the listening experience would have been better with another narrator
Perhaps, but I’m liking the narration so far. I find it about as good as your narration of your book, and even perhaps a bit better.
WilliamKiely
Karma: 876
At the beginning of Chapter 2, Mollick misattributes the idea of a “paperclip maximizer” to Bostrom. Yudkoswky is the actual originator of the idea. Source: https://www.lesswrong.com/tag/squiggle-maximizer-formerly-paperclip-maximizer
Very helpful reply, thank you!
(My salary has always been among the lowest in the organization, mostly as a costly signal to employees and donors that I am serious about doing this for impact reasons)
I appreciate that!
I have completely forfeited my salary, and donated ~$300k to Lightcone at the end of last year myself to keep us afloat
If you had known you were going to do this, couldn’t you have instead reduced your salary by ~60k/year for your first 5 years at Lightcone and avoided paying a large sum in income taxes to the government?
(I’m assuming that your after-tax salary from Lightcone from your first 5-6 years at Lightcone totaled more than ~$300k, and that you paid ~$50k-100k in income taxes on that marginal ~$350k-$400k of pre-tax salary from Lightcone.)
I’m curious if the answer is “roughly, yes” in which case it just seems unfortunately sad that that much money had to be unnecessarily wasted on income taxes.
I originally missed that the “Expected Income” of $2.55M from the budget means “Expected Income of Lighthaven” and consequently had the same misconception as Joel that donations mostly go towards subsidizing Lighthaven rather than almost entirely toward supporting the website in expectation.
Something I noticed:
“Probability that most humans die because of an AI takeover: 11%” should actually read as “Probability that most humans die [within 10 years of building powerful AI] because of an AI takeover: 11%” since it is defined as a sub-set of the 20% of scenarios in which “most humans die within 10 years of building powerful AI”.
This means that there is a scenario with unspecified probability taking up some of the remaining 11% of the 22% of AI takeover scenarios that corresponds to the “Probability that most humans die because of an AI takeover more than 10 years after building powerful AI”.
In other words, Paul’s P(most humans die because of an AI takeover | AI takeover) is not 11%/22%=50%, as a quick reading of his post or a quick look at my visualization seems to imply, but is actually undefined, and is actually >11%/22% = >50%.
For example, perhaps Paul thinks that there is a 3% chance that there is an AI takeover that causes most humans to die more than 10 years after powerful AI is developed. In this case, Paul’s P(most humans die because of an AI takeover | AI takeover) would be equal to (11%+3%)/22%=64%.I don’t know if Paul himself noticed this. But worth flagging this when revising these estimates later, or meta-updating on them.
What exactly is wrong? Could you explicitly show my mistake?
See my top-level comment.
I’m a halfer, but think you did your math wrong when calculating the thirder view.
The thirder view is that the probability of an event happening is the experimenter’s expectation of the proportion of awakenings where the event happened.
So for your setup, with k=2:
There are three possible outcomes: H, HT, and TT.
H happens in 50% of experiments, HT happens in 25% and TT happens in 25%.
When H happens there is 1 awakening, when HT happens there are 2 awakenings, and when TT happens there are 4 awakenings.
We’ll imagine that the experiment is run 4 times, and that H happened in 2 of them, HT happened once, and TT happened once. This results in 2*1=2 H awakenings, 1*2=2 HT awakenings, and 1*4=4 TT awakenings.
Therefore, H happens in 2/(2+2+4)=25% of awakenings, HT happens in 25% of awakenings, and TT happens in 50% of awakenings.
The thirder view is thus that upon awakening Beauty’s credence that the coin came up heads should be 25%.What is you [sic] credence that in this experiment the coin was tossed k times and the outcome of the k-th toss is Tails?
Answering your question, the thirder view is that there was a 6/8=75% chance the coin was tossed twice, and a 4⁄6 chance that the second toss was a tails conditional on it being the case that two tosses were made.
Unconditionally, the thirder’s credence is 4/8=50% chance that it is both true that the coin was tossed two times and that the second toss was a tails.
I have time-space synesthesia, so I actually picture some times as being literally farther away than others.
I visualize the months of the year in a disc slanted away from me, kind of like a clock with New Years being at 6pm, and visualize years on a number line.
I thought of the reason independently: it’s that if the number before 66 is not odd, but even instead, it must be either 2 or 4, since if it was 6 then the sequence would have had a double 6 one digit earlier.
150 or 151? I don’t have a strong intuition. I’m inclined to trust your 150, but my intuition says that maybe 151 is right because 100+99/2+almost1 rounds up to 151. Would have to think about it.
(By the way, I’m not very good at math. (Edit: Ok, fair. Poorly written. What I meant is that I have not obtained certain understandings of mathematical things that those with formal educations in math have widely come to understand, and this leads me to being lower skilled at solving certain math problems than those who have already understood certain math ideas, despite my possibly having equal or even superior natural propensity for understanding math ideas.). I know high school math plus I took differential equations and linear algebra while studying mechanical engineering. But I don’t remember any of it well and don’t do engineering now or use math in my work. (I do like forecasting as a hobby and think about statistics and probability in that context a lot.) I wouldn’t be able to follow your math in your post without a lot of effort, so I didn’t try.)
Re the almost1 and a confusion I noticed when writing my previous comment:
Re my:
E.g. For four 100s: Ctrl+f “100,100,100,100” in your mind. Half the time it will be proceeded by an odd number for length 4, a quarter of the time it will be length 5, etc.
Since 1/2+1/4+1/8...=1, the above would seem to suggest that for four 100s in a row (or two 6s in a row) the expected number of rolls conditional on all even is 5 (or 3). But I saw from your post that it was more like 2.72, not 3, so what is wrong with the suggestion?
My intuition was that B is bigger.
The justification was more or less the following: any time you roll until reaching two in a row, you will have also hit your second 6 at or before then. So regardless what the conditions are, must be larger than .
This seems obviously wrong. The conditions matter a lot. Without conditions that would be adequate to explain why it takes more rolls to get two 6s in a row than it does to get two 6s, but given the conditions that doesn’t explain anything.
The way I think about it is that you are looking at a very long string of digits 1-6 and (for A) selecting the sequences of digits that end with two 6s in a row going backwards until just before you hit an odd number (which is not very far, since half of rolls are odd). If you ctrl+f “66” in your mind you might see that it’s “36266″ for a length of 4, but probably not. Half of your “66”s will be proceeded by an odd number, making half of the two-6s-in-a-row sequences length 2.
For people that didn’t intuit that B is bigger, I wonder if you’d find it more intuitive if you imagine a D100 is used rather than a D6.
While two 100s in a row only happens once in 10,000 times, when they do happen they are almost always part of short sequences like “27,100,100” or “87,62,100,100″ rather than “53,100,14,100,100”.
On the other hand, when you ctrl+f for a single “100” in your mind and count backwards until you get another 100, you’ll almost always encounter an odd number first before encountering another “100“ and have to disregard the sequence. But occasionally the 100s will appear close together and by chance there won’t be any odd numbers between them. So you might see “9,100,82,62,100” or “13,44,100,82,100” or “99,100,28,100” or “69,12,100,100″.
Another way to make it more intuitive might be to imagine that you have to get several 100s in a row / several 100s rather than just two. E.g. For four 100s: Ctrl+f “100,100,100,100” in your mind. Half the time it will be proceeded by an odd number for length 4, a quarter of the time it will be length 5, etc. Now look for all of the times that four 100s appear without there being any odd numbers between them. Some of these will be “100,100,100,100″, but far more will be “100,32,100,100,88,100” and similar. And half the time there will be an odd number immediately before, a quarter of the time it will be odd-then-even before, etc.
EDIT: I did as asked, and replied without reading your comments on the EA forum. Reading that I think we are actually in complete agreement, although you actually know the proper terms for the things I gestured at.
Cool, thanks for reading my comments and letting me know your thoughts!
I actually just learned the term “aleatory uncertainty” from chatting with Claude 3.5 Sonnet (New) about my election forecasting in the last week or two post-election. (Turns out Claude was very good for helping me think through mistakes I made in forecasting and giving me useful ideas for how to be a better forecaster in the future.)
I then ask, knowing what you know now, what probability you should have given.
Sounds like you might have already predicted I’d say this (after reading my EA Forum comments), but to say it explicitly: What probability I should have given is different than the aleatoric probability. I think that by becoming informed and making a good judgment I could have reduced my epistemic uncertainty significantly, but I would have still had some. And the forecast that I should have made (or what market prices should have been is actually epistemic uncertainty + aleatoric uncertainty. And I think some people who were really informed could have gotten that to like ~65-90%, but due to lingering epistemic uncertainty could not have gotten it to >90% Trump (even if, as I believe, the aleatoric uncertainty was >90% (and probably >99%)).
Ah, I think I see. Would it be fair to rephrase your question as: if we “re-rolled the dice” a week before the election, how likely was Trump to win?
Yeah, that seems fair.
My answer is probably between 90% and 95%.
Seems reasonable to me. I wouldn’t be surprised if it was >99%, but I’m not highly confident of that. (I would say I’m ~90% confident that it’s >90%.)
That’s a different question than the one I meant. Let me clarify:
Basically I was asking you what you think the probability is that Trump would win the election (as of a week before the election, since I think that matters) now that you know how the election turned out.
An analogous question would be the following:
Suppose I have two unfair coins. One coin is biased to land on heads 90% of the time (call it H-coin) and the other is biased to land on tails 90% of the times (T-coin). These two coins look the same to you on the outside. I choose one of the coins, then ask you how likely it is that the coin I chose will land on heads. You don’t know whether the coin I’m holding is H-coin or T-coin, so you answer 50% (50%=0.5*.90=+0.5*0.10). I then flip the coin and it lands on heads. Now I ask you, knowing that the coin landed on heads, now how likely do you think it was that it would land on heads when I first tossed it? (I mean the same question by “Knowing how the election turned out, how likely do you think it was a week before the election that Trump would win?”).
(Spoilers: I’d be interested in knowing your answer to this question before you read my comment on your “The value of a vote in the 2024 presidential election” EA Forum post that you linked to to avoid getting biased by my answer/thoughts.)
That makes sense, thanks.
Knowing how the election turned out, how likely do you think it was a week before the election that Trump would win?
Do you think Polymarket had Trump-wins priced too high or too low?
Foreign-born Americans shifted toward Trump
Are you sure? Couldn’t it be that counties with a higher percentage of foreign-born Americans shifted toward Trump because of how the non-foreign-born voters in those counties voted rather than how the foreign-born voters voted?
The title “How I Learned To Stop Trusting Prediction Markets and Love the Arbitrage” isn’t appropriate for the content of the post. That there is a play-money prediction market in which it costs very little to make the prices on conditional questions very wrong does not provide significant reasons to trust prediction markets less. That this post got 193 karma leading me to see it 2 months later is a sign of bad-voting IMO. (There are many far better, more important posts that get far less karma.)
“[A] boy born on a day that I’ll tell you in 5 minutes” is ambiguous. There are two possible meanings, yielding different answers.
If “a boy born on a day that I’ll tell you in 5 minutes” means “a boy, and I’ll tell you the name of a boy I have in 5 minutes” then the answer is 1⁄3 as Liron says.
However, if “a boy born on a day that I’ll tell you in 5 minutes” means “a boy born on a particular singular day that I just wrote down on this piece of paper and will show you in 5 minutes”, then this is equivalent to saying “a boy born on a Tuesday” and the answer is 13⁄27.
The reason why the second meaning is equivalent to “a boy born on a Tuesday” is because it’s a statement that at least one of the children is a particular kind of boy that only 1/7th of boys are, just like how “a boy born on a Tuesday” is a statement that at least one of the children is a particular kind of boy that only 1/7th of boys are. (Conversely, for the first interpretation: “a boy born on a day that I’ll tell you in 5 minutes” is a statement that at least one of the children is a a boy, period.)
Another way to notice the difference if it’s still not clear:
When told “I have two children, at least one of whom is [a boy born on a particular singular day that I just wrote down on this piece of paper and will show you in 5 minutes]”, you assign a 1/7th credence to the paper showing Sunday, 1/7th to Monday, 1/7th to Tuesday, etc.
Then, conditional on the paper showing Tuesday, you know that the parent just told you “I have two children, at least one of whom is [a boy born on [Tuesday and I will show you the paper showing Tuesday in 5 minutes]]”, which is equivalent to the parent saying “I have two children, at least one of whom is a boy born on Tuesday”.
So you then have a 1/7th credence that the paper shows Tuesday, and if it’s Tuesday, your credence that both children are boys is 13⁄27. So your overall credence, reflecting your uncertainty about what day the paper shows is (13/27)*(1/7)+(13/27)*(1/7)+(13/27)*(1/7)+(13/27)*(1/7)+(13/27)*(1/7)+(13/27)*(1/7)+(13/27)*(1/7)=13/27.