In math, "measure" is a way of assigning a "volume" (or length, area, probability) to infinite sets. The "cardinality" of the set of numbers between 0 and 1 is the same infinity as that of the set of numbers between 0 and 2, but the standard "measure" of the latter set is twice as big. You can sum up certain types of functions defined on a continuum by "integrating" those functions values using the appropriate measure.

If there turns out to be a continuum of possible universes created by, say, a particle decay, then there's also a natural physical measure that corresponds to the probabilities we observe; the set of universes in which the particle decays before 1 half-life would be "twice as big" in some sense as the set of universes created in which the decay occurs between one half-life and two half-lifes. If someone offers to do something for you if-and-only-if a particle decays before one half life elapses, you should figure out the expected utility of a 50-50 bet, even if the reality might be that your decision is affecting two different infinities of subsequent universes.

There's a lot I'm glossing over and/or don't understand myself here (why is the probability measure the only ethical measure? lots of different-but-self-consistent measures can always be mathematically well-defined) but hopefully that at least explains the vocabulary a bit.

it depends on if the altruist believes in eugenics

I don't think the answer of a question about the morality of actions depends of the beliefs of the people involved (you can probably construct edge cases where it does); the answer to "Is it okay for bob to rape babies?" doesn't depend on bob's beliefs about baby-rape.

The issue with an index fund that based on something like the SAP 500 is that the SAP 500 changes over time.

If a company loses their SAP 500 all the index funds that are based on the SAP 500 dump their stocks on the market. On average that's not going to be a good trade. The same goes for the trade of buying the companies that just made it into the SAP 500. On average you are going to lose some money to hedge funds or investment banks who take the other side on those trades.

In general you can expect that if you invest money into the stockmarket big powerful banks have some way to screw you. But they won't take all your money and index funds are still a good choice if you don't want to invest too much time thinking about investing.

(I am in the midst of reading the EY-RH "FOOM" debate, so some of the following may be less informed than would be ideal.)

From a purely technical standpoint, one problem is that if you permit self-modification, and give the baby AI enough insight into its own structure to make self-modification remotely a useful thing to do (as opposed to making baby repeatedly crash, burn, and restore from backup), then you cannot guarantee that utility() won't be modified in arbitrary ways. Even if you store the actual code implementing utility() in ROM, baby could self-modify to replace all references to that fixed function with references to a different (modifiable) one.

What you need is for utility() to be some kind of fixed point in utility-function space under whatever modification regime is permitted, or... something. This problem seems nigh-insoluble to me, at the moment. Even if you solve the theoretical problem of preserving those aspects of utility() that ensure Friendliness, a cosmic-ray hit might change a specific bit of memory and turn baby into a monster. (Though I suppose you could arrange, mathematically, for that particular possibility to be astronomically unlikely.)

Could someone explain the reasoning behind answer A being the correct choice in Question 4? My analysis was to assume that, since 30 migraines a year is still pretty terrible (for the same reason that the difference in utility between 0 and 1 migraines per year is larger than the difference between 10 and 11), I should treat the question as asking "Which option offers more migraines avoided per unit money?"

Option A: $350 / 70 migraines avoided = $5 per migraine avoided
Option B: $100 / 50 migraines avoided = $2 per migraine avoided

And when I did the numbers in my head I thought it was obvious that the answer should be B. What exactly am I missing that led the upper tiers of LWers to select option A?

Explain what the point of the wiki is. Right now it seems like a bunch of links to LW posts with short descriptions, which is neither worthwhile nor interesting to edit.

I agree that George definitely does know more information overall, since he can concentrate his probability mass more sharply over the 4 hypotheses being considered, but I'm fairly certain you're wrong when you say that Sarah's distribution is 0.33-0.33-0-0.33. I worked out the math (which I hope I did right or I'll be quite embarassed), and I get 0.25-0.25-0-0.5.

Good point. I was treating the description of Sarah's encounter with the man as a proxy for "Sarah knows one of the man's children is a boy, but not which one." That seems to be the way it's usually intended when the problem is presented, but you're right that in the problem as described, Sarah has an additional relevant piece of information -- that the man is out with a boy. I think this is an unintended artifact of the way the problem is presented, though. The people presenting the problem are usually trying to get at something different. The usual intent of the puzzle is captured by "Sarah knows that one of Brian's two children is a boy, and George knows that his eldest child is a boy. What are the probabilities according to Sarah and George that Brian's other child is a boy?".

I think your analysis in terms of required message lengths is arguably wrong, because the purpose of the question is to establish the genders of the children and not the order in which they were born. That is, the answer to the question "What gender is the child at home?" can always be communicated in a single bit, and we don't care whether they were born first or second for the purposes of the puzzle.

Again, I think this is an unintended artifact of the way the puzzle is stated. The fact that Sarah sees one of the kids and doesn't see the other one gives her a way of individuating the kids other than their birth order. If we don't assume she has this method of individuation (as in the restated puzzle above) then the birth order is relevant.

Thanks. I see why the probability of H1|o and H2|o need to be taken as 25% each. In that case, it seems like Sarah can say that it is 50% likely a boy and 50% likely a girl (at home). Why is the answer to the question then given as 66%?

EDIT: As wgd points out below, my answer here is wrong in its particulars (I didn't take into account all of the information available to Sarah and George in the puzzle as stated). The general principles invoked are sound, though.

George does actually know more. I think you're getting thrown by the fact that his 50-50 probability distribution seems more equivocal (less concentrated) than Sarah's 66-33 distribution. But remember that these distributions are defined over a space of four elements (boy-boy, boy-girl, girl-boy and girl-girl), so the actual distributions are 0.5-0.5-0-0 for George and 0.33-0.33-0.33-0 for Sarah. When you see it this way, it becomes a bit more plausible that Sarah's distribution is actually more "spread out", more equivocal.

To be more precise, suppose you have been given the task of conveying information about the genders of this man's children. You decide that you will transmit a 0 to represent a boy and a 1 to represent a girl. If the receiver has absolutely no information about the man's children, apart from the fact that there are two of them and neither is genderqueer, you will need to send two bits of information -- one for the elder child's gender, and one for the younger one's -- in order to convey full information about the genders. On the other hand, if the receiver already knows the children's genders, you have to send zero bits of information in order to convey full information. So you can think of the number of bits you need to transmit as a measure of the lack of knowledge of the receiver. The fewer bits you need to send, the more the receiver knows.

Now let's compare George and Sarah. George already knows the elder child's gender, so you only need to send one further bit, representing the younger child's gender, in order to convey full information. Sarah's case is trickier. She knows that one of the children is a boy, but she doesn't know which one. If it turns out that both children are boys, then your task is easy: you need to send just one bit of information, a 0, representing the gender of the child she hasn't seen. Once she gets this bit, Sarah will know the genders of both children. But if the other child is a girl, you can't just say that. You will also need to tell Sarah whether the order of birth is boy-girl or girl-boy. So besides sending a 1 to represent a girl, you'll need to send one more bit of information in order to distinguish between boy-girl and girl-boy. This means that the number of bits you will have to send Sarah is either 1 or 2, depending on whether the other child is a boy or a girl. If you did this experiment over and over again, with a bunch of different groups of siblings, the average number of bits you send Sarah will be greater than 1 but less than 2.

So with George you only need 1 bit to convey full information, while with Sarah you need (on average) more than 1 bit. This means Sarah does indeed know less about the situation, and there is no paradox. All of this can be made a lot more rigorous using the concept of Shannon entropy, if you're interested.

I'll just note in passing that this puzzle is discussed in this post, so you may find it or the associated comments helpful.

I think the specific issue is that in the first case, you're assuming that each of the three possible orderings yields the same chance of your observation (the son out walking with him is a boy). If you assume that his choice of which child to go walking with is random, then the fact that you see a boy makes the (girl, boy) possibilities each less likely, so together they are equally likely to the (boy, boy) one.

Let's define (imagining, for the sake of simplicity, that Omega descended from the heavens and informed you that the man you are about to meet has two children who can both be classified into ordinary gender categories):

h1 = "Boy then Girl"
h2 = "Girl then Boy"
h3 = "Girl then Girl"
h4 = "Boy then Boy"
o = "The man is out walking with a boy child"

Our initial estimates for each should be 25% before we see any evidence. Then if we make the aforementioned assumption that the man doesn't like one child more than the other:

P(o | h1) = 0.5
P(o | h2) = 0.5
P(o | h3) = 0.0
P(o | h4) = 1.0

And then we can apply bayes theorem to figure out the posterior probability of each hypothesis:

P(h1 | o) = P(h1) * P(o | h1) / P(o)
P(h2 | o) = P(h2) * P(o | h2) / P(o)
P(h3 | o) = P(h3) * P(o | h3) / P(o)
P(h4 | o) = P(h4) * P(o | h4) / P(o)
(where P(o) = P(o | h1)*P(h1) + P(o | h2)*P(h2) + P(o | h3)*P(h3) + P(o | h4)*P(h4))

The denominator is a constant factor which works out to 0.5 (meaning "before making that observation I would have assigned it 50% probability"), and overall the math works out to:

P(h1 | o) = P(h1) * P(o | h1) / 0.5 = 0.25
P(h2 | o) = P(h2) * P(o | h2) / 0.5 = 0.25
P(h3 | o) = P(h3) * P(o | h3) / 0.5 = 0.0
P(h4 | o) = P(h4) * P(o | h4) / 0.5 = 0.5

So the result in the former case is the same as in the latter, seeing one child offers you no information about the gender of the other (unless you assume that the man hates his daughter and never goes walking with her, in which case you get the original 1/3 chance of it being a boy).

The lesson to take away here is the same lesson as the usual bayesian vs frequentist debate, writ very small: if you're getting different answers from the two approaches, it's because the frequentist solution is slipping in unstated assumptions which the bayesian approach forces you to state outright.