Instead, start with a prior chosen somehow, then update it according to the evidence that a being such as you exists in the universe.

This seems very similar to Radford Neal’s full non-indexical conditioning:

I will here consider what happens if you ignore such indexical information, conditioning only on the fact that someone in the universe with your memories exists. I refer to this procedure as “Full Non-indexical Conditioning” (FNC).

Note however that Neal says this idea implies that the thirder answer is correct in Sleeping Beauty:

In this regard, note that the even though the experiences of Beauty upon wakening on Monday and upon wakening on Tuesday (if she is woken then) are identical in all “relevant” respects, they will not be subjectively indistinguishable. On Monday, a fly on the wall may crawl upwards; on Tuesday, it may crawl downwards. Beauty’s physiological state (heart rate, blood glucose level, etc.) will not be identical, and will affect her thoughts at least slightly. Treating these and other differences as random, the probability of Beauty having at some time the exact memories and experiences she has after being woken this time is twice as great if the coin lands Tails than if the coin lands Heads, since with Tails there are two chances for these experiences to occur rather than only one. This computation assumes that the chance on any given day of Beauty experiencing exactly what she finds herself experiencing is extremely small, as will be the case in any realistic version of the experiment.

Assuming your idea is the same as FNC, I think from a decision theory perspective it's still worse than just not updating. See this comment [LW(p) · GW(p)] of mine under a previous post about FNC.

Thus Boltzmann brains can mess things up from a probability theory standpoint, but we should ignore them from a decision theory standpoint.

Is that true? Imagine you have this choice:

1) Spend the next hour lifting weights

2) Spend the next hour eating chocolate

Lifting weights pays off later, but eating chocolate pays off right away. If you believe there's a high chance that, conditional on surviving the next hour, you'll dissolve into Boltzmann foam immediately after that - why not eat the chocolate?

I'm not a fan of Shut up and Multiply where it means taking a maths equation and then just applying it without stopping to think about the assumptions that it is based upon and whether it is appropriate for this context. You can certainly catch errors by writing up a formal proof, but we need to figure out whether our formalisation is appropriate first.

Indeed, Said Achmiz was able to obtain a different answer [LW(p) · GW(p)] by formalising the problem differently. So the key question ends up being what is the appropriate formalisation. As I explain on my comment [LW(p) · GW(p)], the question is whether p(survival) is

a) the probability that a pre-war person will survive until after the cold war and observe that he didn't die, if they will survive apart from a nuclear holocaust (following Stuart Armstrong)

b) the probability that a post-war person will observe that they survived (following Said Achmiz)

Stuart Armstrong is correct because probability problems usually implicitly assume that the agent knows the problem, so a post-war person is already assumed to know that they survived. In other words, b) involves asking someone who already knows that they survived to update on the fact that they survived again. Of course they aren't going to update!

Concrete example

Anyway, it'll be easier to understand what is happening here if we make it more concrete. On a gameshow, if a coin comes up heads, the contestants face a dangerous challenge that only a 1/3 survive, otherwise they face a safe(r) challenge that 1/2 survive. We will assume there are two lots of 6 people and that those who are "eliminated" aren't actually killed, but just fail to make it to the next round.

This leads us to expect that one group faces the dangerous challenge and one the safe challenge. So overall we expect: Survivors (3 safe, 2 dangerous), Eliminated (2 safe, 4 unsafe). This leads to the following results:

• If we survey everyone: The survivors have a higher ratio of people from the safe challenge than those who were eliminated
• If we only survey survivors: A disproportional number of survivors come from the safer world

So regardless of which group we consider relevant, we get the result that we claimed above. I'll consider the complaint that "dead people don't get asked the question". If we are asked a conditional probability question like, "What is the chance of scoring at least 10 with two dice if the first dice is a 4?", then the systematic way to answer that is to list all the (36) possibilities and eliminate all the possibilities that don't have a 4 for the first dice. Applying this to "If we survived the cold war, what is the probability..." we see that we should begin by eliminating all people who don't survive the cold war from the set of possibilities. Since we've already eliminated the people who die, it doesn't matter that we can't ask them questions. How could it? We don't even want to ask them questions! The only time we need to handle this is when the if statement is correlated with our ability to be asking the question, but doesn't guarantee it. An example would be if a few people end up in a coma; then we might want to update separately on our ability to be asking the question.

Boltzman Brains

I think you've engaged in something of a dodge here. Yes, all of our predictions would be screwed if we are a Boltzman Brain; so if we want to get physics correct we have to hope that that this isn't the case. However, your version of anthropics require us to hope much harder. In other theories, we just have to hope that the calculations indicating that Boltzman Brains are the most likely scenario are wrong. However, in your anthropics, if the probability of at least one Boltzman Brain with our sensations approaches 1, then we can't update on our state at all. This holds even if we know that the vast majority of beings with that state aren't Boltzmann brains. That makes the problem much, much worse than it is under other theories.

Clarifying this with the Tuesday Problem

I think you've made the same mistake that I've identified here [LW · GW]:

A man has two sons. What is the chance that both of them are born on the same day if at least one of them is born on a Tuesday?

Most people expect the answer to be 1/7, but the usual answer is that 13/49 possibilities have at least one born on a Tuesday and 1/49 has both born on Tuesday, so the chance in 1/13. Notice that if we had been told, for example, that one of them was born on a Wednesday we would have updated to 1/13 as well.

The point is that there is a difference between the following:

a) meeting a random son and noting that he was born on Tuesday

b) discovering that one of the two sons (you don't know which) was born on a Tuesday

Similarly there is a difference between:

a) Discovering a random consciousness is experiencing a stream of events

b) Discovering that at least one consciousness is experiencing that stream of events

The only reason why this gives the correct answer for the first problem is that (in the simplification) we assume all consciousnesses before the war either survive or all of them die. This makes a) and b) coincide, so that it doesn't matter which one is used.

… But for the question at the beginning of the post, we aren’t asking about the probability of survival (conditional on other factors), but the probability of the cold war being safe (conditional on survival). And that’s something very different:

• P(cold war safe | survival) = P(cold war safe)*P(survival | cold war safe)/​P(survival).

Now, P(survival | cold war safe) is greater that P(survival) by definition—that’s what “safe” means—hence P(cold war safe | survival) is greater than P(cold war safe). Thus survival is positive evidence for the cold war being safe.

Note that this doesn’t mean that the cold war was actually safe—it just means that the likelihood of it being safe is increased when we notice we survived.

No. Here is the correct formalization:

• S = we observe that we have survived
• Ws = Cold War safe
• Wd = Cold War dangerous

We want P(Ws|S)—probability that the Cold War is safe, given that we observe that we have survived:

P(Ws|S) = P(Ws) × P(S|Ws) / P(S)

Note that P(S) = 1, because the other possibility—P(we observe that we haven’t survived)—is impossible. (Is it actually 1 minus epsilon? Let’s say it is; that doesn’t materially change the reasoning.)

This means that this—

… P(survival | cold war safe) [i.e., P(S|Ws). —SA] is greater that P(survival) [P(S)]** by definition …

… is false. P(S|Ws) is 1, and P(S) is 1. (Cogito ergo sum and so on.)

So observing that we have survived is not positive evidence for the Cold War being safe.

Sorry for the stupid question, but what is P(x) in the Fermi Paradox section. It's a prior given x (the probability of life appearing on a well-situated terrestrial-like planet around a long-lived star). But the prior of what?

It feels like we are leaning extremely hard on the reasonable priors condition. I am deeply confused by the relationship between anthropics and priors; it feels like as soon as we start considering anthropics it should race away into the past, because it's not like we are suddenly alive now and weren't before when we got our priors. This makes me suspect that updating on the fact that I am alive is some kind of updating on old evidence.

I don't think you're correct.

P(cold war safe | survival) = P(cold war safe)*P(survival | cold war safe)/P(survival). [...]

That's it. No extra complications, or worries about what to do with multiple copies of yourself (dealing with multiple copies comes under the purview of decision theory, rather than probability).

Copies seem to me to matter. In a quantum universe where you share consciousness with all of your copies, you have P(survival|$\bullet$) = 1, and therefore $P\left(\text{cws | s}\right)=P\left(\text{s | cws}\right)\cdot \frac{P\left(\text{cws}\right)}{P\left(\text{s}\right)}=1\cdot \frac{P\left(\text{cws}\right)}{1}=P\left(\text{cws}\right)$

You might not have P(survival | $\bullet$) = 1 for your copy's survival, but you have it for the survival of any of your copies, and that's the probability which matters, because that's the observation you make.

Imagine a dumbbell shaped space-time, two hyper-spheres, each the size of the observable universe connected by a thin wormhole. While that wormhole remains open, both pieces are part of the same universe, when the wormhole collapses,the universe is suddenly smaller, and the probability we should assign to a planet producing intelligent life grows. If we knew the wormhole would collapse next week, we would break conservation of expected evidence.

I think that the fact that "I exist" presents zero evidence for anything, as - most likely - all possible observers exist. However, if I observer some random variable, like distance to the Sun from the center of the galaxy, it should be distributed randomly, and thus I am most likely found it the middle of its interval. And, not surprisingly, the Sun is approximately in the middle between the center of the Galaxy and its edges (ignoring here for simplicity the difference of star density and impossibility to live in the galactic center).

The same way my birthday is non-surpassingly located somewhere in the middle of the year.

The only random variable which is not obey this rule is my position in supposedly very long future civilisation: I found my date of birth surprisingly early. And exactly this surprise is the nature of the Doomsday argument: the contradiction between expectation that we will live very long as civilisation and my early position.

Speaking about the Cold war survival, the fact of the survival provide zero evidence about nuclear war extinction probability. But, as Bostrom showed in his article about surviving space catastrophes, if Cold war survival is unlikely, a random observer is more likely to find herself earlier in time, maybe in 1950s.

The argument at the start just seems to move the anthropics problem one step back - how do we know whether we "survived"* the cold war?

*Not sure how to succinctly state this better; I mean if Omega told me that the True Probability of surviving the Cold War was 1%, I would update on the safety of the Cold War in a different direction than if it told me 99%, even though both entail me, personally, surviving the Cold War.