Re: Doomsday Example

Nick Bostrom (
Sun, 23 Aug 1998 01:17:44 +0000

Robin Hanson writes:

> Nick B. writes:
> >> But nothing else could have been me exactly. The only thing that
> >> could be me exactly is something born when I was, and which then
> >> did everything I did afterward, including writing this message.
> >
> >How does that support your claim that it makes sense to say that you
> >could have been a rock?
> We are talking counterfactuals here. I am male, but could I
> imagine being female? Yes. Could I imagine being born in 1400?
> Yes. Could I imagine being a chimp? Yes. Can I imagine being a rock?
> Yes. I see no boundary beyond which I can't imagine the variation.
> They all seem like reasonable counterfactuals to me. What reasons
> can you offer for rejecting some counterfactuals and accepting others?

Well, suppose there were two possible worlds, A and B, that are a priori equally probable. In A there are a hundred humans and nothing else. In B there are a hundred humans and a million stones. If you know this and nothing more except that you exist and are a human, what would you say the posterior probabilities are for the two worlds?

I would say 1/2. Finding myself being a human would give me no information as to the number of stones.

But in order to get this result we need to assume that there was a zero chance that you would have been a stone.

> >> >> What prior would you assign to world * in my example?
> >> >
> >> >That depends on such things as simplicity etc. If the three worlds
> >> >are equal in these repects, I would say P(*) = 1/3. This would be the
> >> >absolute prior. Then you take account of the fact that you exist, and
> >> >you rule out world * (though I'm not sure what do about the monkeys).
> >> >Then you renormalize and get P(#) = P(@) = 1/2.
> >>
> >> But your *prior* is before you take information into account, and you
> >> seem to prefer p(*) = 1/3, p(#) = 7/20, p(@) = 3/20. I instead prefer
> >> all three being 1/3. You say:
> >
> >But I too would say all absolute priors three are 1/3. But these are
> >the probabilities you should assign if you knew absolutely nothing,
> >not even that you existed. In practice, you will always know you
> >exist. If you conditionalize on this information then you get
> >p'(*)=0, p'(#)=p'(@)=1/2. Now if you want you can call this new
> >probability distribution, p'=p( _ | "I exist."), your "prior".
> We seem to disagree on the math here. And since one of the reasons
> for precise models like this is so we can agree on such things, we
> really aught to get this clear. I say that since the probability
> of existing conditional on being in universes # and @ is different,

which I deny!

> you can't have equal probabilities for these universes BOTH for an
> absolute prior AND after you condition on existing. If they are
> equal to 1/2 after conditioning on existing, they cannot be both
> equal to 1/3 before conditioning. Could you please recheck your
> calculations?

Let's take it slowly and write it out in full:

Let * be "Universe * exists." And similarly for # and @. Let Me be "I exist."


I think we have agreed about that. That's what I call the absolute priors. Then I say that

P(* | Me)=0
P (# | Me)=1/2
P (@ | Me)=1/2

By Bayes' theorem, that implies:

P(Me | *) = P(* | Me)P(Me)/P(*)
= (0*P(Me))/(1/3) = 0


P(Me | #) = P(# | Me)P(Me)/P(#)
= ((1/2)*P(Me))/(1/3) = (3/2)P(Me)


P(Me | @) = P(@ | Me)P(Me)/P(@)
= ((1/2)*P(Me))/(1/3) = (3/2)P(Me)

Since we have P(Me)=P(Me|*)P(*)+P(Me|#)P(#)+P(Me|@)P(@),

we get P(Me)=0+(3/2)P(Me)(1/3)+(3/2)P(Me)(1/3) = P(Me)

I don't see any inconsistency.

> >> "I'm not sure how many stones a universe has, but I expect universes
> >> that have more stones in one region to have more stones in other regions
> >> as well. If I look in one spot in this universe and I find a stone, that
> >> suggests I'm more likely to find stones at other spots in this universe
> >> as well."
> >
> >Sure, but this reasoning only works for things that may or may not be
> >present where there are observers. Stones are a good example. But
> >many other phenomena are correlated with the presense of observers
> >and then this reasoning doesn't work. For instance, as you yourself
> >has argued in your "Must early life be easy?"-paper, it would be
> >wrong to use the observation that evolutionary processes here on
> >Earth were fairly quick to infer that quick evolutionary processes
> >are common in the universe; for we would not have existed if the
> >particular evolutionary process we observe had not been quick enough
> >to happen before the sun becomes a red gigant.
> But my paper does things exactly the way I'm reccomending here.
> I choose a prior to have nice natural independence properties,
> without assuming that I exist. I then condition on my existing and

But when you are conditioning on your existing, what you do is increase the probability of those worlds with many observers (SIA). [And with the views you have expressed about the reference class problem, you even want to increase the estimated number of stones??? Let's set that to one side for the moment.] And I say, if you look in one place, and find an X there, then that gives you reason to believe there are many Xs; but *only* if the reason why you looked where you looked was not that there was an X there. Otherwise independence fails. That's why you can't infer "There are many quick evolutionary processes." (I assume you agree with that?) But why, then, do you want to infer from finding that you are an observer that there are many observers? (That's what you want to do, right?) This looks like the paradigmatic case where independence fails. Being an observer and finding that you are an observer are surely not statistically independent events. (Right?) So, since independence fails completely in this case, you should not infer from finding that you are an observer that there are many observers. (Do you accept this? If not, exactly where do you begin to disagree?)

Nick Bostrom
Department of Philosophy, Logic and Scientific Method London School of Economics