability is proportional to the number
*

*>of copies. So do you, in fact, agree with that rule?
*

With RSSA? Yes. And you are correct, the "naive" probability remains 90%.

Even without immortality. With mortality the probability at Brussels that

I will die at Moscow in 2001 is 9/10 * 8/10. The multiplication of the

Washingtonian will not change the fact that I have had a bigger probabilty

to appear in Moscow. There will be a bigger set of Moscovian experiences

(including the fear to die with a chance of 8/10). The multiplication of

the one in Washington will not consolate those in Moscow.

With immortality it is even simpler, except that for a real computation

you should take into account the way you kill the Moscovians and you

must bet on the level of substitution.

Look at this: suppose you are annihilated at Brussels again.

And again reconstituted at M and W. Suppose now that at W, after

one day, you are

simply duplicate in you-W1 and you-W2, the day after, you-W2 is killed

and *at the same time* you-W1 is duplicated again, and the day after,

same scenario, all this lasting one year.

With ASSA you could say that at Brussels there is a bigger chance

to find yourself in W, because during all the year there is always two

examplars of you in Washington and only one at Moscow.

With RSSA there is 1/2 chance for W and M. But for the one in Washington

there is 1/2^365 to remain alive (supposing mortality) or healthy

(supposing

immortality) after one year.

OK? Are we agreeing (at least) on the meaning of ASSA and RSSA ?

Bruno

PS I am wrong, wrong, wrong ! Isn't it ? What do you think ?

The problem is that an observer-moment is really defined (with comp)

with the set of maximal consistent continuations going through that

moment, and we don't know really how these sets are structured. This

is our very much discussed measure problem.

Received on Sat Jan 15 2000 - 10:39:56 PST

Date: Sat Jan 15 10:39:56 2000

This post is my answer to Jacques Mallah. I have send it yesterday

(saturday) to Jacques only, but that was due to an error of manipulation.

(apology to jacques).

But I realise also that my answer (last paragraph) is most probably

false. I think Christopher Maloney is plausibly right concerning the

matter.

My 'real' opinion is that such probability computations (which I do

not use in UDA!) are so

much conceptually difficult that it is necessary to approach the whole

problem through some conceptual tools (like Goedel's arithmetisation

and modal logics, for exemple).

I nevertheless send you my whole answer, but please keep in mind this

proviso.

Note also the link between the SSA (ASSA, or RSSA) and the need to better

defined or represent notion like worlds, observer-moment, memories, etc.

Here is my not quite good answer to Jacques Mallah:

=========================================================

Jacques M. Mallah wrote:

Fair enough.

1st person expectation = expectation by an agent of his own

personal subjective experience.

Exemple: - a child who expect a nightmare.

- a computationalist practioner who expect to be somewhere

after a self-duplication.

- a dog who expect his "master" giving him a bone.

"myself" : well, for a rough third person description, look at my

signature. For a first person description I'm afraid the best

approximation is you (from your first person point of view).

In fine, I modelise the first person by the knower, and I

modelise knowledge by the modal system S4 and S4Grz.

Nice.

With RSSA? Yes. And you are correct, the "naive" probability remains 90%.

Even without immortality. With mortality the probability at Brussels that

I will die at Moscow in 2001 is 9/10 * 8/10. The multiplication of the

Washingtonian will not change the fact that I have had a bigger probabilty

to appear in Moscow. There will be a bigger set of Moscovian experiences

(including the fear to die with a chance of 8/10). The multiplication of

the one in Washington will not consolate those in Moscow.

With immortality it is even simpler, except that for a real computation

you should take into account the way you kill the Moscovians and you

must bet on the level of substitution.

Look at this: suppose you are annihilated at Brussels again.

And again reconstituted at M and W. Suppose now that at W, after

one day, you are

simply duplicate in you-W1 and you-W2, the day after, you-W2 is killed

and *at the same time* you-W1 is duplicated again, and the day after,

same scenario, all this lasting one year.

With ASSA you could say that at Brussels there is a bigger chance

to find yourself in W, because during all the year there is always two

examplars of you in Washington and only one at Moscow.

With RSSA there is 1/2 chance for W and M. But for the one in Washington

there is 1/2^365 to remain alive (supposing mortality) or healthy

(supposing

immortality) after one year.

OK? Are we agreeing (at least) on the meaning of ASSA and RSSA ?

Bruno

PS I am wrong, wrong, wrong ! Isn't it ? What do you think ?

The problem is that an observer-moment is really defined (with comp)

with the set of maximal consistent continuations going through that

moment, and we don't know really how these sets are structured. This

is our very much discussed measure problem.

Received on Sat Jan 15 2000 - 10:39:56 PST

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