bigjeff5

Karma: 774

bigjeff5 17 Oct 2023 20:32 UTC
3 points
2
in reply to: lc’s comment on: Arguments for radical optimism on AI Alignment
This is exactly right. To put it more succinctly: Memory corruption is a known vector for exploitation, therefore any bug that potentially leads to memory corruption also has the potential to be a security vulnerability. Thus memory corruption should be treated with similar care as a security vulnerability.

bigjeff5 17 Oct 2023 20:27 UTC
7 points
0
in reply to: Zach Furman’s comment on: Arguments for radical optimism on AI Alignment
The reason Person A in scenario 2 has the intuition that Person B is very wrong is because there are dozens, if not hundreds of examples where people claimed no vulnerabilities and were proven wrong. Usually spectacularly so, and often nearly immediately. Consider the fact that the most robust software developed by the most wealthy and highly motivated companies in the world, who employ vast teams of talented software engineers, have monthly patch schedules to fix their constant stream vulnerabilities, and I think it’s pretty easy to immediately discount anybody’s claim of software perfection without requiring any further evidence.
All the evidence Person A needs is the complete and utter lack of anybody having achieved such a thing in the history of software to discount Person B’s claims.
I’ve never heard of an equivalent example for AI. It just seems to me like Scenario 2 doesn’t apply, or at least it cannot apply at this point in time. Maybe in 50 years we’ll have the vast swath of utter failures to point to, and thus a valid intuition against someone’s 9-9′s confidence of success, but we don’t have that now. Otherwise people would be pointing out examples in these arguments instead of vague unease regarding problem spaces.

bigjeff5 21 Jan 2015 4:50 UTC
3 points
0
in reply to: skeptical_lurker’s comment on: Markets are Anti-Inductive
That was four years ago, but I’m pretty sure I was using hyperbole. Pros don’t bluff often, and when they do they are only expecting to break even, but I doubt it’s as low as 2% (the bluff will fail half the time).

I’d also put in a caveat that the best hand wins among hands that make it all the way to the river. There are plenty of times where a horrible hand like a 6 2, which is an instant fold if you respect the skills of your fellow players, ends up hitting a straight by the river and being the best hand but obviously didn’t win. Certainly more often than 1%, and there are plenty of better hands that you still almost always fold pre-flop that are going to hit more often.

So, at best it was poorly stated (i.e. hyperbole without saying so), at worst it’s just wrong.

bigjeff5 10 Jul 2014 17:16 UTC
1 point
0
in reply to: Lumifer’s comment on: Rationality Quotes July 2014
Pretty much.

bigjeff5 10 Jul 2014 16:41 UTC
2 points
0
in reply to: mwengler’s comment on: Rationality Quotes July 2014
At this point you have to ask what you mean by “theory” and “learning”.

The original method of learning was “those that did it right didn’t die”—i.e. natural selection. Those that didn’t die have a pattern of behavior (thanks to a random mutation) that didn’t exist in previous generations, which makes them more successful gene spreaders, which passes that information on to future generations.

There is nothing in there that requires one to ask any questions at all. However, considering that there is information gained based on past experience, I think the definition of learning could be stretched to cover it. Obviously there is no individual learning, but there is definitely species learning going on there.

Since the vast majority of creatures that use this method of learning as their primary method of learning don’t even have brains, it seems obvious that there is no theory there. However, if we stretch the definition of theory to include any pattern of information that attempts to reflect reality (regardless of how well it does that job), well then even the lowliest bacteria have theories about how their world is supposed to work, and act accordingly.

That same broader definition of “theory” would cover wedrifid’s theoryless algorithms as well, as all you care about are patterns of information attempting to reflect reality, and they certainly have those.

All that said, the point of the quote is that in order for you as an individual to learn, then you as an individual must have an underlying theory of how things are supposed to be that can be challenged when faced with reality, in order to learn.

I have no idea if it’s actually true, I’m no psychologist or human learning expert or anything even remotely related, but it sounds like it has to be true even under the strict sense. It seems like it’s practically a tautology to me. Even wedfrid’s algorithms have a starting framework that attempts to reflect reality, however simplistic it may be. The algorithm itself is the theory there; it didn’t come from nothing.

bigjeff5 22 Dec 2013 23:29 UTC
0 points
0
in reply to: trist’s comment on: Bayes’ Theorem Illustrated (My Way)
I see that now, it took a LOT for me to get it for some reason.

bigjeff5 22 Dec 2013 23:29 UTC
1 point
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
Wow.

I’ve seen that same explanation at least five times and it didn’t click until just now. You can’t distinguish between the two on tuesday, so you can only count it once for the pair.

Which means the article I said was wrong was absolutely right, and if you were told that, say one boy was born on January 17th, the chances of both being born on the same day are 1-(364/365)^2 (ignoring leap years), which gives a final probability of roughly 49.46% that both are boys.

Thanks for your patience!

ETA: I also think I see where I’m going wrong with the terminology—sampling vs not sampling, but I’m not 100% there yet.

bigjeff5 22 Dec 2013 22:52 UTC
−1 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
How can that be? There is a ¹⁄₇ chance that one of the two is born on Tuesday, and there is a ¹⁄₇ chance that the other is born on Tuesday. ¹⁄₇ + ¹⁄₇ is ²⁄₇.

There is also a ¹⁄₄₉ chance that both are born on tuesday, but how does that subtract from the other two numbers? It doesn’t change the probability that either of them are born on Tuesday, and both of those probabilities add.

bigjeff5 22 Dec 2013 21:54 UTC
0 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)

This statement leads me to believe you are still confused. Do you agree that if I know a family has two kids, I knock on the door and a boy answers and says “I was born on a Tuesday,” that the probability of the second kid being a girl is 1/2? And in this case, Tuesday is irrelevant? (This the wikipedia called “sampling”)

I agree with this.

Do you agree that if, instead, the parents give you the information “one of my two kids is a boy born on a Tuesday”, that this is a different sort of information, information about the set of their children, and not about a specific child?

I agree with this if they said something along the lines of “One and only one of them was born on Tuesday”. If not, I don’t see how the Boy(tu)/Boy(tu) configuration has the same probability as the others, because it’s twice as likely as the other two configurations that that is the configuration they are talking about when they say “One was born on Tuesday”.

Here’s my breakdown with 1000 families, to try to make it clear what I mean:

1000 Families with two children, 750 have boys.

Of the 750, 500 have one boy and one girl. Of these 500, ¹⁄₇, or roughly 71 have a boy born on Tuesday.

Of the 750, 250 have two boys. Of these 250, ²⁄₇, or roughly 71 have a boy born on Tuesday.

71 = 71, so it’s equally likely that there are two boys as there are a boy and a girl.

Having two boys doubles the probability that one boy was born on Tuesday compared to having just one boy.

And I don’t think I’m confused about the sampling, because I didn’t use the sampling reasoning to get my result*, but I’m not super confident about that so if I am just keep giving me numbers and hopefully it will click.

*I mean in the previous post, not specifically this post.

bigjeff5 22 Dec 2013 18:54 UTC
0 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
The answer I’m supporting is based on flat priors, not sampling. I’m saying there are two possible Boy/Boy combinations, not one, and therefore it takes up half the probability space, not ¹⁄₃.

Sampling to the “Boy on Tuesday” problem gives roughly 48% (as per the original article), not 50%.

We are simply told that the man has a boy who was born on tuesday. We aren’t told how he chose that boy, whether he’s older or younger, etc. Therefore we have four possibilites, like I outlined above.

Is my analysis that the possibilities are Boy (Tu) /Girl, Girl / Boy (Tu), Boy (Tu)/Boy, Boy/Boy (Tu) correct?

If so, is not the probability for some combination of Boy/Boy 1/2? If not, why not? I don’t see it.

BTW, contrary to my previous posts, having the information about the boy born on Tuesday is critical because it allows us (and in fact requires us) to distinguish between the two boys.

That was in fact the point of the original article, which I now disagree with significantly less. In fact, I agree with the major premise that the tuesday information pushes the odds of Boy/Boy closer 50%, I just disagree that you can’t reason that it pushes it to exactly 50%.

bigjeff5 22 Dec 2013 18:11 UTC
0 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
For the record, I’m sure this is frustrating as all getout for you, but this whole argument has really clarified things for me, even though I still think I’m right about which question we are answering.

Many of my arguments in previous posts are wrong (or at least incomplete and a bit naive), and it didn’t click until the last post or two.

Like I said, I still think I’m right, but not because my prior analysis was any good. The ¹⁄₃ case was a major hole in my reasoning. I’m happily waiting to see if you’re going to destroy my latest analysis, but I think it is pretty solid.

bigjeff5 22 Dec 2013 17:48 UTC
0 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
Yes, and we are dealing with the second question here.

Is that not what I said before?

We don’t have 1000 families with two children, from which we’ve selected all families that have at least one boy (which gives ¹⁄₃ probability). We have one family with two children. Then we are told one of the children is a boy, and given zero other information. The probability that the second is a boy is ¹⁄₂, so the probability that both are boys is ¹⁄₂.

The possible options for the “Boy born on Tuesday” are not Boy/Girl, Girl/Boy, Boy/Boy. That would be the case in the selection of 1000 families above.

The possible options are Boy (Tu) / Girl, Girl / Boy (Tu), Boy (Tu) / Boy, Boy / Boy (Tu).

There are two Boy/Boy combinations, not one. You don’t have enough information to throw one of them out.

This is NOT a case of sampling.

bigjeff5 22 Dec 2013 17:34 UTC
1 point
0
in reply to: nshepperd’s comment on: Bayes’ Theorem Illustrated (My Way)
Yeah, probably the biggest thing I don’t like about this particular question is that the answer depends entirely upon unstated assumptions, but at the same time it clearly illustrates how important it is to be specific.

bigjeff5 22 Dec 2013 17:23 UTC
−2 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
The relevant quote from the Wiki:

The paradox arises because the second assumption is somewhat artificial, and when describing the problem in an actual setting things get a bit sticky. Just how do we know that “at least” one is a boy? One description of the problem states that we look into a window, see only one child and it is a boy. This sounds like the same assumption. However, this one is equivalent to “sampling” the distribution (i.e. removing one child from the urn, ascertaining that it is a boy, then replacing). Let’s call the statement “the sample is a boy” proposition “b”. Now we have: P(BB|b) = P(b|BB) P(BB) / P(b) = 1 ¹⁄₄ / ¹⁄₂ = ¹⁄₂. The difference here is the P(b), which is just the probability of drawing a boy from all possible cases (i.e. without the “at least”), which is clearly 0.5. The Bayesian analysis generalizes easily to the case in which we relax the ⁵⁰⁄₅₀ population assumption. If we have no information about the populations then we assume a “flat prior”, i.e. P(GG) = P(BB) = P(G.B) = ¹⁄₃. In this case the “at least” assumption produces the result P(BB|B) = ¹⁄₂, and the sampling assumption produces P(BB|b) = ²⁄₃, a result also derivable from the Rule of Succession.

We have no general population information here. We have one man with at least one boy.

bigjeff5 22 Dec 2013 17:17 UTC
0 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
Re-read it.

bigjeff5 22 Dec 2013 17:08 UTC
−1 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

I know it’s not the be all end all, but it’s generally reliable on these types of questions, and it gives P = ¹⁄₂, so I’m not the one disagreeing with the standard result here.

Do the math yourself, it’s pretty clear.

Edit: Reading closer, I should say that both answers are right, and the probability can be either ¹⁄₂ or ¹⁄₃ depending on your assumptions. However, the problem as stated falls best to me in the ¹⁄₂ set of assumptions. You are told one child is a boy and given no other information, so the only probability left for the second child is a 50% chance for boy.

bigjeff5 22 Dec 2013 17:04 UTC
0 points
0
in reply to: Jiro’s comment on: Bayes’ Theorem Illustrated (My Way)
How is it different? In both cases I have two independent coin flips that have absolutely no relation to each other. How does knowing which of the two came up heads make any difference at all for the probability of the other coin?

If it was the first coin that came up heads, TT and TH are off the table and only HH and HT are possible. If the second coin came up heads then HT and TT would be off the table and only TH and HH are possible.

The total probability mass of some combination of T and H (either HT or TH) starts at 50% for both flips combined. Once you know one of them is heads, that probability mass for the whole problem is cut in half, because one of your flips is now 100% heads and 0% tails. It doesn’t matter that you don’t know which is which, one flip doesn’t have any influence on the probability of the other. Since you already have one heads at 100%, the entire probability of the remainder of the problem rests on the second coin, which is a ⁵⁰⁄₅₀ split between heads and tails. If heads, HH is true. If tails, HT is true (or TH, but you don’t get both of them!).

Tell me how knowing one of the coins is heads changes the probability of the second flip from 50% to 33%. It’s a fair coin, it stays 50%.

bigjeff5 22 Dec 2013 16:50 UTC
0 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
No, it’s the exact same question, only the labels are different.

The probability that any one child is boy is 50%. We have been told that one child is a boy, which only leaves two options—HH and HT. If TH were still available, then so would TT be available because the next flip could be revealed to be tails.

Here’s the probability in bayesian:

P(BoyBoy) = 0.25 P(Boy) = 0.5 P(Boy|BoyBoy) = 1

P(BoyBoy|Boy) = P(Boy|BoyBoy)*P(BoyBoy)/P(Boy)

P(BoyBoy|Boy)= (1*0.25) / 0.5 = 0.25 / 0.5 = 0.5

P(BoyBoy|Boy) = 0.5

It’s exactly the same as the coin flip, because the probability is 50% - the same as a coin flip. This isn’t the monty hall problem. Knowing half the problem (that there’s at least one boy) doesn’t change the probability of the other boy, it just changes what our possibilities are.

bigjeff5 22 Dec 2013 10:12 UTC
0 points
0
in reply to: EHeller’s comment on: Bayes’ Theorem Illustrated (My Way)
Lets add a time delay to hopefully finally illustrate the point that one coin toss does not inform the other coin toss.

I have two coins. I flip the first one, and it comes up heads. Now I flip the second coin. What are the odds it will come up heads?

bigjeff5 22 Dec 2013 6:06 UTC
0 points
0
in reply to: Jiro’s comment on: Bayes’ Theorem Illustrated (My Way)
The only relevant information is that one of the children is a boy. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl. Since you already know that one of the children is a boy, the posterior probability that they are both boys is 50%.

Rephrase it this way:

I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?

Now, to see why Tuesday is irrelevant, I’ll re-state it thusly:

I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?

The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born. There is a ¹⁄₇ chance that child 1 was born on each day of the week, and there is a ¹⁄₇ chance that child 2 was born on each day of the week. There is a ¹⁄₄₉ chance that both children will be born on any given day (1/7*1/7), for a ⁷⁄₄₉ or ¹⁄₇ chance that both children will be born on the same day. That’s your missing ¹⁄₇ chance that gets removed inappropriately from the Tuesday/Tuesday scenario.