PHYS771 Lecture 17: Fun With the Anthropic Principle

This is a lecture about the Anthropic Principle, and how you apply Bayesian reasoning where you have reason about the probability of your own existence, which seems like a very strange question. It's a fun question, though---which you can almost define as a question where it's much easier to have an opinion than to have a result. But then, we can try to at least clarify the issues, and there are some interesting results that we can get.

There's a central proposition that many people interested in rationality believe they should base their lives around---even though they generally don't in practice. This is Bayes's Theorem.

If you talk to philosophers, this is often the one mathematical fact that they know. (Kidding!) As a theorem, Bayes' Theorem is completely unobjectionable. The theorem tells you how to update your probability of a hypothesis H being true, given some evidence E.

The term P[E|H] describes how likely you are to observe the evidence E in the case that the hypothesis H holds. The remaining terms on the right-hand side, P[H] and P[E], are the two tricky ones. The first one describes the probability of the hypothesis being true, independent of any evidence, while the second describes the probability of the evidence being observed, averaged over all possible hypotheses. Here, you're making a commitment that there are such probabilities in the first place---in other words, that it makes sense to talk about what the Bayesians call a prior. When you're an infant first born into the world, you estimate there's some chance that you're living on the third planet around the local star, some other chance you're living on the fourth planet and so on. That's what a prior means: your beliefs before you're exposed to any evidence about anything. You can see already that maybe that's a little bit of a fiction, but supposing you have such a prior, Bayes' theorem tells you how to update it given new knowledge.

The proof of the theorem is trivial. Multiply both sides by P[E], and you get that P[H|E] P[E] = P[E|H] P[H]. This is clearly true, since both sides are equal to the probability of the evidence and the hypothesis together.

So if Bayes' Theorem seems unobjectionable, then I want to make you feel queasy about it. That's my goal. The way to do that is to take the theorem very, very seriously as an account of how we should reason about the state of the world.

I'm going to start with a nice thought experiment which is due to the philosopher Nick Bostrom. This is called God's Coin Toss. Last lecture, I described the thought experiment as a puzzle.

Imagine that at the beginning of time, God flips a fair coin (one that lands heads or tails with equal probability). If the coin lands heads, then God creates two rooms: one has a person with a red hair and the other has a person with green hair. If the coin lands tails, then God creates one room with a red-haired person.

Q: Are these rooms the entire universe?
Scott: Yes, and these are the only people in the universe, in either case.

We also imagine that everyone knows the entire situation and that the rooms have mirrors. Now suppose that you wake up, look in the mirror and learn your hair color. What you really want to know is which way the coin landed. Well, in one case, this is easy. If you have green hair, then the coin must have landed heads. Suppose that you find you have red hair. Conditioned on that, what probability should you assign to the coin having landed heads?

Q: So there's no door to the other room?
Scott: That's right. You can't see the other room. All you can see is the room that you're in. So does anyone have a guess?
A: One half.

One half is the first answer that someone could suggest. You could just say, "look, we know the coin was equally likely going to land heads or tails, and we know that in either case that there was going to be a red-haired person, so being red-haired doesn't really tell you anything about which way it landed, and therefore it should be a half." Can anyone defend a different answer?

A: It seems more likely for it to have landed tails, since in the heads case, the event of waking up with red hair is diluted with the other possible event of waking up with green hair. The effect would be more dramatic if there were a hundred rooms with green hair.
Scott: Exactly.
Q: It isn't clear at all to me that the choice of whether you're the red- or the green-haired person in the heads case is at all probabilistic. We aren't guaranteed that.
Scott: Right. This is a question.
Q: It could have been that before flipping the coin, God wrote down a rule saying that if the coin lands heads, make you red-haired.
Scott: Well, then we have to ask, what do we mean by "you?" Before you've looked in the mirror, you really don't know what your hair color is. It really could be either, unless you believe that it's really a part of the "essence" of you to have red hair. That is, if you believe that there is no possible state of the universe in which you had green hair, but otherwise would have been "you."
Q: Are both people asked this question, then?
Scott: Well, the green-haired person certainly knows how to answer, but you can just imagine that the red-haired people in the heads and tails cases are both asked the question.

To make the argument a little more formal, you can just plop things into Bayes's Theorem. We want to know the probability P[H|R] that the coin landed heads, given that you have red hair. We could do the calculation, using that the probability of us being red haired given that the coin lands heads is ½, all else being equal. There are two people, and you aren't a priori more likely to be either the red-haired or the green-haired person. Now, the probability of heads is also ½ — that's no problem. What's the total probability of your having red hair? That's just given by P[R|H] P[H] + P[R|T] P[T]. As we've said before, if the coin lands tails, you certainly have red hair, so P[R|T] = 1. Moreover, we've already assumed that P[R|H] = ½. Thus, what you get out is that P[H|R] = ¼ / ¾ = ⅓. So, if we do the Bayesian calculation, it tells us that the probability should be ⅓ and not ½.

Does anyone see an assumption that you could make that would get the probability back to ½?

A: You could make the assumption that whenever you exist, you have red hair.
Scott: Yeah, that's one way. But is there a way to do it that doesn't require a prior commitment about your hair color?

Well, there is a way to do it, but it's going to be a little bit strange. One way is to point out that in the heads world, there are twice as many people in the first place. So you could say that you're a priori twice as likely to exist in a world with twice as many people. In other words, you could say that your own existence is a piece of evidence you should condition upon. I guess if you want to make this concrete, the metaphysics that this would correspond to is that there's some sort of a warehouse full of souls, and depending on how many people there are in the world, a certain number of souls get picked out and placed into bodies. You should say that in a world with more people, it would be more likely that you'd be picked at all.

If you do make that assumption, then you can run through the same Bayesian ringer. You find that the assumption precisely negates the effect of reasoning that if the coin landed heads, then you could have had green hair. So you get back to ½.

Thus, we see that depending on how you want to do it, you can get an answer of either a third or a half. It's possible that there are other answers that could be defended, but these seem like the most plausible two.

That was a fairly serene thought experiment. Can we make it more dramatic?

Q: Part of the reason that this feels like philosophy is that there aren't any real stakes here.
Scott: Exactly. Let's get closer to something with real stakes in it.

The next thought experiment is, I think, due to the philosopher John Leslie. Let's call it the Dice Room. Imagine that there's a very, very large population of people in the world, and that there's a madman. What this madman does is, he kidnaps ten people and puts them in a room. He then throws a pair of dice. If the dice land snake-eyes, then he simply murders everyone in the room. If the dice do not land snake-eyes, then he releases everyone, then kidnaps 100 people. He now does the same thing: he rolls two dice; if they land snake-eyes then he kills everyone, and if they don't land snake-eyes, then he releases them and kidnaps 1,000 people. He keeps doing this until he gets snake-eyes, at which point he's done. So now, imagine that you've been kidnapped. You can assume either that you do or do not know how many other people are in the room.

Q: Have you been watching the news?
Scott: You have been watching the news and you know the entire situation. That's always a convenient assumption.

So you're in the room. Conditioned on that fact, how worried should you be? How likely is it that you're going to die?

A: 1/36.
Scott: OK. That would be one guess.

One answer is that the dice have a 1/36 chance of landing snake-eyes, so you should be only a "little bit" worried (considering). A second reflection you could make is to consider, of people who enter the room, what the fraction is of people who ever get out. Let's say that it ends at 1,000. Then, 110 people get out and 1,000 die. If it ends at 10,000, then 1,110 people get out and 10,000 die. In either case, about 8/9 of the people who ever go into the room will die.

Q: But that's not conditioning on the full set of information. That's just conditioning on the fact that I'm in the room at some point.
Scott: But you'll basically get the same answer, no matter what time you go into the room. No matter when you assume the process terminates, about 8/9 of the people who ever enter the room will be killed. For each termination point, you can imagine being a random person in the set of rooms leading up to that point. In that case, you're much more likely to die.
Q: But aren't you conditioning on future events?
Scott: Yes, but the point is that we can remove that conditioning. We can say that we're conditioning on a specific termination point, but that no matter what that point is, we get the same answer. It could be 10 steps or 50 steps, but no matter what the termination point is, almost all the people who go into the room are going to die, because the number of people is increasing exponentially.

If you're a Bayesian, then this kind of seems like a problem. You could see this as a bizarre madman-related thought experiment, or if you preferred to, you could defend the view that this is the actual situation the human species is in. We'd like to know what the probability is that we're going to suffer some cataclysm or go extinct for some reason. It could be an asteroid hitting the earth, nuclear war, global warming or whatever else. So there are kind of two ways of looking at it. The first way is to say that all of these risks seem pretty small---they haven't killed us yet! There have been many generations, and in each one, there have been people predicting imminent doom and it never materialized. So we should condition on that and assign a relatively small probability to our generation being the one to go extinct. That's the argument that conservatives like to make; I'll call it the Chicken Little Argument.

Against that, there's the argument that the population has been increasing exponentially, and that if you imagine that the population increases exponentially until it exhausts the resources of the planet and collapses, then the vast majority of the people who ever lived will live close to the end, much like in the dice room. Even supposing that with each generation there's only a small chance of doom, the vast majority of the people ever born will be there close to when that chance materializes.

Q: But it still seems to me like that's conditioning on a future event. Even if the answer is the same no matter which future event you choose, you're still conditioning on one of them.
Scott: Well, if you believe in the axioms of probability theory, then if p = P[A|B] = P[AB], then P[A] = p.
Q: Yes, but we're not talking about B and ¬B, we're talking about an infinite list of choices.
Scott: So you're saying the infiniteness makes a difference here?
Q: Basically. It's not clear to me that you can just take that limit, and not worry about it. If your population is infinite, maybe the madman gets really unlucky and is just failing to roll snake-eyes for all eternity.
Scott: OK, we can admit that the lack of an upper bound on the number of rolls could maybe complicate matters. However, one can certainly give variants of this thought experiment that don't involve infinity.

The argument that I've been talking about goes by the name of The Doomsday Argument. What it's basically saying is that you should assign to the event of a cataclysm in the near future a much higher probability that you might naïvely think, because of this sort of reasoning. One can give a completely finitary version of the Doomsday Argument. Just imagine for simplicity that there are only two possibilities: Doom Soon and Doom Late. In one, the human race goes extinct very soon, whereas in the other it colonizes the galaxy. In each case, we can write down the number of people who will ever have existed. For the sake of discussion, suppose 80 billion people will have existed in the Doom Soon case, as opposed to 80 quadrillion in the Doom Late case. So now, suppose that we're at the point in history where almost 80 billion people have lived. Now, you basically apply the same sort of argument as in God's Coin Toss. You can make it stark and intuitive. If we're in the Doom Late situation, then the vast, vast majority of the people who will ever have lived will be born after us. We're in the very special position of being in the first 80 billion humans---we might as well be Adam and Eve! If we condition on that, we get a much lower probability of being in the Doom Late case than of being in the Doom Soon case. If you do the Bayesian calculation, you'll find that if naïvely view the two cases are equally likely, then after applying the Doomsday reasoning, we're almost certainly in the Doom Soon case. For conditioned on being in the Doom Late case, we almost certainly would not be in the special position of being amongst the first 80 billion people.

Maybe I should give a little history. The Doomsday Argument was introduced by a astrophysicist named Brandon Carter in 1974. The argument was then discussed intermittently throughout the 80s. Richard Gott, who was also an astrophysicist, proposed the "mediocrity principle": if you view the entire history of the human race from a timeless perspective, then all else being equal we should be somewhere in the middle of that history. That is, the number of people who live after us should not be too much different from the number of people who lived before us. If the population is increasing exponentially, then that's very bad news, because it means that humans are not much longer for the world. This argument seems intuitively appealing, but has been largely rejected because it doesn't really fit into the Bayesian formalism. Not only is it not clear what the prior distribution is, but you may have special information that indicates that you aren't likely to be in the middle.

So the modern form of the Doomsday Argument, which was formalized by Bostrom, is the Bayesian form where you just assume that you have some prior over the possible cases. Then, all the argument says is that you have to take your own existence into account and adjust the prior. Bostrom has a book about this where he concludes that the resolution of the Doomsday Argument really depends on how you'd resolve the God's Coin Toss puzzle. If you give ⅓ as your answer to the puzzle, that corresponds to the Self-Sampling Assumption (SSA) that you can sample a world according to your prior distribution and then sample a person within that world at random. is called the Self-Sampling Assumption (SSA). If you make that assumption about how to apply Bayes's Theorem, then it seems very hard to escape the doomsday conclusion.

If you want to negate that conclusion, then you need an assumption he calls the Self-Indication Assumption (SIA). That assumption says that you are more likely to exist in a world with more beings than one with less beings. You would say in the Doomsday Argument that if the "real" case is the Doom Late case, then while it's true that you are much less likely to be one of the first 80 billion people, it's also true that because there's so many more people, you're much more likely to exist in the first place. If you make both assumptions, then they cancel each other out, taking you back to your original prior distribution over Doom Soon and Doom Late, in exactly the same way that making the SIA led us to get back to ½ in the coin toss puzzle.

On this view, it all boils down to which of the SSA and SIA you believe. There are some arguments against Doomsday that don't accept these presuppositions at all, but those arguments themselves are open to different objections. One of the most common arguments that you hear against Doomsday is that cavemen could have made the same argument, but that they would have been completely wrong. The problem with that argument is that the Doomsday Argument doesn't at all ignore that effect. Sure, some people who make the argument will be wrong; the point is that the vast majority will be right.

Q: It seems as though there's a tension between wanting to be right yourself and wanting to design policies that try to maximize the number of people who are right.
Scott: That's interesting.
Q: Here's a variation I want to make on the red room business: suppose that God has a biased coin such that with 0.9 probability, there's one red and many, many greens. With 0.1 probability, there's just a red. In either case, in the red-haired person's room, there's a button. You have the option to push the button or not. If you're in the no-greens case, you get a cookie if you press the button, whereas if you're in the many-greens case, you get punched in the face if you press the button. You have to decide whether to press the button. So now, if I use the SSA and find that I'm in a red room, then probably we're in the no-greens case and I should press the button.
Scott: Absolutely. It's clear that what probabilities you assign to different states of the world can influence what decisions you consider to be rational. That, in some sense, is why we care about any of this.

There's also an objection to the Doomsday Argument that denies that it's valid at all to talk about you being drawn from some class of observers. "I'm not a random person, I'm just me." The response to that is that there are clearly cases where you think of yourself as a random observer. For example, suppose there's a drug that kills 99% of the people who take it, but such that 1% are fine. Are you going to say that since you aren't a random person, the fact that it kills 99% of the people is completely irrelevant? Are you going to just take it anyway? So for many purposes, you do think of yourself as being drawn over some distribution of people. The question is when is such an assumption valid and when isn't it?

Q: I guess to me, there's a difference between being drawn from a uniform distribution of people and a uniform distribution of time. Do you weight the probability of being alive in a given time by the population at that time?
Scott: I agree; the temporal aspect does introduce something unsettling into all of this. Later, we'll get to puzzles that don't have a temporal aspect. We'll see what you think of those.
Q: I've also sometimes wondered "why am I a human?" Maybe I'm not a random human, but a random fragment of consciousness. In that case, since humans have more brain matter than other animals, I'm much more likely to be a human.
Scott: Another question is if you're more likely to be someone who lives for a long time. We can go on like this. Suppose that there's lots of extraterrestrials. Does that change the Doomsday reasoning? Almost certainly, you wouldn't have been a human at all.
Q: Maybe this is where you're going, but it seems like a lot of this comes down to what you even mean by a probability. Do you mean that you're somehow encoding a lack of knowledge, or that something is truly random? With the Doomsday Argument, has the choice of Doom Soon or Doom Late already been fixed? With that drug argument, you could say, "no, I'm not randomly chosen—I am me—but I don't know this certain property of me."
Scott: That is one of the issues here. I think that as long as you're using Bayes's Theorem in the first place, you may have made some commitment about that. You certainly made a commitment that it makes sense to assign probabilities to the events in question. Even if we imagine the world was completely deterministic, and we're just using all this to encode our own uncertainty, then a Bayesian would tell you that's what you should do anytime you're uncertain about anything, no matter what the reason. You must have some prior over the possible answers, and you just assign probabilities and start updating them. Certainly, if you take that view and try to be consistent about it, then you're led to strange situations like these.

As the physicist John Baez has pointed out, anthropic reasoning is kind of like science on the cheap. When you do more experiments, you can get more knowledge, right? Checking to see if you exist is always an experiment that you can do easily. The question is, what can you learn from having done it? It seems like there are some situations where it's completely unobjectionable and uncontroversial to use anthropic reasoning. One example is asking why the Earth is 93 million miles away from the Sun and not some other distance. Can we derive 93 million miles as some sort of a physical constant, or get the number from first principles? It seems clear that we can't, and it seems clear that to the extent that there's an explanation, it has to be that if Earth were much closer, it would be too hot and life wouldn't have evolved, whereas if it were much further, it'd be too cold. This is called the "Goldilocks Principle": of course, life is going to only arise on those planets that are at the right temperature for life to evolve. It seems like even if there's a tiny chance of life evolving on a Venus or a Mars, it'd still be vastly more likely that life would evolve on a planet roughly our distance from the Sun, and so the reasoning still holds.

Then there are much more ambiguous situations. This is actually a current issue in physics, which the physicists argue over. Why is the fine structure constant roughly 1/137 and not some other value? You can give arguments to the effect that if it were much different, we wouldn't be here.

Q: Is that the case like with the inverse-square law of gravity? If it weren't r⁸², but just a little bit different, would the universe be kind of clumpy?
Scott: Yes. That's absolutely right. In the case of gravity, though, we can say that general relativity explains why it's an inverse square and not anything else, as a direct consequence of space having three dimensions.
Q: But we wouldn't need that kind of advanced explanation if we just did science on the cheap and said "it's gotta be this way by the Anthropic Principle."
Scott: This is exactly what people who object to the Anthropic Principle are worried about—that people will just get lazy and decide that there's no need to do any experiment about anything, because the world just is how it is. If it were any other way, we wouldn't be us; we'd be observers in a different world.
Q: But the Anthropic Principle doesn't predict, does it?
Scott: Right, in many cases, that's exactly the problem. The principle doesn't seem to constrain things before we've seen them. The reductio ad absurdum that I like is where a kid asks her parents why the moon is round. "Clearly, if the moon were square, you wouldn't be you, but you would be the counterpart of you in a square-moon universe. Given that you are you, clearly the moon has to be round." The problem being that if you hadn't seen the moon yet, you couldn't make a prediction. On the other hand, if you knew that life was much more likely to evolve on a planet that's 93 million miles away from the Sun than 300 million miles away, then even before you'd measured the distance, you could make such a prediction. In some cases, it seems like the principle really does give you predictions.
Q: So apply the principle exactly when it gives you a concrete prediction?
Scott: That would be one point of view, but I guess one question would be what if the prediction comes out wrong?

Like a student said before, this does feel like it's just philosophy. You can set things up, though, so that real decisions depend on it. Maybe you've heard of the surefire way of winning the lottery: buy a lottery ticket and if it doesn't win, then you kill yourself. Then, you clearly have to condition on being alive to ask the question of whether you are alive or not, and so because you're asking the question, you must be alive, and thus must have won the lottery. What can you say about this? You can say that in actual practice, most of us don't accept as a decision-theoretic axiom that you're allowed to condition on remaining alive. You could jump off a building and condition on there happening to be a trampoline or something that will rescue you. You have to take into account the possibility that your choices are going to kill you. On the other hand, tragically, some people do kill themselves. Was this in fact what they were doing? Were they eliminating the worlds where things didn't turn out how they wanted?

Of course, everything must come back to complexity theory at some point. And indeed, certain versions of the anthropic principle should have consequences for computation. We've already seen how it could be the case with the lottery example. Instead of winning the lottery, you want to do something even better: solve an NP-complete problem. You could use the same approach. Pick a random solution, check if it's satisfying, and if it isn't, kill yourself. Incidentally, there is a technical problem with that algorithm. Does anyone see what it is?

A: Is there a solution at all?
Scott: Right. If there's no solution then you'd seem to be in trouble. On the other hand, there's actually a very simple fix to this problem.
A: Add some dummy string like *n that acts as a "get out of jail free" card.
Scott: Exactly.

So we say that there are these 2n possible solutions, and that there's also this dummy solution you pick with some tiny probability like 2-2n. If you pick the dummy solution, then you do nothing. Otherwise, you kill yourself if and only if the solution you picked is unsatisfying. Conditioned upon their being no solution and your being alive, then you'll have picked the dummy solution. Otherwise, if there is a solution, you'll almost certainly have picked a satisfying solution, again conditioned on your being alive.

As you might expect, you can define a complexity class based on this principle: BPPpath. Recall the definition of BPP: the class of problems solvable by a probabilistic polynomial time algorithm with bounded error. That is, if the answer to a problem is "yes," at least 2/3 of the BPP machine's paths must accept, whereas if the answer is "no," then at most 1/3 of the paths must accept. So BPPpath is just the same, except that the computation paths can all have different lengths. They have to be polynomial, but they can be different. A simple argument shows that BPPpath is equivalent to a class which I'll call PostBPP (BPP with post selection). PostBPP is again the set of problems solvable by a polynomial time probabilistic algorithm where again you have the 2/3-1/3 acceptance condition, but now if you don't like your choice of random bits, then you can just kill yourself. You can condition on having chosen random bits such that you remain alive. Physicists call this postselection. You can postselect on having random bits with some very special property. Conditioned on having that property, a "yes" answer should cause 2/3 of the paths to accept, and a "no" answer should cause no more than 1/3 to accept. Can you see why this is equivalent to BPPpath?

Q: First, I'm a little unclear on the difference between BPP and BPPpath. You said that the computation paths are of different lengths. The way I've always seen BPP, you're making random choices, but you can make a different number on each path.
Scott: OK. Here's the point: in BPPpath, if a choice leads to more different paths, then it can get counted more. Let's say for example that in 2n − 1 branches, we just accept or reject---that is, we just halt---but in one branch, we're going to flip more coins and do something more. In BPPpath, we can make one branch completely dominate all the other branches. I've shown an example of this in the figure at right–suppose we want branch colored in red to dominate everything else. Then we can hang a whole tree from that path, and it will dominate the paths we don't want (colored in black).
A: So you just find one path you like, and then just sit there flipping a coin for a while?
Scott: Exactly, that's basically a proof that PostBPP ⊆ BPPpath. Given an algorithm with postselection, you make a bunch of random choices and if you like them, you make a bunch more random choices, and those paths overwhelm the paths where you didn't like your random bits. What about the other direction? BPPpath ⊆ PostBPP?

The point is, in BPPpath we've got this tree of paths that have different lengths. What we could do is complete it to make a balanced binary tree. Then, we could use postselection to give all these ghost paths suitably lower probabilities than the true paths, and thereby simulate BPPpath in PostBPP.

Completing a BPPpath tree.

Q: Can you go over the definition of PostBPP again?
Scott: Yes. PostBPP is the class of all languages L for which there exist polynomial time Turing machines A and B (A being the thing that decides whether to accept or reject and B being the thing that decides to postselect) such that for every xL, Prr[A(x,r)|B(x,r)] ≥ 2/3 and such that for every xL, Prr[A(x,r)|B(x,r)] ≤ 1/3. As a technical issue, we also require that Pr[B(x,r)] > 0.

Now that we know that PostBPP = BPPpath, we can ask how big BPPpath is. By the argument we gave before, NP ⊆ BPPpath.

Q: The argument we gave before picked the dummy solution with exponentially small, but non-zero probability.
Scott: That's right, but remember that's allowed here. It still satisfies our bounded-error conditions.

On the other hand, is NP = BPPpath? Certainly, that's going to be hard to show, even if it's true. One reason is that BPPpath is closed under complement. Another reason is that it contains BPP. In fact, you can show that BPPpath contains MA and P||NP (P with parallel queries to an NP oracle). I'll leave that as an exercise. In the other direction, it's possible to show that BPPpath is contained in BPP||NP, and thus in the polynomial hierarchy. Under a derandomization hypothesis, then, we find that the anthropic principle gives you the same computational power as P||NP.

Q: How does BPPpath relate to PP?
Scott: That's a great question. BPPpath ⊆ PP.
Q: PP is so huge.
Scott: Why is BPPpath ⊆ PP? Deciding whether to accept or reject is kind of this exponential summation problem. You can say that each of the paths which is a dummy path contributes both an accept and a reject, while each of the accepting paths contributes two accepts and each of the rejecting paths contributes two rejects. Then, just ask if there are more accepts than rejects. That will simulate it in PP.

Of course, none of this would be complete if we didn't consider quantum postselection. That's what I wanted to end with. In direct analogy to PostBPP, we can define PostBQP as the class of decision problems solvable in polynomial time by a quantum computer with the power of post selection. What I mean is that this is the class of problems where you get to perform a polynomial-time quantum computation and then you get to make some measurement. If you don't like the measurement outcome, you get to kill yourself and condition on still being alive.

Q: Right, but is the B in the definition of PostBPP classical or quantum?
Scott: In PostBQP, we're going to have to define things a bit differently, because there's no analog of r. Instead, we'll say that you perform a polynomial-time quantum computation, make a measurement that accepts with probability greater than 0, and then condition on the outcome of that measurement. Finally, you perform a further measurement on the reduced quantum state that tells you whether to accept or reject. If the answer to the problem is "yes," then the second measurement should accept with probability at least 2/3, conditioned on the first measurement accepting. Likewise, if the answer to the problem is "no," the second measurement should accept with probability at most 1/3, conditioned on the first measurement accepting.

Then, we can ask how powerful PostBQP is. One of the first things you can say is that certainly, PostBPP ⊆ PostBQP. That is, we can simulate a classical computer with post selection. In the other direction, we have PostBQP ⊆ PP. There's this proof that BQP ⊆ PP due to Adleman, DeMorrais and Huang. In that proof, they basically do what physicists would call a Feynman path integral, where you sum over all possible contributions to each of the final amplitudes. It's just a big PP computation. From my point of view, Feynman won the Nobel Prize in physics for showing that BQP ⊆ PP, though he didn't state it that way. Anyway, the proof easily generalizes to show that PostBQP ⊆ PP, because you just have to restrict your summation to those paths where you end up in one of those states you postselect on. You can make all the other paths not matter by making them contribute an equal number of pluses and minuses.

Q: Once you postselect, can you postselect later on as well?
Scott: Can you simulate multiple postselections with one postselection? That's another great question. The answer is yes. We get to that by using the so-called Principle of Deferred Measurement, which tells us that in any quantum computation, we can assume without loss of generality that there's only one measurement at the end. You can simulate all the other measurements using controlled-NOT gates, and then just not look at the qubits containing the measurement outcomes. The same thing holds for postselection. You can save up all the postselections until the end.

What I showed three years ago is that the other direction holds as well: PP ⊆ PostBQP. In particular, this means that quantum postselection is much more powerful than classical postselection, which seems kind of surprising. Classical postselection leaves you in the polynomial hierarchy while quantum postselection takes you up to the counting classes, which we think of as much larger.

Let's run through the proof. So we've got some Boolean function f : {0,1}n→{0,1} where f is efficiently computable. Let s be the number of inputs x for which f(x) = 1. Our goal is to decide whether s ≥ 2n-1. This is clearly a PP-complete problem. For simplicity, we will assume without loss of generality that s > 0. Now using standard quantum computing tricks (which I'll skip), it's relatively easy to prepare a single-qubit state like:

This also means that we can prepare the state:
That is, essentially a conditional Hadamard applied to |ψ⟩ for some α and β to be specified later. Let's write out explicitly what H|ψ⟩ is:
So now, I want to suppose that we take two-qubit state above and post-select on the second qubit being 1, then look at what that leaves in the first qubit. You can just do the calculation and you'll get the following state, which depends on what values of α and β we chose before:
Using post selection, we can prepare a state of that form for any fixed α and β that we want. So now, given that, how do we simulate PP? What we're going to do is keep preparing different versions of that state, varying the ratio β/α through {2-n, 2-(nn+1), ..., ½, 1, 2, ..., 2n}. Now, there are two cases: either s < 2n-1 or s ≥ 2n-1. Suppose that the first case holds. Then, s and 2n−2s have the same sign. Since α and β are real, the states |ψα, β⟩ lie along the unit circle:
If s<2n − 1, then as we vary β/α, the state |ψα, β⟩ will always have a positive amplitude for both |0⟩ and |1⟩ (it will lie in the upper-right quadrant). It's not hard to see that at some point, the state is going to become reasonably balanced. That is, the amplitudes of |0⟩ and |1⟩ will come within a constant of each other, as is shown by the solid red vector in the figure above. If we keep measuring these states in the {|+⟩, |−⟩} basis, then one of the states will yield the outcome |+⟩ with high probability.

In the second case, where s ≥ 2n-1, the amplitude of |1⟩ is never positive, no matter what α and β are set to, while the amplitude of |0⟩ is always positive. Hence, the state always stays in the lower-right quadrant. Now, as we vary β/α through a polynomial number of values, |ψα, β⟩ never gets close to |+⟩. This is a detectable difference.

So I wrote this up, thinking it was a kind of cute proof. A year later, I realized that there's this Beigel-Reingold-Spielman Theorem which showed that PP is closed under intersection. That is, if two languages are both in PP, then the AND of those two languages is also in PP. This solved a problem that was open for 20 years. What I noticed is that PostBQP is trivially closed under intersection, because if you want to find the intersection of two PostBQP languages, just run their respective PostBQP machines and postselect on both computations giving a valid outcome, and then see if they both accept. You can use amplification to stay within the right error bounds. Since PostBQP is trivially closed under intersection, it provides an alternate proof that PP is closed under intersection that I think is much simpler than the original proof. The way to get this simpler proof really is by thinking about quantum anthropic postselection. It's like a higher-level programming language for constructing the "threshold polynomials" that Beigel-Reingold-Spielman needed for their theorem to work. It's just that quantum mechanics and postselection give you a much more intuitive way of constructing these polynomials.

I just had a couple puzzles to leave you with. You asked about the temporal aspect and how it introduced additional confusion into the Doomsday Argument. There's one puzzle that doesn't involve any of that, but which is still quite unsettling. This puzzle—also due to Bostrom—is called the Presumptuous Philosophers. Imagine that physicists have narrowed the possibilities for a final theory of physics down to two a priori equally likely possibilities. The main difference between them is that Theory 1 predicts that the universe is a billion times bigger than Theory 2 does. In particular, assuming the universe is relatively homogenous (which both theories agree about), Theory 2 predicts that there are going to be about a billion times as many sentient observers in the universe. So the physicists are planning on building a massive particle accelerator to distinguish the two theories—a project that will cost many billions of dollars. Now, the philosophers come along and say that Theory 2 is the right one to within a billion-to-one confidence, since conditioned on Theory 2 being correct, we're a billion times more likely to exist in the first place. The question is whether the philosophers should get the Nobel Prize in Physics for their "discovery."

Of course, what the philosophers are assuming here is the Self-Indication Assumption. So here's where things stand with the SSA and SIA. The SSA leads to the Doomsday Argument, while the SIA leads to Presumptuous Philosophers. It seems like whichever one you believe, you get a bizarre consequence.

Finally, if we want to combine the anthropic computing idea with the Doomsday Argument, then there's the Adam and Eve puzzle. Suppose that Adam and Eve are the first two observers, and that what they'd really like is to solve an instance of an NP-complete problem, say 3SAT. To do so, they pick a random assignment, and form a very clear intention beforehand that if the assignment happens to be satisfying, then they won't have any kids, whereas if the assignment is not satisfying, then they will go forth and multiply. Now, let's assume the SSA. Then, conditioned on having chosen an unsatisfying assignment, how likely is it that they would be an Adam and Eve in the first place, as opposed to one of the vast number of future observers? If we assume that they'll ultimately have (say) 22n descendants, then the probability would seem to be at most 2-2n + 1. Therefore, conditioned upon the fact that they are the first two observers, the SSA predicts that with overwhelming probability, they will pick a satisfying assignment. If you're a hardcore Bayesian, you can take your pick between SSA and SIA and swallow the consequences either way!