justinpombrio

Karma: 655

justinpombrio 17 Jun 2017 2:42 UTC
0 points
on: Open thread, June. 12 - June. 18, 2017
I’d like to ask a question about the Sleeping Beauty problem for someone that thinks that ¹⁄₂ is an acceptable answer.

Suppose the coin isn’t flipped until after the interview on Monday, and Beauty is asked the probability that the coin has or will land heads. Does this change the problem, even though Beauty is woken up on Monday regardless? It seems to me to obviously be equivalent, but perhaps other people disagree?

If you accept that these problems are equivalent, then you know that P(Heads | Monday) = P(Tails | Monday) = ¹⁄₂, since if it’s Monday then a fair coin is about to be flipped. From this we can learn that P(Monday) = 2 * P(Heads), by the calculation below.

This is inconsistent with the halfer position, because if P(Heads) = ¹⁄₂, then P(Monday) = 2 * ¹⁄₂ = 1.

EDIT: The calculation is that P(Heads) = P(Monday) P(Heads | Monday) + P(Tuesday) P(Heads | Tuesday) = ¹⁄₂ P(Monday) + 0 P(Tuesday), so P(Monday) = 2 * P(Heads).

justinpombrio 17 Jun 2017 17:40 UTC
0 points
in reply to: Jiro’s comment on: Open thread, June. 12 - June. 18, 2017
I didn’t say it did. I said that P(Heads | Monday) = P(Tails | Monday) = ¹⁄₂, because it’s determined by a fair coin flip that’s yet to happen. This is in contrast to the standard halfer position, where P(Heads | Monday) > ¹⁄₂, and P(Tails | Monday) < ¹⁄₂. Everyone agrees that P(Heads | Monday) + P(Tails | Monday) = 1.

Or are you disagreeing with the calculation?

P(Heads) = P(Monday) P(Heads | Monday) + P(Tuesday) P(Heads | Tuesday) is just Baye’s theorem.

P(Heads | Tuesday) = 0, because if Beauty is awake on Tuesday then the coin must have landed tails.

P(Heads | Monday) = ¹⁄₂ by the initial reasoning.

Then P(Monday) = 2 * P(Heads) by a teeny amount of algebra.

justinpombrio 17 Jun 2017 23:32 UTC
0 points
in reply to: Jiro’s comment on: Open thread, June. 12 - June. 18, 2017
Thank you for walking me through this; I’m having a very hard time seeing the other perspective here.

I understand that P(Monday) is ambiguous. I meant to refer to “the probability that the current day, as Beauty is currently being interviewed, is Monday”. Regardless of Beauty’s perspective, she can ask weather the current day is Monday or Tuesday, and she does know that it is not currently both. And she can ask what the probability that the coin landed tails given that the current day is Tuesday, etc. Yes? Given that, I’m not seeing what part of my reasoning doesn’t work if you replace each instance of “Monday” with “IsCurrentlyMonday”.

justinpombrio 18 Jun 2017 3:12 UTC
0 points
in reply to: Jiro’s comment on: Open thread, June. 12 - June. 18, 2017
Sorry, you lost me completely. I didn’t prove that P(Heads | Monday) > ¹⁄₂ at all.

Could you say which step (1-6) is wrong, if I am Beauty, and I wake up, and I reason as follows?
1. The experiment is unchanged by delaying the coin flip until Monday evening.
2. If the current day is Monday, then the coin is equally likely to land heads or tails, because it is a fair coin that is about to be flipped. Thus P(Heads | CurrentlyMonday) = ¹⁄₂.
3. By Bayes’ theorem, which is applicable because it cannot currently be both Monday and Tuesday:
  
  P(Heads) = P(CurrentlyMonday) P(Heads | CurrentlyMonday) + P(CurrentlyTuesday) P(Heads | CurrentlyTuesday)
4. P(Heads | CurrentlyTuesday) = 0, because if it is Tuesday then the coin must have landed tails.
5. Thus P(CurrentlyMonday) = 2 * P(Heads) by some algebra.
6. It may not currently be Monday, thus P(CurrentlyMonday) != 1, thus P(Heads) < ¹⁄₂.

justinpombrio 19 Jun 2017 0:26 UTC
0 points
in reply to: entirelyuseless’s comment on: Open thread, June. 12 - June. 18, 2017

and in my opinion SSA is right, and SIA is wrong.

I’m curious: is this grounded on anything beyond your intuition in these cases?

SSI is grounded on frequency. In the Incubator situation, the SSI probabilities are:

P(1st cell) = ²⁄₃

P(2nd cell) = ¹⁄₃

P(H | 1st cell) = ¹⁄₂

P(H | 2nd cell) = 0

(FYI, I find this intuitive, and find SSA in this situation unintuitive.)

These agree with the actual frequencies, in terms of expected number of people in different circumstances, if you repeat this experiment. And frequencies seem very important to me, because if you’re a utilitarian that’s what you care about. If we consider torturing anyone in the first cell vs. torturing anyone in the second cell, the former is twice as bad in expectation (please tell me if you disagree, because I would find this very surprising).

So your probabilities aren’t grounded in frequency&utility. Is there something else they’re grounded in that you care about? Or do you choose them only because they feel intuitive?

justinpombrio 19 Jun 2017 0:59 UTC
0 points
in reply to: Jiro’s comment on: Open thread, June. 12 - June. 18, 2017

“CurrentlyMonday” as you have defined it is a per-awakening probability

The halfer position assumes that the probability that is meaningful is a per-experiment probability.

To be clear, you’re saying that, from a halfer position, “the probability that, when Beauty wakes up, it is currently Monday” is meaningless?

Neither of your links to the halfer position shows anyone claiming that.

Sorry, I wrote that without thinking much. I’ve seen that position, but it’s definitely not the standard halfer position. (It seems to be entirelyuseless’ position, if I’m not mistaken.)

The per-experiment probabilities you give make perfect sense to me: they’re the probabilities you have before you condition on the fact that you’re Beauty in an interview, and they’re the probabilities from which I derived the “per-awakening” probabilities myself (three indistinguishable scenarios: HM, TM, TT, each with probability ¹⁄₂; thus they’re all equally likely, though that’s not the most rigorous reasoning).

I’m confused why anyone would want not to condition on the fact that Beauty is awake when the problem states that she’s interviewed each time she wakes up. If instead, on Heads you let Beauty live and on Tails you kill her, then no one would have trouble saying that Beauty should say P(Heads) = 1 in an interview. Why is this different?

Thanks again for the discussion.

justinpombrio 19 Jun 2017 18:24 UTC
0 points
in reply to: Jiro’s comment on: Open thread, June. 12 - June. 18, 2017

It’s meaningless in the sense that it doesn’t have a meaning that matches what you’re trying to use it for. Not that it literally has no meaning.

What I’m trying to use it for is to compute P(Heads), from a halfer position, while carrying out my argument.

So in other words, P(per-experiment-heads | it-is-currently-Monday) is meaningless? And a halfer, who interpreted P(heads) to mean P(per-experiment-heads), would say that P(heads | it-is-currently-Monday) is meaningless?

justinpombrio 19 Jun 2017 20:15 UTC
0 points
in reply to: entirelyuseless’s comment on: Open thread, June. 12 - June. 18, 2017

This is a SIA idea, and it’s wrong.

Please don’t make statements like this unless you really understand the other person’s position (can you guess how I will respond?). For instance, notice that I haven’t ever said that the halfer position is wrong.

There’s nothing to condition on because there’s no new information

This is just a restatement of SSA. By SIA there is new information, since you’re more likely to be one of a larger set of people.

just as there’s no new information when you find that you exist

Sure there is! Flip a coin and kill Beauty on tails. Now ask her what the coin flip said: she learns from the fact that she’s alive that it landed heads.

I understand that SSA is a consistent position, and I understand that it matches your intuition if not mine. I’m curious how you’d respond to the question I asked above. It’s in the post with “So your probabilities aren’t grounded in frequency&utility.”

justinpombrio 22 Jun 2017 2:23 UTC
0 points
in reply to: entirelyuseless’s comment on: Open thread, June. 12 - June. 18, 2017
I’d recommend distinguishing between the probability that the coin landed heads (which happens exactly once), and the probability that, if you were to plan to peak you would see heads (which would happen on average 250,000 times).

justinpombrio 22 Jun 2017 2:50 UTC
0 points
in reply to: Jiro’s comment on: Open thread, June. 12 - June. 18, 2017
THANK YOU. I now see that there are two sides of the coin.

However, I feel like it’s actually Heads, and not P, that is ambiguous. There is the probability that the coin would land heads. The coin lands exactly once per experiment, and half the time it will land heads. If you count Beauty’s answer to the question “what is the probability that the coin landed heads” once per awakening, you’re sometimes double-counting her answer (on Tails). It’s dishonest to ask her twice about an event that only happened once.

On the other hand, there is the probability that if Beauty were to peek, she would see heads. If she decided to peek, then she would see the coin once or twice. Under SIA, she’s twice as likely to see tails. If you count Beauty’s answer to the question “what is the probability that the coin is currently showing heads” once per experiment, you’re sometimes ignoring her answer (on Tuesdays). It would be dishonest to only count one of her two answers to two distinct questions.

(Being more precise: suppose the coin lands tails, and you ask Beauty “What is the probability that the coin is currently showing heads?” on each day, but only count her answer on Monday. Well, you’ve asked her two distinct questions, because the meaning of “currently” changes between the two days, but only counted one of them. It’s dishonest.)

Thus, this question isn’t up for interpretation. The answer is ¹⁄₂, because the question (on Wikipedia, at least) asks about the probability that the coin landed heads. There are two interpretations—per experiment and per awakening—but the interpretation should be set by the question. Likewise, setting a bet doesn’t help settle which interpretation to use: either interpretation is perfectly capable of figuring out how to maximize expectation for any bet; it just might consider some bets to be rigged.

Although this is subtle, and maybe I’m still missing things. For one, why is Baye’s rule failing? I now know how to use it both to prove that P(Heads) < ¹⁄₂ and to prove that P(Heads) = ¹⁄₂, by marginalizing on either CurrentlyMonday/CurrentlyTuesday or on WillWakeUpOnTuesday/WontWakeUpOnTuesday. When you use
```
P(X) = P(X | A) * P(A) + P(X | B) * P(B)
```
you need that A and B are mutually exclusive. But this seems to be suggesting that there’s some other subtle requirement as well that somehow depends on what X is.

It could be, as you say, that P is different. But P should only depend on your knowledge and priors. All the priors are fixed here (it’s a fair coin, use SIA), so what are the two sets of knowledge?

justinpombrio 25 Jun 2017 4:20 UTC
0 points
in reply to: Jiro’s comment on: Open thread, June. 12 - June. 18, 2017

In other words, when you say “P(per-experiment event)” the “per-experiment” is really describing the P, not just the event.

My understanding is that P depends only on your knowledge and priors. If so, what is the knowledge that differs between per-experiment and per-awakening? Or am I wrong about that?

That doesn’t help. “Coin landed heads” can still be used to describe either a per-experiment or per-awakening situation:

Ok, yes, agreed.

justinpombrio 1 Jul 2017 22:14 UTC
0 points
in reply to: pengvado’s comment on: The Absent-Minded Driver
And then you can correct for the double-counting. When would you like to count your chickens? It’s safe to count them at X or Y.

If you count them at X, then how much payoff do you expect at the end? Relative to when you’ll be counting your payoff, the relative likelihood that you are at X is 1. And the expected payoff if you are at X is p^2 + 4p(1-p). This gives a total expected payoff of P(X) E(X) = 1 (p^2 + 4p(1-p)) = p^2 + 4p(1-p).

If you count them at Y, then you much payoff do you expect at the end? Relative to when you’ll be counting your payoff, the relative likelihood that you are at Y is p. And the expected payoff if you are at Y is p + 4(1-p). This gives a total expected payoff of P(Y) E(Y) = p (p + 4(1-p)) = p^2 + 4(1-p).

I’m annoyed that English requires a tense on all verbs. “You are” above should be tenseness.

EDIT: formatting

justinpombrio 26 Jul 2017 22:31 UTC
4 points
on: Type Theory quick question
What is your goal? Type theory is at the intersection of programming languages and logic. If you care about programming languages and type systems, read TAPL:

https://www.cis.upenn.edu/~bcpierce/tapl/

If you care about type theory purely as a logic, I don’t have an obvious recommendation, but could point you at some material.

(Programming Languages researcher)

justinpombrio 27 Jul 2017 22:14 UTC
0 points
in reply to: Faustus2’s comment on: Type Theory quick question
Ah, then you’ll want to read about the [https://hal.inria.fr/inria-00076024/document](Calculus of Constructions):
Yeah, TAPL is the book on type systems. I’m not aware of competition.

justinpombrio 7 Jan 2018 7:18 UTC
3 points
on: Demon Threads
Where did the term “demon thread” come from? A lot more people are going to be familiar with the term “flame war”. (A search engine will confirm it is a common phrase, and that it is essentially synonymous with what you are calling “demon thread”.). If you want to distinguish one from the other, you should have a good reason to do so, and you should tell the reader. Otherwise I’d just say “flame war”.
EDIT: Maybe “malignant demon thread” = “flame war”?
https://www.urbandictionary.com/define.php?term=flame%20war

justinpombrio 28 Mar 2018 4:25 UTC
6 points
on: Defining the ways human values are messy
Scott A. wrote an article about axiology, morality, and law that distinguishes between some of these concepts. For instance, after reading it I now see a qualitative distinction between morality and preference. (Which isn’t to say that no one ever uses those words in ambiguous ways; rather that I see two clearly distinct concepts that largely agree with the way the words are typically used.)

justinpombrio 13 May 2018 21:22 UTC
6 points
on: Fun With DAGs

This is called a utility function. This utility function U is defined by U(chicken) = 1, U(pork) = 2, U(steak) = 3. We prefer things with higher utility — in other words, we want to maximize the utility function!

Just because you have a DAG of your preferences, does not mean that you have a utility function. There are many utility functions that are consistent with any particular DAG, so if all you have is a DAG then your utility function is under-specified.

First, maybe you don’t care about the difference between pork and steak, and remove that arrow from your DAG. In that case, the toposort is going to produce either U(chicken) = 1, U(pork) = 2, U(steak) = 3 or U(chicken) = 1, U(steak) = 2, U(pork) = 3, arbitrarily. Toposort is guaranteed to produce an ordering that matches the DAG, but there might be several of them. (A slight variant on toposort could produce U(chicken) = 1, U(pork) = 2, U(steak) = 2 instead, but that’s still a bit questionable because it’s conflating incomparability with equality.)

Second, the magnitude of these numbers is pretty arbitrary in a toposort (they’re just sequential integers), but very important in a utility function. For example, maybe you really like steak, so that the more appropriate utility function is U(chicken) = 1, U(pork) = 2, U(steak) = 10. You’re not going to get this from a toposort.

I just wanted to clarify this, because the post could be read as saying DAG==utility-function.

justinpombrio 13 May 2018 22:14 UTC
2 points
in reply to: johnswentworth’s comment on: Fun With DAGs
And if we assume that the nodes are whole worlds, rather than pieces of worlds.

For example, if I’m also ordering a soda, and prefer Pepsi to Coke, then the relative magnitudes become important. (There’s an implicit assumption here that the utility of the whole is the sum of the utilities of the parts.) However, if the node includes the entire meal, so that there are six nodes (chicken, pepsi), (chicken, coke), (pork, pepsi), (pork, coke), (steak, pepsi), (steak, coke), then the magnitude doesn’t matter. Are utility functions generally assumed to be whole-world like this?

justinpombrio 16 Jun 2018 5:05 UTC
15 points
on: The Curious Prisoner Puzzle
What, exactly, would the guard would say in different situations? Using the standard, utterly unrealistic, interpretation of probability problems like this, the guard is supposed to say this:
```
VM: If you are on Vulcan, you are in the Mountain
VD: I have nothing to tell you.
EM: If you are on Vulcan, you are in the Mountain
ED: If you are on Vulcan, you are in the Mountain
```
in which case the probability is ¹⁄₃. But I have a hard time believing that the guard is willing to talk to you here, but wouldn’t be willing to talk if you were in the Vulcan Desert.
Since the guard refused to talk at first, but then told you something later, it seems pretty clear that they’re trying to help you out. The most obvious way for them to communicate to you where you are is like this:
```
VM: If you are on Vulcan, you are in the Mountain
VD: If you are on Vulcan, you are in the Desert
EM: If you are on Earth, you are in the Mountain
ED: If you are on Earth, you are in the Desert
```
But there are other possibilities. Perhaps there’s a policy of executing guards that reveal information about where you are, so the guard wants plausible deniability by lying to you:
```
VM: If you are on Vulcan, you are in the Desert
VD: If you are on Vulcan, you are in the Mountain
EM: If you are on Earth, you are in the Desert
ED: If you are on Earth, you are in the Mountain
```
It seems that you’ve ruled that out in the problem statement, though.
Altogether, as Dacyn says, “it depends on what you know about the psychology of the guard.”
Somewhere in Rationality, there’s a post about this.

justinpombrio 12 Sep 2018 0:15 UTC
3 points
on: Track-Back Meditation
I like this idea. I think I’ll try it. It’s similar to what I’ve been doing while meditating recently: every time I notice I’m distracted, I’ll say the current thought that was distracting me, and count up by 1.
In more detail, for the past 10 days, I’ve spent an hour or so meditating while walking. I’ll let my object of focus vary pretty freely between my footsteps or sounds. When I notice that I’ve been distracted, I’ll say the immediate cause (without tracing back) and count up by 1. I’ll count up to around 50 distractions in an hour or so (this is with practice: on the first day I don’t think I got past 30 before things became… indistinct?). This has been fairly effective, to the point that today some of my distractions were short enough that I realized their original cause without even thinking about it. I’m interested to see what happens when I try to trace back my distractions.
Something like half of my distractions are thoughts about how awesome I’m going to be once I’m enlightened or whatever. It’s ironic that this may be the main thing holding back my meditation practice right now.
I find walking much more effective than sitting. If I try to meditate while sitting, I’ll inevitably get groggy. This isn’t unique to meditating—I’ll get groggy while sitting for a lecture too—I just generally focus better while standing. (Unless I’m coding, in which case sitting is fine? I’m not sure why it’s different.)
Question for more experienced meditators: is it ok that I let my object of focus vary? Should I be more concerned about having a consistent object of focus?
Advice for less experienced meditators: “focus on the object of meditation” does not mean what I intuitively thought it did. When trying to do that, I try to focus on that to the exclusion of all else, and it goes poorly. Instead, it means what I would have described as “be present and aware of things in general, and also be aware of the object of meditation in particular”. That is, focus on it, but not to the exclusion of other sensations. I learned this from The Mind Illuminated.