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From: Jacques M Mallah <jqm1584.domain.name.hidden>

Date: Sat, 17 Apr 1999 17:03:16 -0400

On Sat, 17 Apr 1999, Nick Bostrom wrote:

*> Jacques M Mallah wrote:
*

*> > The interesting thing is that if Adam believed in the MWI, he
*

*> > could calculate (roughly) the distribution of observers, and then he would
*

*> > realize that the effective probability of him seeing himself have a child,
*

*> > and therefore of any correlated coin flips or deer crossings, was the same
*

*> > as a third party observer would calculate.
*

*>
*

*> Presumably Adam does not know the quantum mechanical chance of seing
*

*> himself have a child. For all he knows, it could be
*

*> 0.99999999999999999 or 0.00000000000000001. If there is in fact no
*

*> deer in the region, only a quantum freak event would make one
*

*> materialize outside his cave. Now, what is the chance of Adam
*

*> observing what he does if there would be many subsequent people in
*

*> the world? Much, much smaller than if there is only himself and Eve.
*

*> So he should consequently think that the q.m chance that he has a
*

*> child is very, very small.
*

If he has no other information, of course he should. But the

difference is that if he does observe deer in the region, he would realize

that over the ensemble of worlds, the effective probability of seeing a

deer is the same as would be the case if the number of subsequent people

didn't depend on it.

The point is that the quantum 'probability' can be extropolated

from past events, even if you start with a prior that is heavily skewed

in the wrong direction, but the outcome of one particular event in a

non-MWI can't be. Suppose he'd have kids if on a certain measurement he

saw spin up.

If you draw many spins from a sample and they are found to be 50%

up and 50% down, you can be reasonably sure that the quantum probability

p to measure up is about 50%. In this case the sample size matters; to

overcome Adam's prior expectation of 10^-9 or 2^-30, to get his median

expectation of p to >= q, he'd need to measure over M spins. What's M(q)?

It might be interesting to look more closely at that: it depends on the

shape of his prior f(p), which we can obtain from the usual Bayesian

proceedure starting with a uniform f(p) on (0,1) and using the conditional

effective probability of an observer being Adam as a function of p.

Let's look at a rough estimate of what's going on. If he measures

3 spins and they're all up, he knows p can't be 10^-9! He probably will

think it's about 10^-3, so that there is a balance between unlikely

events. Actually I think his estimate would be even higher.

After about O(30) random spins he should be fairly close to .5.

- - - - - - -

Jacques Mallah (jqm1584.domain.name.hidden)

Graduate Student / Many Worlder / Devil's Advocate

"I know what no one else knows" - 'Runaway Train', Soul Asylum

My URL: http://pages.nyu.edu/~jqm1584/

Received on Sat Apr 17 1999 - 14:04:44 PDT

Date: Sat, 17 Apr 1999 17:03:16 -0400

On Sat, 17 Apr 1999, Nick Bostrom wrote:

If he has no other information, of course he should. But the

difference is that if he does observe deer in the region, he would realize

that over the ensemble of worlds, the effective probability of seeing a

deer is the same as would be the case if the number of subsequent people

didn't depend on it.

The point is that the quantum 'probability' can be extropolated

from past events, even if you start with a prior that is heavily skewed

in the wrong direction, but the outcome of one particular event in a

non-MWI can't be. Suppose he'd have kids if on a certain measurement he

saw spin up.

If you draw many spins from a sample and they are found to be 50%

up and 50% down, you can be reasonably sure that the quantum probability

p to measure up is about 50%. In this case the sample size matters; to

overcome Adam's prior expectation of 10^-9 or 2^-30, to get his median

expectation of p to >= q, he'd need to measure over M spins. What's M(q)?

It might be interesting to look more closely at that: it depends on the

shape of his prior f(p), which we can obtain from the usual Bayesian

proceedure starting with a uniform f(p) on (0,1) and using the conditional

effective probability of an observer being Adam as a function of p.

Let's look at a rough estimate of what's going on. If he measures

3 spins and they're all up, he knows p can't be 10^-9! He probably will

think it's about 10^-3, so that there is a balance between unlikely

events. Actually I think his estimate would be even higher.

After about O(30) random spins he should be fairly close to .5.

- - - - - - -

Jacques Mallah (jqm1584.domain.name.hidden)

Graduate Student / Many Worlder / Devil's Advocate

"I know what no one else knows" - 'Runaway Train', Soul Asylum

My URL: http://pages.nyu.edu/~jqm1584/

Received on Sat Apr 17 1999 - 14:04:44 PDT

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