So what if we are uncertain about the size of the universe (so that its size depends on which possible world we are in)? Then we are faced with the same question as before: Should we treat finding ourselves in bigger universes as more probable a priori, or not?

Formally, the question we face is, if we have a prior P0 over possible worlds, what should our prior over (possible world, space-time-Everett location) pairs be?

Physical self-sampling without self-indication. P((w,x)) = P0(w) / number of possible locations in world w

Physical self-sampling with physical self-indication. P((w,x)) = alpha * P0(w), where alpha is a normalization constant (alpha = 1 / Sum_w'. P0(w') * number of possible locations in world w')

(As before, we may want to weigh Everett branches in the obvious way.) Both of these definitions give us the weak principle of self-indication (defined in the previous comment), since they agree with the previous comment's definition when all possible worlds contain the same number of locations. So they both support thirding in Sleeping Beauty.

But which of the definitions should we adopt? Note that sampling without self-indication has the property that P(w) = P0(w), i.e., before we condition on any evidence (including the fact that we are conscious observers), the probability of finding ourselves in world w is exactly the probability of that world, according to P0. On the face of it, this sounds exactly like what we mean by having a prior P0 over the possible worlds.

I think we may mean different things with P0 depending on how we arrive at P0, though. But for the moment, let me note that while the principle of weak self-indication forces me to accept the presumptuous philosopher's position in both the Case of the Twin Stars and the Case of the Death Rays, I may still have a good reason to reject the conclusion that the cosmos is infinite with probability one.

In my previous comment, I mentioned my worry that accepting observer self-sampling without self-indication means that you've been suckered into taking conscious observation as an ontological primitive. (Also, I've been careful not to use examples that involve the size of the cosmos.) I would like to suggest that instead of a prior over observer-moments in possible worlds, we start with a prior over space-time-Everett locations in possible worlds. If all possible worlds we consider have the same set of space-time-Everett locations, and we have a prior P0 over possible worlds, then I suggest that we adopt the prior over (world, location) pairs:

P((w,x)) = P0(w) / number of possible locations

(Actually, that's not necessarily quite right: If the "amplitude as degree of reality" interpretation is true, Everett branches should of course be weighted in the obvious way.)

As with observer-moments, we then condition on all the evidence we have about our actual space-time-Everett location in our actual possible world, and call the result our "subjective probability" distribution.

Isn't anthropic reasoning about taking into account the observer selection effects related to the fact that we are conscious observers? Sure, but it seems to me that any non-mysterious anthropic reasoning is taken care of just fine by the conditioning step. Any possible worlds, Everett branches and cosmic regions that don't support intelligent life will automatically be ruled out, for example.

The above definition trivially implies the following weak principle of self-indication:

If all possible worlds we consider have the same set of locations, worlds that contain more locations consistent with our evidence will tend to be more likely after conditionalization. (To be precise, the probability of each world w is weighted by P0(w) * number of locations in w consistent with our evidence).

This principle is enough to support being a thirder in the Sleeping Beauty problem, for example (which was what originally suggested it to me, when I was wondering what prior Beauty should update when she observes herself to be awake).

So if I think that (something like) the Self-Indication Assumption is correct, what about Nick's standard thought experiment in which the silly philosopher thinks she can derive the size of the cosmos from the fact she's alive?

Well, the experiment does worry me, but I'd like to note that self-sampling without self-indication produces, in fact, a very similar result (if the reference class is all conscious observers, which Nick's version of the experiment seem to assume). I give you The Presumptuous Philosopher and the Case of the Twin Stars:

Physicists have narrowed down the search for the Theory of Everything to T1 and T2, between which considerations of super-duper-symmetry are indifferent. We know that the cosmos is very big, to the tune of containing a trillion trillion galaxies, most of which are expected to contain life. But there's a twist: According to T1, all but one in a trillion galaxies should consist entirely of twin star systems. T2 does not make this prediction. Physicists are preparing to do a simple test that would decide between the theories. Enter the Presumptuous Philosopher: "Guys, our galaxy has lots of single-star systems. The conditional probability of this if T1 is true and we're a random sample from all conscious observers is only one in a trillion! Stop doing this silly experiment and do something else instead!"

If you accept this thought experiment (which requires only self-sampling) but reject a variation where T1 is ruled out because it predicts that cosmological death rays will make life impossible in all galaxies but one in a trillion (which requires self-sampling), then I think you've allowed yourself to be suckered into implicitly assuming that conscious observation is something ontologically fundamental. Though I accept that you may not be convinced of this yet :-)

(Side note: Lest you be biased against the philosopher just because she dares to apply probability theory, do also consider the case where T1 predicts that Mars had a chance of 4/5 per year of flying out of the solar system since it came into existence -- and beat those odds by random chance every single time. Of course, in that case, the physicists would already be convinced that her reasoning is sound, to the tune that they would already have applied it itself.)

It may be silly to continue this here, since I'm not sure anybody's still reading, but at least I'm writing it down at all this way, so... here's "Nick's Sleeping Beauty can be Dutch Booked" (by Nick's own rules)

In his Sleeping Beauty paper, Nick considers the ordinary version of the problem: Beauty is awakened on Monday. An hour later, she is told that it is Monday. Then she is given an amnesia drug and put to sleep. A coin is flipped. If the coin comes up tails, she is awakened again on Tuesday (and can't tell the difference to Monday). Otherwise, she sleeps through to Wednesday.

Nick distinguishes five possible observer-moments: Beauty wakes up on Monday (h1 and t1, depending on heads/tails); Beauty is told that it's Monday (h1m and t1m); Beauty wakes up on Tuesday (t2). Let P-(x) := P(x | h1 \/ t1 \/ t2), and P+(x) := P(x | h1m \/ t1m).

There are two possible worlds, heads-world (h1,h1m) and tails-world (t1,t1m,t2). Within each of the groups (h1,t1,t2) and (h1m,t1m), Nick assigns equal probabilities to each observer-moment in a given possible world. This gives:

P-(h1) = 1/2; P-(t1) = 1/4; P-(t2) = 1/4
P+(h1m) = 1/2; P+(t1m) = 1/2

In his paper, Nick considers the following Dutch book, suggested by a referee (I'm quoting from the paper):

Upon awakening, on both Monday and Tuesday, before either knows what day it is, the bookie offers Beauty the following bet: Beauty gets $10 if HEADS and MONDAY. Beauty pays $20 if TAILS and MONDAY. (If TUESDAY, then no money changes hands.) On Monday, after both the bookie and Beauty have been informed that it is Monday, the bookie offers Beauty a further bet: Beauty gets $15 if TAILS. Beauty pays $15 if HEADS. If Beauty accepts these bets, she will emerge $5 poorer.

Nick dismisses this argument because if the coin falls tails, Beauty will accept the first bet twice, once on Monday and once on Tuesday. Now, on Tuesday no money changes hands, so what's the difference? Well, Nick thinks it's very interesting that it could make a difference, but clearly it does, you see, because otherwise Sleeping Beauty could be Dutch booked if she accepts his probability assignments!

Instead of trying to argue that it makes no difference, let me just exhibit a variation where Beauty only accepts every bet at most once in every possible world.

Before Beauty is put to sleep, we throw a second fair coin, labelled A and B. If it comes up A, then on Monday, we tell Beauty, "It's day A!" And if we wake her up on Tuesday, we tell her, "It's day B!" If the coin comes up B, Monday is B, and Tuesday is A.

We now have doubled the number of worlds and observer-moments. The worlds are HA, HB, TA, and TB, each with probability 1/4; the observer-moments are ha1, ha1m; hb1, hb1m; ta1, ta1m, ta2; tb1, tb1m, tb2. P- and P+ are defined analogously to before, and again, we assign equal probability to each of the awakenings in every possible world (and make them sum to the probability of that world). This gives:

P-(ha1) = P-(hb1) = 1/4
P-(ta1) = P-(ta2) = P-(tb1) = P-(tb2) = 1/8
P+(ha1m) = P+(hb1m) = P+(ta1m) = P+(tb1m) = 1/4

The sets of observer-moments that Beauty cannot distinguish are: {ha1,ta1,tb2}; {hb1,tb1,ta2}; {ha1m,ta1m}; {hb1m,tb1m}. (E.g., on {ha1,ta1,tb2}, Beauty just knows that she's been awakened and that it's "Day A." In world B, Tuesday is Day A, thus tb2 is in this set.)

Note well that in none of these sets, there is more than one observer-moment from the same possible world. I exhibit the following variation of the above Dutch Book.

On {ha1,ta1,tb2}, the Bookie offers Beauty the first bet above: Beauty gets $10 if HEADS and MONDAY. Beauty pays $20 if TAILS and MONDAY. (If TUESDAY, then no money changes hands.)

On {ha1m,ta1m}, the Bookie offers the second bet: Beauty gets $15 if TAILS. Beauty pays $15 if HEADS.

Beauty now loses $5 if the day-label-coin comes up A, and breaks even if it comes up B. Every bet is accepted exactly once in every possible world in which it is offered at all. We could add symmetrical additional bets to make sure that Beauty also loses money in B worlds, but I think I've made my point. Nick can create his priors over observer-moments without violating the axioms of probability, but if it worries him if Beauty can be Dutch-booked in the way he discusses in his paper, I do believe he needs to be worried...

Okay, after reading several of Nick Bostrom's papers and mulling about the problem for a while, I think I may have sorted out my position enough to say something interesting about it. But now I'm finding myself suffering from a case of writer's block in explaining it, so I'll try to pull a small-scale Eliezer and say it in a couple of hiccups, rather than one fell swoop :-)

I have been significantly wrong at least twice in this thread, the first time when I thought everybody was reasoning from the same definitions as me, but getting their math wrong, and the second time when I said I held my view because I was "pretty sure I [was] applying my probability theory right". I had an intuition and a formal argument, but then I found that the two disagree in some edge cases, and I decided to retain the intuition, so my formal argument was not the solid rock I thought it was. All of which is a long-winded way of saying, it's about time that I concede that I may still be wrong about this, and if so, please do help me figure it out...

We all seem to agree that the issue depends on whether we accept self-indication, and that self-indication is equivalent to being a thirder in the Sleeping Beauty problem. When I first learned about this problem from Robin's post, I was very convinced that the halfer view was right -- to the tune of having been willing to bet money on it -- for about fifteen minutes. Then I thought about something like the following variation of it:

Beauty is put to sleep on Sunday, and a fair coin is tossed. Beauty is awakened twice, once on Monday and once on Tuesday; in between, she is given an amnesia-inducing drug, so that when she wakes up, she cannot tell whether she has been woken up before. One minute after Beauty wakes up, a light flashes. If it is Tuesday, and the coin came up heads, the light is red; otherwise, it is blue.

When Beauty wakes up, before the light flashes, what is her subjective probability that (h1) the coin came up heads, and it's Monday; (h2) heads, Tuesday; (t1) tails, Monday; (t2) tails, Tuesday?

I cannot conceive of a reason not to assign the probability 1/4 to each of these propositions, and in my opinion, when Beauty sees the light flash red, she must update her subjective probability in the obvious way (or the notion of subjective probability no longer makes much sense to me). Then, of course, after seeing the light flash blue, Beauty's probability that the coin fell heads is 1/3.

Short of assigning special ontological status to being consciously awake, I don't see a way to distinguish between the original Sleeping Beauty and my variation after the light flashes blue, so I'm a thirder now. My new view is that observing the random variable (color=blue) can change my probability in non-mysterious ways, so observing the random variable (awake=yes) can, too.

In his paper on the problem, Nick argues for a "solution" that would apply to my version, too. He would reject my view of how Beauty must update her probabilities if she sees a blue light. His argument goes something like this:

What I really need to consider is all of Beauty's observer-moments in all possible worlds; Beauty has a prior over these moments, considers the evidence she has for which moment she is in, and does a Bayesian update. The moment when Beauty wakes up is different from the moment when the light flashes, so she needs to consider at least eight possible moments: (h1-) heads, Monday, she wakes up; (h1+) heads, Monday, the light flashes; and so on. Nothing in the axioms of probability theory requires the probability of (h1+) to be related in any way to the probability of (h1-)! In fact, Nick would argue, we should simply assign probabilities like this:

p(xx- | h1- \/ h2- \/ t1- \/ t2-) = 1/4 (for xx in {h1,h2,t1,t2})
p(h1+ | h1+ \/ t1+ \/ t2+) = 1/2
p(xx+ | h1+ \/ t1+ \/ t2+) = 1/4 (for xx in {t1,t2})

I agree that this is formally consistent with the axioms of probability, but in order for Beauty to be rational, in my opinion she must still update her probability estimate in the "normal" way when the light flashes blue. Nick's approach strikes me as saying, "I'm a completely new observer-moment now, why should I care about my probability estimates a minute ago?" If our formalism allows us to do that, I think our formalism isn't strong enough. In this case, I'd require that

p(xx- | h1- \/ h2- \/ t1- \/ t2-)
= p(xx+ | h1+ \/ h2+ \/ t1+ \/ t2+)

--i.e., before conditioning on the actual colors she sees, Beauty's probability estimates when the light flashes must be the same as when she wakes up. I don't know how well this generalizes, but if we accept it in this case, it blocks Nick's proposal.

Anybody here who finds Nick's solution intuitively right?

...Allan, sorry for the delay in replying. Hopefully tomorrow. (In my defense, I've spent the whole day seriously thinking about the problem ;-))

Allan: I don't believe that's what I've been saying; the question is whether the LHC failing is evidence for the LHC being dangerous, not whether surviving is evidence for the LHC having failed.

I was trying to restate in different terms the following argument for failure to be considered evidence:

The intuition on my side is that, if you consider yourself a random observer, it's amazing that you should find yourself in one of the extremely few worlds where the LHC keeps failing, unless the LHC is dangerous, in which case all observers are in such a world.

For "observer" I substituted "surviving observer," because when doing the math I find it more helpful to consider all potential observers and then say that some of them are dead and thus can't observe anything. So my "surviving observer" is the same as your "observer," right?

So I read your argument as: If the LHC is benign, and you're a random (surviving) observer, then it's amazing if (i.e., there is a low probability that) you find yourself in one of the few worlds where the LHC keeps failing. If the LHC is dangerous, and you're a random observer, then it's non-amazing (i.e., there is a high probability that) you find yourself in a world where the LHC keeps failing. Therefore, if you're a random observer, and you find yourself in a world where the LHC keeps failing, then the LHC is probably dangerous (because then, we don't need to assume something amazing going on). Am I misunderstanding something?

If I understand you right, what I'm saying is that both the if's are clearly correct, but I believe that the 'therefore' doesn't follow.

To me, the problem is essentially the same as the following: You are one of 10,000 people who have been taken to a prison. Nobody has explained why. Every morning, the guards randomly select 9/10 of the remaining prisoners and take them away, without explanation. Among the prisoners, there are two theories: one faction thinks that the people taken away are set free. The other faction thinks that they are getting executed.

It is the fourth morning. You're still in prison. The nine other people who remained have just been taken away. Now, if the other people have been executed, then you are the only remaining observer, so if you're a random observer, it's not surprising that you should find yourself in prison. But if the other people have been set free, then they're still alive, so if you're a random observer, there is only a 1/10,000 chance that you are still in prison. Both of these statements are correct if you are a random (surviving) observer. But it doesn't follow that you should conclude that the other people are getting shot, does it? (Clearly you learned nothing about that, because whether or not they get shot does not affect anything you're able to observe.)

Now, I get that you probably think something makes this line of reasoning not apply when we consider the anthropic principle (although I do think that you're wrong then :)). But my point is that, unless I'm missing something, the probabilistic reasoning is the same as in my restatement of your argument, so if the laws of probability don't make the conclusion follow in this scenario, they don't make the conclusion follow in your argument, either.

I should say that I don't reject "the" anthropic principle. I wholeheartedly embrace the version of it that I can derive from the kind of reasoning as above. For example: If our theory of evolution seems to suggest that there is one very improbable step in the evolution of intelligent life -- so improbable that it's not likely to have happened even a single time in the history of the universe -- should we then take that as a reason to conclude that something is wrong with our theory? If we are pretty sure that there is only a single universe, yes. If we have independent evidence that all possible Everett branches exist, no. (If something like mangled worlds is true, maybe -- but let's not get into that now...)

Why should we reject our theory in a single universe, but not if all Everett branches exist? Consider again the prison analogy. You observed how the guards chose the prisoners to take away, and it sure looked random. But now you are the only surviving prisoner. Should you conclude that the guards' selection process wasn't really random? There's no reason to: If the guards used a random process, one prisoner had to remain on the fourth day, and this may just as well have been you -- nothing surprising going on. This corresponds to the scenario where all possible Everett branches exist.

But suppose that you were the only prisoner to begin with (and you know this), and every morning the guards threw a ten-sided die which is marked "keep in prison" on one side and "take away" on the nine others -- and it came up "keep in prison" every morning. In this case, it seems to me that you do have a reason to start suspecting that the die is fixed (i.e., that your original theory, that the "keep in prison" outcome had only a 10% chance of happening, was wrong). This corresponds to the scenario where there is only a single universe.

This is how I always understood the anthropic principle when reading about it, and this version of it I embrace. The other version I'm pretty sure is wrong.

That said, if you have the energy to do so, please do keep arguing with me! :-) I don't really understand this "other anthropic principle," and I'm rejecting it simply because it disagrees with my calculations and I'm really pretty sure that I'm applying my probability theory right here. If I'm wrong, that will be humbling, but I would still rather know than not know, please :-)

Allan, oh ****, the elementary math in my previous comment is completely wrong. (In the scenario I gave, the probability that you have breast cancer is 1%, not 10%, before taking the test.) My argument doesn't even approximately work as given: if having breast cancer makes it more likely that you get a positive mammography, then *indeed* getting a positive mammography must make it more likely that you have breast cancer. Sorry!

(I'm still convinced that my argument re the LHC is correct, but I realize that I'm just looking stupid right now, so I'll just shut up for now :-))

simon, that's right, of course. The reason I'm dragging branches into it is that for the (strong) anthropic principle to apply, we would need some kind of branching -- but in this case, the principle doesn't apply [unless you and I are both wrong], and the math works the same with or without branching.

Eliezer, huh? Surely if F => S, then F is the same event as (F /\ S). So P(X | F) = P(X | F, S). Unless P(X | F, S) means something different from P(X | F and S)?

Allan, you are right that if the LHC would destroy the world, and you're a surviving observer, you will find yourself in a branch where LHC has failed, and that if the LHC would not destroy the world and you're a surviving observer, this is much less likely. But contrary to mostly everybody's naive intuition, it doesn't follow that if you're a suriving observer, LHC has probably failed.

Suppose that out of 1000 women who participate in routine screening, 10 have breast cancer. Suppose that out of 10 women who have breast cancer, 9 have positive mammographies. Suppose that out of 990 women who do not have breast cancer, 81 have a positive mammography.

If you do have breast cancer, getting a positive mammography isn't very surprising (90% probability). If you do not have breast cancer, getting a positive mammography is quite surprising (less than 10% probability).

But suppose that all you know is that you've got a positive mammography. Should you assume that you have breast cancer? Well, out of 90 women who get a positive mammography, 9 have breast cancer (10%). 81 do not have breast cancer (90%). So after getting a positive mammography, the probability that you have breast cancer is 10%...

...which is the same as before taking the test.

The intuition behind the math: If the LHC would not destroy the world, then on date X, a very small number of Everett branches of Earth have the LHC non-working due to a string of random failures, and most Everett branches have the LHC happily chugging ahead. If the LHC would destroy the world, a very small number of Everett branches of Earth have the LHC non-working due to a string of random failures -- and most Everett branches have Earth munched up into a black hole.

The very small number of Everett branches that have the LHC non-working due to a string of random failures is the same in both cases.

Thus, if all you know is that you are in an Everett branch in which the LHC is non-working due to a string of random failures, you have no information about whether the other Everett branches have the LHC happily chugging ahead, or dead.