Did you plan to elaborate? Otherwise this looks a lot like an applause light.
Vanilla_cabs
Karma: 728
I don’t think that’s what happened. I think evolution put us on the path to understanding the world around us. We became better at it that any other living being. Then at some point (recently from our point of view), culture became the driving force behind our progress. Without culture, maybe we’d have evolved to be better at avoiding logical fallacies, like you say, or maybe not. But evolution is too slow compared to culture, so from our historical point of view, our genome is stopped halfway towards rationality and if we want to get better we have to learn it individually. That doesn’t mean being halfway rational is an evolutionary optimum.
I want to establish a distinction.
There are ways that a complicated theory can be simplified while keeping all its value. The shift from Roman numerals to Indo-Arabic numerals is all benefits, it’s why kids nowadays can easily learn arithmetics that back in the time took a lifetime to master. Likewise, converting probabilities to odds when doing some bayesian calculations is a really neat trick with only upsides.
These tricks are rare, and require a deep understanding of the matter. Apart from these, any simplification is a shortcut that loses some of the message. The problem here is that at the center of rationality is precisely the act of catching your brain doing some shallow work and trying to go deeper instead. As such, there is a contradiction in trying to convey rationality through a flyer. The message (thinking is a serious matter) is at odds with the medium (here is the essence of dozens of articles of reflection in only 8 pages made of one-liners). I suspect that the medium would win.
But rationality is not one terminal value among love, harmony and many other. It is instrumental to reaching almost any value. If your terminal value is love or harmony, you still have to make tricky decisions on a regular basis, and without an accurate map of reality you risk taking the wrong decisions.
Truth is not some feel-good notion, it’s the thing that you ignore at your own risks.
Just to be sure: in your original formulation, the bet counted only once for the two wake-up events in the Tails case, and it was canceled if she didn’t accept it on Monday and Tuesday.
That is not the scheme I described in previous comment. Each wake-up event is independent. She is awoken then asked to pay the fee now and gets the rewards now. If she pays twice she gets the rewards twice. It’s a bet that can be done, it is no less valid than yours.
In this bet (when asked during a wake-up event), her gain expectation is $18. Accepting for $19 will lose money on average. Her decision-making process on the other day is irrelevant. Test it if you don’t believe it.
Let’s be sure we agree on the parameters of the bet. If SB agrees to bet, she pays a fee, then:
she gains $30 if the coin came up heads.
she gains $12 if the coin came up tails.
And the question is: How much should she be willing to pay to bet? (assuming she’s ok to break even)
Answer: depends on when you ask SB.
If you propose the bet to SB on Sunday or Wednesday, she should be willing to pay $21.
If you propose the bet during a wake-up event, she should be willing to pay only $18. That’s the meaning of the $18. The difference comes from the fact that when she wakes up, the coin has only 1⁄3 chance of having landed heads. That’s where the thirder position comes from. Test it if you don’t believe it checks out.
Ok, well it seems that most of our disagreement has disappeared, which should allow us to focus on what remains.
That’s part and parcel to the whole Halfer position that in waking, Beauty hasn’t learned anything, and without learning anything, there’s nothing to update.
The weirdness in the thirder answer is not exactly there. If on Sunday, before the whole sleeping/amnesia thing, you asked SB what’s her credence that the coin came Head, she’d say 1⁄2. But if you asked her what would be her credence that the coin came Head when she would be woken up later, she’d answer 1⁄3. She doesn’t have to update when she wakes up, because she already has a different answer for the two questions.
And I agree that this only moves the problem: why is her answer different for the two questions? That’s where I’m still confused, but I can only say: the math checks up.
Say a crazed Riddler invades the experiment and happens upon Beauty. He hadn’t expected to find her, but he’s immediately captivated by her allure, and decides to make her the target of his games. He demands of Beauty: “You must correctly tell me the day of the week, or I will blow up Gotham!” Some fans of Gotham’s criminal underworld might know that the Riddler is usually more active on Mondays versus Tuesdays, but Beauty doesn’t know anything about him. His appearance gives her no additional nudge towards one day or another. She should reason that there was a 1⁄2 chance of Heads, therefore it’s Monday, and a 1⁄2 chance of Tails, therefore it could equally be Monday or Tuesday, and thus she should answer “Monday” with a confidence of 3⁄4.
I don’t see the difference between this version and my 1) version where the Riddler picks a day at random, and nothing happens if he picks Tuesday and the coin came Heads.
Let’s say this version happens over 4 parallel worlds, 2 where the Riddler picks Monday and 2 where he picks Tuesday. We know we’re not in the one world where he picked Tuesday and the coin came Heads (Tuesday ^ Heads), so 3 worlds remain: Monday ^ Heads, Monday ^ Tails, Tuesday ^ Tails. 2 out of these 3 worlds are Monday worlds.
Ok, I read it, and I still don’t understand what you call the day 1 Objection, or the argument that you’re trying to make. What’s most puzzling is how you came to use 3⁄4 for the probability that it’s Monday. Which you attribute to thirders if I understood correctly? (Edit after finishing writing: now I think that your belief? I’m even more confused about what you argue and what you assume thirders argue). Anyway, I’ll just list what I disagree with.
I’m fine with most of A “two-faced” example, except here:You check your name tag; what was the hench-name they’d given you? Ah, right: “Ralph-wrecker”. A quick check with Claude reveals that exactly 3⁄4 of hench-names are alphabetically ahead of yours.
You reason: The chance of getting paired up with someone like Alan and seeing them get Safe was 3⁄4, meaning P(obs) = 3⁄4. On a Heads flip, both you and the other recruit would always be Safe, meaning P(obs | H) = 1. And that means your original calculation was correct:
P(H | obs) = P(obs | H)*P(H)/P(obs) = (1)*(1/2)/(3/4) = 2⁄3
No, P(obs | H) = 3⁄4. In other words, obs (the other applicant has a name before yours and gets a Safe uniform) and H are independent.
And that gives: P(H | obs) = P(obs | H)*P(H)/P(obs) = (3/4)*(1/2)/(3/4) = 1⁄2 which is the result you initially came up with and is correct.
I think you know that the 2⁄3 reasoning is incorrect, since you disprove it correctly right in the next paragraph. But you seem to think that the mistake has a different cause (you talk about double counting...). I don’t know about that, I just know that you don’t need any argument to disprove it because the math is wrong, so you’re probably attacking an argument that no one makes.Next paragraph is The day 1 objection.
Let’s also say that the room always has a calendar Beauty can check to learn the current day.
Well then (like I said many times at this point), the whole experiment is useless, the result is trivial. I notice that you seem to be ignoring that point (that your versions of SB problem that have 1⁄2 as a solution work exactly the same without the whole sleep and amnesia thing) every time I make it. The whole point of the sleep and amnesia is that SB doesn’t know which day it is. Of course, if she sees Monday on her calendar she has a credence of 1⁄2 for Heads. If she saw Tuesday, it would be 0! (not factorial 0, just 0)
The “Day 1” objection argues that this set of statements is impossible:
P(obs Mon) = 3⁄4
P(obs Mon | H) = 1
P(H | obs Mon) = P(obs Mon | H)*P(H)/(P(obs Mon) = (1)*(1/2)/(3/4) = 2⁄3
Again, I don’t understand this objection, but I disagree with P(obs Mon) = 3⁄4 (also, no idea where it comes from), so I’m pretty sure at least I am not arguing this.
If Beauty were to be asked on every waking, “What’s the chance today is Monday?” and she wanted to minimize the margin of error of her responses, then she would answer 3⁄4. This is because 3⁄4 is the proportion of time she would spend awake in Monday during many repetitions of the experiment.
Uh? The proportion of time she would spend awake on Monday during many repetitions of the experiment is 2⁄3. She should answer 2⁄3. After finishing writing this reply, I think there’s a chance that the crux of our disagreement is here.
But here’s the difference: In these scenarios, when we describe P(obs Mon) for an event “today is Monday”, the “today” refers to something specific.
In the first example, “today” refers to the average waking across repeated experiments.
In the second and third examples, “today” becomes identified as “the random day that the Riddler chose to invade”.
Yeah, SB’s answer really depends on what she believes the Riddler used as an heuristic for invading the experiment. In particular:
1) If she’s confident that the Riddler picked Monday or Tuesday at random, then was ready to cancel the invasion if SB was sleeping (in the case of Heads and Tuesday), then there’s a 2⁄3 chance that it’s Monday, and a 1⁄3 that the coin came Heads.
2) If she’s confident that the Riddler checked the result of the coin toss, then picked Monday in case of Heads and picked Monday or Tuesday with equal probability in case of Tails, then there’s a 3⁄4 chance that it’s Monday, and a 1⁄2 chance that the coin came Heads.
Anyway, sorry for not really engaging with the core points of your blog post, but I just don’t understand them. I hope this reply at least brings some insight for you.
This is the piece I don’t understand about thirders (apologies if that sounds like aggressive wording): If you have a credence for the coin, why can’t you use it?
No problem, I am often in that situation too of being puzzled by the other side of an argument :)
And, it’s not like I’m confident that I fully understand the SB problem.
I’m guessing that the crux is this: for me, if (let’s call it experiment 1) you interrogate SB once after the coin landed tails, she should say her credence (of heads) is 1⁄2. But if (experiment 2) you interrogate her twice (under amnesia), she should say her credence is 1⁄3. You seem to disagree with that, on the grounds that her answer means something else than what the “SB problem” is about? But if you run the experiments many times, in experiment 1 half of SB’s answers will happen after a heads outcome, while in experiment 2 it’s only a third, so it means something.
I don’t think it’s accurate to say that SB made $12 on Mon and also say that SB made $12 on Tues. So in other words, I think this scheme is doing double counting. Which also means to me that the $18 value doesn’t really correspond to anything relevant and useful to the scenario?
The $18 value corresponds to how much money SB expects to make during this iteration on average.
I’m curious about your answer to the following experiment:No coin. SB is put to sleep on Sunday, awoken on Monday, interrogated, and put to sleep+amnesia. Then the same happens on Tuesday: awoken, interrogated, and put to sleep+amnesia.
When interrogated (on Monday and Tuesday), they tell her: “Here is $10, how much do you think you will have gained at the end of the experiment on Wednesday?”
What should be her answer?
If you answer $20, do you see how the same reasoning is part of my $18 answer on the previous problem?
I went to check Adam Elga’s paper. It’s very short, only 5 pages written with a big font. He does ask
When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?
But he doesn’t mean that you (the person experimented upon) know that it’s your first awakening. Rather, it’s a way to single out Monday’s awakening for the sake of the reader. It’s obvious later in the paper:
If (upon first awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either T1 or T2.
So the 2 versions of the problem on wikipedia are consistent. Still, if I was confused by the formulation of the first, I bet I’m not the only one.
I took a look at your table, and I have to say, it was a great variant. It took me a while to de-confuse my thoughts about it, and made me realize I wasn’t as comfortable with the SB problem as I thought.
Initially I was a bit uncertain what you meant by “round”, whether it was a wake-up event of a full iteration of the experiment. After rereading a few times, I became more certain that it was the latter.
So, you gain $30 if the coin lands heads and $6 + $6 = $12 if it lands tails. So your expectation of gain per iteration of the experiment is 1⁄2 * $30 + 1⁄2 * $12 = $21
There are 3 equally likely wake-up events, so the expectation of gain per wake-up event is 1⁄3 * $30 + 1⁄3 * $6 + 1⁄3 * $6 = $14
We can notice that there are on average 1.5 wake-up events per iteration, and $14 * 1.5 = $21, so that checks out.
Now, let’s say I experience a wake-up event. I believe that there is 1⁄3 chance that the coin landed heads. You think that my calculation for the expected gain per iteration will be 1⁄3 * $30 + 2⁄3 * $6 = $14
However, I reason: if the coin landed tails, I won’t gain $6, I will gain $6 * 2, once for Monday and once for Tuesday. So my expected gain per iteration is 1⁄3 * $30 + 2⁄3 * $6 * 2 = $18
That’s the point where I was confused. What did that $18 mean? What actually helped me figure it out was the betting scheme that you added in your post.
Sleeping Beauty is put to sleep like normal with only the usual information supplied to her about the original SB problem. When she awakes, there is a surprise message left for her:
You may agree to play the following game: If the coin had flipped Heads, your bank account will receive $30 right now. If the coin had flipped Tails, your bank account will receive $6 right now. However, once this round of experiment is over and you awake, you will pay $18 for having played this game. (If you are inconsistent between wakings about whether you agree to play the game, the game is considered defunct and all bank transfers will be reverted.)
This betting scheme is a based on a version of the SB problem where the Monday guess and the Tuesday guess are counted as a single one. So, Tuesday is unnecessary, so (see previous comment) it boils down to the trivial prediction of the result of a fair coin toss. $21 - $18 = $3 gain expectation.
Next I tried a few variations:
if the bets are per wake-up (SB can bet once on Monday and another time on Tuesday), but SB is only paid for her current situation (if she bets on Monday she gains $6 and pay $18, same for Tuesday), then her gain expectation is 1/3($30 - $
if the bets are per wake-up, but SB is paid for both Monday and Tuesday when she bets on either Monday or Tuesday, then her gain expectation is 1/3($30 - $
So, cool variation, but ultimately each interpretation finds the result that it expected.
I have read your blog post. You are aware of the traditional arguments for both sides. You also agree that halfers and thirders answer slightly different questions.
If you want to maximize the number of times you answer correctly, go Thirder.
If you want to maximize the number of flips you guess correctly, go Halver.
You even say yourself that the halfers answer a version of the problem where Tuesday doesn’t matter:
To test Beauty’s accuracy at guessing coin flips, we should see what she says for every Heads, and see what she says for every Tails. It doesn’t make sense to ask her the same question again for every Tails waking, because [...] we already know her answer.
And since Tuesday doesn’t matter, the whole protocol (sleeping, waking up, the amnesia) is unnecessary: it comes down to guessing the result of a fair coin toss. Which is my point: halfers answer a trivial version of the problem. Same for your “unfair coin” variant.
I’ll use what I read from you post to speculate a little, so feel free to correct me. If I had to guess, I think our disagreement might come from 2 points:
- the version of the SB problem that you favor is Adam Elga’s, which asks “When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?” Which indeed has 1⁄2 as its unique answer, but is trivial.
- maybe you view the fair coin has having an intrinsic 1⁄2 chance of landing heads. I view that all probabilities are in a person’s head (nothing original, I just absorbed the standard Bayesian view as taught in the Sequences on LW, see https://www.lesswrong.com/posts/f6ZLxEWaankRZ2Crv/probability-is-in-the-mind for example). A probability expresses a state of knowledge at a given moment, not an intrinsic property of a physical object.
You said it yourself in your blog post when examining the thirder version:
The questions ask about degrees of belief.
I am guessing that to you, it’s an exception. To me, there is nothing else: all probabilities are degrees of belief.
I discovered the SB problem through the version deemed canonical by wikipedia, which is different from Elga’s:
Any time Sleeping Beauty is awakened and interviewed she will not be able to tell which day it is or whether she has been awakened before. During the interview Sleeping Beauty is asked: “What is your credence now for the proposition that the coin landed heads?”
In that version, each day is treated equally.
And to that version, my answer is: when SB wakes up, the coin has a 1⁄3 chance to have landed up heads. You can verify it simply (and you did in your post) by remarking that if you iterate the experiment a large number of times, 1⁄3 of the wake-up events happen after a heads toss.
I might be unfair, but I have noticed that, faced with such a version that does not trivially reduce to a coin toss, you and other halfers make this weird dance where you say that SB (waking up) ought to believe that the coin has a 1⁄3 chance to have come up heads, but really, really, in the real world the coin has a 1⁄2 chance to have come up heads. As if the point of forming beliefs wasn’t to accurately model the real world.
Adding a betting structure, at best, doesn’t change anything (at worst it can give wrong intuitions): the reason why betting 1⁄3 is the best (under a well-formed betting scheme) is because it is the actual probability.
You already know a lot about the arguments of the thirders. This here is just my point of view, but hopefully, it helps fill some of the gaps.
I agree with your point that the answer depends on the exact formulation of the question. But I think that the two questions that you submitted don’t work.
First, there’s no difference between them: the degree I ought believe that my guess will be correct if I guess Heads is the same as the degree I ought believe that the outcome of the toss is Heads (implied: knowing that I am in a wake-up event).
Second, if you preface your question with “when you are first awakened”, then the answer will always be 1⁄2 whatever comes next. The reason is that you ask to consider only Monday, and in that case the existence of Tuesday is irrelevant so it’s just Monday^heads VS Monday^tails.
It is hard finding a formulation that unambiguously yields 1⁄2 as an answer without being equally trivial. The reason is that the answer is always 1⁄3 as long as you consider the context of being in a wake-up event and take every answer into account separately. If you somehow group the answers into a single one or ignore all answers but one (by asking “when you are first awakened”, or by discarding all answers except one on a random day, etc.) then you get 1⁄2, but again trivially, making the existence of Tuesday and the induced amnesia completely pointless.
In the fission experiment, how do you know in advance that you will experience about 1⁄2 heads and 1⁄2 tails, if you have no method of determining which of the resulting clones you will be? After all, on average a random clone will have experienced 1⁄3 heads and 2⁄3 tails, so you seem to know that you will be more likely to be in a subpart of all clones that experiences heads more often than average.
Here is a model that might interest halfers. You participate in this experiment: the experimenter tosses a fair coin, if Heads nothing happens, you sleep through the night uneventfully. If Tails they will split you in the middle into two halves, completing each half by cloning the missing part onto it. The procedure is accurate enough that the memory is preserved in both copies. Imagine yourself waking up the next morning: you can’t tell if anything happened to you, if either of your halves is the same physical piece yesterday, or if there is another physical copy in another room. But regardless, you can participate in the same experiment again. The same thing happens when you find yourself waking up the next day. and so on.....As this continues, you will count about an equal number of Heads and Tails in the experiments you have subjective experiences of...
Counting subjective experience does not necessarily lead to Thirderism.
Does your experiment really make a difference with the incubator experiment? I still think you will subjectively count witnessing about twice as many tails than heads. Say you run your experiment for a month, forcing all copies to undergo the coin toss and split in case of tails every day. Then at the end of the month you sample one copy at random. Well I think that copy will report seeing about twice as many tails than heads.
Intuitively, you’re more likely to end up being a copy who has seen more tails than heads. And I think that if you count total tails:total heads (totaled over all the clones), you’ll get around 2:1.
Just to check, I ran the following code that returns a ratio of total tails, ran it 20 times and it usually returned something between 0.6 and 0.7.def experiment_day(clone):
tails = random.random() < 0.5
if tails:
return [clone + [“tails”], clone + [“tails”]]
else:
return [clone + [“heads”]]
def run_experiment(nb_days):
clone_pool = [[]]
for i in range(nb_days):
new_clone_pool = []
for a_clone in clone_pool:
new_clone_pool += experiment_day(a_clone)
clone_pool = new_clone_pool
return clone_pool
days = 30
final_clone_pool = run_experiment(days)
nb_tails = 0
nb_heads = 0
for a_clone in final_clone_pool:
nb_tails += a_clone.count(“tails”)
nb_heads += a_clone.count(“heads”)
print(“Proportion of tails %r” %(nb_tails/(nb_tails+nb_heads)))
Edit: formatting.
I didn’t fully understand OP’s argument, but I used a different approach that feels simpler and more intuitive to me.
First, I agree that there is no paradox: if, upon waking up in a green room, you are offered to bet $1:$1 on whether the coin came “heads”, then you should take the bet: after pooling your gains and losses, you and your clones will have gained (1/2 * 18 * 1$) - (1/2* 2 * $1) = $8. There is no paradox with this bet.
So what’s different in Eliezer’s bet? That a single bet is made for the entirety of the clones who wake up in a green room. When the coin comes “heads”, there might be 18 clones in green rooms, but there’s not 18 bets, only a single one.
To check my intuition, I kept Eliezer’s version, but simplified the bet structure: if you and your other clones in green rooms unanimously agree, I will give $1 to your collective of clones if the coin came “heads” and take $1 if it came “tails”. Well then it’s a fair bet, the expected value is (1/2 * $1) - (1/2 * $1) = $0.
The way I see it, since all clones will behave the same way, it is no different from asking a single one chosen at random among green rooms to decide.
Imagine the following variant: after waking up the clones, I pick one at random among those who are in a green room, and I offer that one to decide whether to pay all clones in green rooms $1 and take $3 from all clones in red rooms. I don’t contact any other clone.Well now the gain in probability mass from waking up in a green room is exactly counterbalanced by the loss of probability mass of being chosen: you have 9 times more green rooms in the “heads” case than in the “tails” case, but a clone in a green room is 9 times less likely to be picked in the “heads” case than in the “tails” case. Accepting the bet in this variant gives the same general utility as in Eliezer’s.
Houellebecq visits Silicon Valley.
I’m not an expert, but assuming that by revolution you mean something close to “an attempt to change government through non-legal means”, then I agree with your points, but I’ll also note that revolt and revolution only partially overlap. Revolts are typically less organised and with more modest goals than a government overthrow. They are also mostly initiated and fueled by the resentment and desperation of a lower class.
My tentative model is “Starving peasants revolt. Kings don’t like revolts.” Not “Starving peasants lead successful revolutions.”
To take a modern day example that I have experience with, the yellow vest movement in France was a revolt from the working poor outside big cities because the rise in the gas price made their life impossible in a context where they needed cars to work and purchase essential goods. They were leaderless and actually opposed attempts at vertical organisation. In their early stages, they would have been content with gas prices returning to their previous levels. Nonetheless, they were a thorn in the side of the government, and were even a threat to it at some point.
Yes, it seems I read too fast.
I like the idea but I have a hard time identifying what sets it apart from just fantasizing. Maybe the focus on finding the most awesome vision, or letting go of realism?
Maybe the most similar thing I’ve done is how I endeavored to be less bad at romance with the opposite sex. It was at a time when I had a good number of data points (including from fiction), but very little personal success. While in bed before sleeping, I would imagine a challenging, uncomfortable scenario with a girl I liked (e.g.: she noticed that I accidentally left my zipper open). I would attack the problem under dozens of angles, playing variations (would I judge her negatively if the roles were reversed? Even if I didn’t like her? Do I have a memory of something similar? Can I defuse the situation with a joke? Can the solution be generalized to other shameful situations? Can I turn a shameful situation over and actually score points? etc.) I wouldn’t stop until I found a course of action that felt amazing, and a session would last anything from 30 min to 2 hours.