Comment author: Vladimir_M 20 July 2010 05:49:56PM *  5 points [-]

Are there actually any materials on Earth that are so rare and precious (and perhaps in danger of running out in the foreseeable future) that it would make sense to mine them from space?

By the way, the claim about aluminum sounds highly implausible to me. Aluminum accounts for about 8% of the Earth's crust by weight, and even if most of it is difficult to access, I would expect that more than the amount present on Eros would be extractable with methods much easier than any conceivable sort of asteroid mining.

Comment author: Soki 20 July 2010 06:22:48PM 1 point [-]

Helium-3 could be mined from the moon. It would be a good fusion fuel, but it is rare on earth so it makes sense to get it from space.

Comment author: multifoliaterose 19 July 2010 04:26:31PM 4 points [-]

Okay, fine: I currently believe that funding SIAI and FHI has expected value near zero but my belief on this matter is unstable and subject to rapid change with incoming evidence.

Comment author: Soki 19 July 2010 05:10:37PM *  1 point [-]

This video addresses this question : Anna Salamon's 2nd Talk at Singularity Summit 2009 -- How Much it Matters to Know What Matters: A Back of the Envelope Calculation
It is 15 minutes long, but you can take a look at 11m37s

Edit : added the name of the video, thanks for the remark Vladimir.

Comment author: JanetK 15 July 2010 09:27:30AM 6 points [-]

Your idea that the subjects are not taking the question seriously is a good one.

I had a discussion with someone about a very similar real life 'Linda'. It was finally resolved by realizing that the other person didn't think of 'and' and 'or' as defined terms that always differed and was quite put out that I thought he should know that. To put it in 'Linda' terms: he know that Linda was a feminist and doubted that she was a teller. This being the case the 'and' should be thought of as an 'or' and b was more likely than a. Why would anyone think differently? It kind of blew my mind that I was being accused of being sloppy or illogical by using the fixed defined meaning for 'and' and 'or'. I have since that time noticed that people actually often have this vagueness about logical terms.

Comment author: Soki 17 July 2010 05:31:36PM *  0 points [-]

I would not say that this person replaced "and" by "or".
I guess they considered the statement "Lisa is a bank teller and a feminist" to be "50%" true if Lisa turns out to be a feminist but not a bank teller.

The formula used would be something like P(AB)=1/2*(P(A)+P(B))

Comment author: Douglas_Knight 17 July 2010 12:30:26AM 2 points [-]

Shouldn't it be for exchanging the number of Drawn and Good balls? (In the wikipedia example, black is good and white is defective.) If the number of bad balls is 1 and the number drawn is 2, then the probability of getting 2 good balls when drawing 2 is high, but if you exchange drawn and bad, then the probability of getting 2 good balls becomes zero.

Drawn and Good are symmetric. I recommend a Venn diagram. There are two binary conditions: good vs bad and drawn vs left, both of which are effectively random. We are interested in the intersection, which is preserved by the symmetry.

Comment author: Soki 17 July 2010 04:19:40PM *  0 points [-]

What you said is true : you exchange the number of Drawn and Counted marbles.

However, the counted balls are white on wikipedia. (they are called defective indeed, so on wikipedia, people count the number of bad balls)

There was also a mistake in the part about symmetries, I replaced :
"Swapping the roles of black and drawn marbles" by :
"Swapping the roles of white and drawn marbles"
since m is the number of white marbles

Comment author: AlexM 12 July 2010 01:15:24PM 1 point [-]

Cryonics wants to be small, or why should the future want you?

All this technical discussion misses what I see as the major problem of cryonics if it works as advertised - why should the future want us?

Imagine if today were discovered few frozen Homo habilis and had technology to revive them. After, they would spend their lives in comfortable zoo that is paradise by ape men standards ( plentiful food! no dangerous beasts! warm shelter!)

Now try the same scenario, but with few millions of our frozen ancestors. The results will be same - at best, few dozens would be picked to be resurrected and studied, but I cannot see us welcoming millions of new hairy citizens.

Conclusion - to me it seems that if you want to maximize chance of future society resurrecting you, keep cryonics as close guarded secret of tiny elite...

Comment author: Soki 12 July 2010 02:42:11PM *  1 point [-]

why should the future want us?

Someone who knew you may want to bring you back.
If it takes centuries, then the more people frozen the better since it will be more likely that someone you knew would be brought back by someone else. And then he may bring you back too.
This assumes that the government does not prevent people form doing this.

Comment author: Soki 08 July 2010 04:18:39PM 2 points [-]

If you care about cryonics and its sustainability during an economic collapse or worse, chemical fixation might be a good alternative. http://en.wikipedia.org/wiki/Chemical_brain_preservation

The main advantage is that it requires no cooling and is cheap. People might be normally buried after the procedure, so it would seem less weird.
However, a good perfusion of the brain with the fixative is hard to achieve.

Chemical fixation could also be combined with those low maintenance cryonic graves just in case the nitrogen boils off.

Comment author: utilitymonster 03 July 2010 05:28:47PM *  8 points [-]

Here's a puzzle I've been trying to figure out. It involves observation selection effects and agreeing to disagree. It is related to a paper I am writing, so help would be appreciated. The puzzle is also interesting in itself.

Charlie tosses a fair coin to determine how to stock a pond. If heads, it gets 3/4 big fish and 1/4 small fish. If tails, the other way around. After Charlie does this, he calls Al into his office. He tells him, "Infinitely many scientists are curious about the proportion of fish in this pond. They are all good Bayesians with the same prior. They are going to randomly sample 100 fish (with replacement) each and record how many of them are big and how many are small. Since so many will sample the pond, we can be sure that for any n between 0 and 100, some scientist will observe that n of his 100 fish were big. I'm going to take the first one that sees 25 big and team him up with you, so you can compare notes." (I don't think it matters much whether infinitely many scientists do this or just 3^^^3.)

Okay. So Al goes and does his sample. He pulls out 75 big fish and becomes nearly certain that 3/4 of the fish are big. Afterwards, a guy named Bob comes to him and tells him he was sent by Charlie. Bob says he randomly sampled 100 fish, 25 of which were big. They exchange ALL of their information.

Question: How confident should each of them be that 3/4 of the fish are big?

Natural answer: Charlie should remain nearly certain that ¾ of the fish are big. He knew in advance that someone like Bob was certain to talk to him regardless of what proportion of fish were big. So he shouldn't be the least bit impressed after talking to Bob.

But what about Bob? What should he think? At first glance, you might think he should be 50/50, since 50% of the fish he knows about have been big and his access to Al's observations wasn't subject to a selection effect. But that can't be right, because then he would just be agreeing to disagree with Al! (This would be especially puzzling, since they have ALL the same information, having shared everything.) So maybe Bob should just agree with Al: he should be nearly certain that ¾ of the fish are big.

But that's a bit odd. It isn't terribly clear why Bob should discount all of his observations, since they don't seem to subject to any observation selection effect; at least from his perspective, his observations were a genuine random sample.

Things get weirder if we consider a variant of the case.

VARIANT: as before, but Charlie has a similar conversation with Bob. Only this time, he tells him he's going to introduce Bob to someone who observed exactly 75 of 100 fish to be big.

New Question: Now what should Bob and Al think?

Here, things get really weird. By the reasoning that led to the Natural Answer above, Al should be nearly certain that ¾ are big and Bob should be nearly certain that ¼ are big. But that can't be right. They would just be agreeing to disagree! (Which would be especially puzzling, since they have ALL the same information.) The idea that they should favor one hypothesis in particular is also disconcerting, given the symmetry of the case. Should they both be 50/50?

Here's where I'd especially appreciate enlightenment: 1.If Bob should defer to Al in the original case, why? Can someone walk me through the calculations that lead to this?

2.If Bob should not defer to Al in the original case, is that because Al should change his mind? If so, what is wrong with the reasoning in the Natural Answer? If not, how can they agree to disagree?

3.If Bob should defer to Al in the original case, why not in the symmetrical variant?

4.What credence should they have in the symmetrical variant?

5.Can anyone refer me to some info on observation selection effects that could be applied here?

Comment author: Soki 03 July 2010 09:07:30PM *  0 points [-]

First off all, I think that if Al does not see a sample, it makes the problem a bit simpler. That is, Al just tells Bob that he (Bob) is the first person that saw 25 big fishes.

I think that the number N of scientists matters, because the probability that someone will come to see Al depends on that.

Lets call B then event the lake has 75% big fishes, S the opposite and C the event someone comes, which means that someone saw 25 fishes.

Once Al sees Bob, he updates :
P(B/C)=P(B)* P(C/B)/(1/2*P(C/B)+1/2*P(C/S)).
When N tends toward infinity, both P(C/B) and P(C/S) tend toward 1, and P(B/SC) tends to 1/2.
But for small values of N, P(C/B) can be very small while P(C/S) will be quite close to 1.
Then the fact that someone was chosen lowers the probability of having a lake with big fishes.

If N=infinity, then the probability of being chosen is 0, and I cannot use Bayes' theorem.

If Charlie keeps inviting scientists until one sees 25 big fishes, then it becomes complicated, because the probability that you are invited is greater if the lake has more big fishes. It may be a bit like the sleeping beauty or the absent-minded driver problem.

Edited for formatting and misspellings

Comment author: taiyo 01 July 2010 05:44:01PM 1 point [-]

K. S. Van Horn gives a few lines describing the derivation in his PT:TLoS errata. I don't understand why he does step 4 there -- it seems to me to be irrelevant. The two main facts which are needed are step 2-3 and step 5, the sum of a geometric series and the Taylor series expansion around y = S(x). Hopefully that is a good hint.

Nitpicking with his errata, 1/(1-z) = 1 + z + O(z^2) for all z is wrong since the interval of convergence for the RHS is (-1,1). This is not important to the problem since the z here will be z = exp(-q) which is less than 1 since q is positive.

Comment author: Soki 01 July 2010 09:54:39PM 1 point [-]

It is not very important, but since you mentioned it :

The interval of convergence of the Taylor series of 1/(1-z) at z=0 is indeed (-1,1).

But "1/(1-z) = 1 + z + O(z^2) for all z" does not make sense to me.

1/(1-z) = 1 + z + O(z^2) means that there is an M such as |1/(1-z) - (1 + z)| is no greater that M*z^2 for every z close enough to 0. It is about the behavior of 1/(1-z) - (1 + z) when z tends toward 0, not when z belongs to (-1,1).

Comment author: taiyo 29 June 2010 08:56:07AM *  1 point [-]

Yeah. A total derivative. The way I think about it is the dv thing there (jargon: a differential 1-form) eats a tangent vector in the y-z plane. It spits out the rate of change of the function in the direction of the vector (scaled appropriately with the magnitude of the vector). It does this by looking at the rate of change in the y-direction (the dy stuff) and in the z-direction (the dz stuff) and adding those together (since after taking derivatives, things get nice and linear).

I'm not too familiar with the functional equation business either. I'm currently trying to figure out what the heck is happening on the bottom half of page 32. Figuring out the top half took me a really long while (esp. 2.50).

I'm convinced that the inequality in eqn 2.52 shouldn't be there. In particular, when you stick in the solution S(x) = 1 - x, it's false. I can't figure out if anything below it depends on that because I don't understand much below it.

Comment author: Soki 01 July 2010 03:23:58PM 2 points [-]

I could not figure out why alpha > 0 neither and it seems wrong to me too. But this does not look like a problem.

We know that J is an increasing function because of 2-49. So in 2-53, alpha and log(x/S(x)) must have the same sign, since the remaining of the right member tends toward 0 when q tends toward + infinity.

Then b is positive and I think it is all that matters.

However, if alpha = 0, b is not defined. But if alpha=0 then log(x/S(x))=0 as a consequence of 2-53, so x/S(x)=1. There is only one x that gives us this since S is strictly decreasing. And by continuity we can still get 2-56.

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