Re: copying related probability question

Wei Dai (
Sat, 27 Sep 1997 19:14:02 -0700 (PDT)

On Sun, 28 Sep 1997, Nicholas Bostrom wrote:

> >I agree with Nicholas that these two problems are closely related.
> >The nice thing about the copying problem is that the reference class
> >is very clear. It is the class of your (potential) clones.
> If the thought experiment is supposed to occur in the actual world,
> then there will also be other members of the reference class, namely
> all those people not involved in the copying procedure. These are the
> "outsiders". As I argued at length in the Doomsday paper, the
> Doomsday argument only works (at full strength) if we postulate the
> "no-outsider requirement": that there be no outsiders in the world.

I think the "outsiders" is only a distraction. This is true in both
problems, but is especially clear in the copying problem. In the copying
experiment, at no time does the subject place any positive probability on
himself being an outsider, so the number of outsiders cannot possibly be
relevant. As an analogy, consider picking a ball from an urn blindfolded.
If you are told that the ball you picked is not red, then it cannot matter
how many red balls there were in the urn. In other words, the probability
of any statement concerning the ball conditioning on the ball not being
red remains the same if you alter the experiment by changing the number of
red balls originally in the urn. Similarly, the subject's conditional
probabilities given that he is not an outsider cannot change if the number
of outsiders changes.

In your God's coin tosses experiment, the odds should be exactly 10:1 no
matter how many outsiders there are. Let H be "the coin showed head" and C
be "I was created as a result of the coin toss." P(H|C)/P(not
H|C)=P(C|H)P(H)/(P(C|not H)P(not H)). In your calculations, you assume
that P(H)=P(not H). But as I argued in the copying problem, and I think it
is true in this problem also, P(H)!=P(not H) given the subject's state of
knowledge after the coin toss. As a result P(H|C)/P(not H|C) is exactly
10 even though P(C|H)/P(C|not H) only approaches 10 asymptoticly as the
number of outsiders goes to infinity.