From: Vladimir Nesov (email@example.com)
Date: Tue Mar 11 2008 - 23:35:47 MDT
On Tue, Mar 11, 2008 at 6:39 PM, John K Clark <firstname.lastname@example.org> wrote:
> On Tue, 11 Mar 2008 21:14:49 +1100, "Stathis Papaioannou"
> <email@example.com> said:
> > I don't see how you arrive at the 50% figure. Imagine
> > that there are 101 identical copies of you carrying
> > an apparently identical envelope. Inside one envelope
> > is a note saying "torture" while inside the other
> > 100 envelopes is a note saying "no torture". Each
> > of the copies knows this information and has to
> > guess what the note in his envelope says.
> > Assuming getting it right is important, isn't it
> > best to guess that it says "no torture"?
> No, there is still a 50% chance I will be tortured. There are a 101
> identical copies of me running and they are all me, eventually the
> copies open their envelopes and one of them sees something the other 100
> do not, it says torture. He is still me but now he's different from the
> other 100, he has diverged. So now there are 2 of me, the tortured copy
> and the other identical 100.
And if other envelopes contain numbers 1 to 100, chance jumps from 50%
to 1:100? If this difference is 1 atom in each of no-torture 100
envelopes, is it still 1:100? Does it return to 50% once one-atom
difference is removed? Probability is for making decisions, and it's
not right to base high-level decisions on one atom in the envelope.
-- Vladimir Nesov firstname.lastname@example.org
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