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