Steff
>> 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
This disregards that on Tails, the same toss result will result in an extra guess.
>> if you preface your question with “when you are first awakened”, then the answer will always be 1⁄2 whatever comes next
Ah, yes, the Day 1 Objection. I refute this argument here:
https://ramblingafter.substack.com/p/sleeping-beauty-meets-two-face
Sorry for taking so long!
$14 was a mistake, that’s my bad and I’ve updated the post.
>> 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
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?
>> (your second variant) SB is paid for both Monday and Tuesday when she bets on either Monday or Tuesday… if you repeat the experiment a large number of times, and then tag each wake-up event with the profit that SB made during the iteration that included that event, then the average of tags will be $18
Not exactly. I laid out a separate $6 on Mon and $6 on Tues to make this easier to think about. Combining into $12, you end up with the same amount of money being equally shared between two days. 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?
Hey there, I was wondering if you’d be able to formulate why my following counter is wrong?
https://www.lesswrong.com/posts/cqdDiMe6K6WcMAF2R/boltzmann-brains-like-doomsday-require-no-explaining
I’ve been getting a lot of downvotes, so I figure I must have made a mistake somewhere, but nobody has explained it yet (well, I’ve received one comment but not one I understood).
I assume you’re too busy and this isn’t important so I’m not actually expecting any response here—either way, cheers! I’m a fan of all your work
Hey, thanks for the comment.
I’m not understanding the problem that needs fixing.
Firstly, any observation I make provides no evidence to either Boltzmann brains or non-Boltzmann brains.
Secondly, let’s say the first was false and observations do lend credence to non-Boltzmann brains. I can still believe in a theory where Boltzmann brains greatly outnumber non-Boltzmann brains, which is what my post is about.
>> A fair coin is flipped. If it lands heads, nothing happens. If it lands tails, an extremely biased towards tails coin is flipped. If that coin lands tails, you are killed. If that coin lands heads, nothing further happens.
That’s the essence of my post: The question that begs itself. Your setup assumes a coin flip. But we have no reason to think that the universe is acting in such a coin flipping way.
I didn’t think it involved cards at all, I assumed the card symbols were being used as shorthand for something else, haha.
Disagreed—I don’t believe that the idea of Boltzmann brains requires theories to explain “why are we observing ourselves to be structured brains” whatsoever.
Short version is this: If both types of brains must exist, then there must be observers of both types. Observers of the minority type haven’t learned anything to update priors on, since it’s guaranteed that they must exist, and there’s no reason to assume P(structured) = K/(K + M) unless you were privy to some sort of random selection system in place that picked you out ahead of you existing.
Longer version explanation here: https://www.lesswrong.com/posts/cqdDiMe6K6WcMAF2R/boltzmann-brains-like-doomsday-require-no-explaining
It’s the exact same assumption that goes into the Doomsday Argument, but it’s an assumption that requires justification. E.g., if you believed that a God exists and He instantiates souls before inserting them into bodies randomly, then the Bayesian math would flow from there and you would indeed believe on a relatively soon Doomsday or that universes must have an explanation for why-aren’t-I-a-Boltzmann-brain-which-was-overwhelmingly-more-likely.
Boltzmann brains, like Doomsday, require no explaining
To counter the idea of Boltzmann brains, isn’t it enough to simply observe that order is self-replicating? Therefore ordered brains arising from order are vastly more likely than those arising from chaos.
I’m hoping to find the time for a full response later, but right now:
>> First, I’m still waiting for you to fully describe the random day. Is it chosen from a set of two, three, or four?
It’s a random choice among the days she’ll be awoken. If she’s to awake just Mon, then she’ll be offered on Mon. If she’s to awake on Mon and Tues, then a coin is flipped.
The bet must be offered on each waking day, and evaluated each time SB is asked fro a credence, to correctly represent her credence. If you do this, the experiment is a zero-sum game if and only if 2:1 odds are used. But I suspect you will disagree with me. Which is why I do not use betting arguments.
I actually think this might be useful to drill into further. Why couldn’t SB use her credence of the coin for the bet I described? Are you thinking that SB actually has two different credences about the coin, and which one she applies will depend on the exact question asked?
You have ignored how I point out that the need to consider it is demonstrated by letting SB observe it in a way that distinguishes it from the others: Wake her on Tuesday, after Heads, but take her shopping instead of asking about the coin.
Hoping for a clarification with this one: Do you mean that she goes shopping on Tuesday and is told on Tuesday? Or that a random day was chosen for her to go shopping, and she knew that it’d be a random day? I think I would agree with your presented math depending on exactly what she’s learning.
Okay I’m getting tired of fighting the editor, going to quote differently:
>> That when SB is asked the question, there are exactly three possible faces that could be showing, and one of them is green.
I’m not denying that those are the possibilities, but that they share equal probabilities.
>> Did you note that the problem asks about her credence based on her current circumstance, and that there are circumstances that differ? Circumstances that can exist regardless of whether SB is awake.>> So Pr(Awake) is not 1. “Awake” will not happen on Tuesday, after Heads, which is a possible result.
Unfortunately I’m not following this part. P(Awake on Tues) is not 1, but that’s not what she’s observing, she’s observing being awake, which she already knew would happen.
That depends entirely on how you evaluate the”bet.”
I pick a random day to approach Beauty with an offer of a bet. If she guesses right, on Wednesday I’ll give her $1.5, otherwise she’ll give me $2. If she truly believes that the coin flipped Tails with a probability of 2⁄3, then she’ll take this bet.
What is so difficult to see about this?
You’re not making yourself clear. Yes, from the outside perspective of a randomly chosen day, there are different die rolls associated with whether Beauty is asleep or not on that day. How does that contradict the math I presented? Would you like to share some Bayesian equations of your own? Did you read my post about the Day 1 Objection, which I strongly suspect is very relevant?
Apologies if that comes off as rude, I’m just finding it difficult to respond usefully to this latest comment. (“What is so difficult to see about this?” isn’t a helpful question—I could just as easily pose the same question back to you!)
Thanks for taking a look! I’ll be sure to go over what you’ve written—I ran out of time to spend any more time on this stuff for this weekend / this week so I won’t be able to reply soon, just wanted to let you know that your response is appreciated
What observation are you trying to update on?
P(G) = 1⁄2
P(waking) = 1
P(waking | G) = 1
P(G | waking) = P(waking | G)*P(G)/P(waking) = 1⁄2
Let’s say she learns it’s Monday. Half the time, she’ll be awake on a random day. Half the time, she’ll be awake on 2⁄3 of days.
P(M) = P(T) = P(W) = (1/2)(1/3) + (1/2)(2/3) = 1⁄2
(Note that this is distinct from the probability that a randomly chosen waking is M/T/W—more on that in a second)
P(M | G) = 1⁄3
P(G | M) = P(M | G)*P(G)/P(M) = (1/3)(1/2)/(1/2) = 1⁄3
Again, no updates are occurring.
What did I mean by “a randomly chosen waking is M/T/W”? An example would be the chance of Beauty receiving a knock on the door, if an outside experimenter flips a coin to decide whether to knock on her door on her first red waking or the second (or the only choice of green waking, if it was green).
The chance that she gets a knock on a Monday would be (1/2)(1/3) + (1/2)(2/3)(1/2) = 1⁄3.
The Thirder mistake is to treat different wakings as a random draw, rather than as sequential events that we know will both happen. (Also the same mistake made on the Doomsday Argument.) This is exactly analogous to the “Day 1 objection” to the Halfer position of the original problem, wherein Beauty learns that it’s Monday—similar to the scenario you’re describing with green and red sides of a dice, but simpler to reason about, so I think it would be better to discuss.
And I break down exactly why that objection fails here:
https://ramblingafter.substack.com/p/sleeping-beauty-meets-two-face
—
Meta-note for anyone else reading this: As far as I’ve seen, the Thirder logic results in credences of expected value that cannot be used without leading to losing bets (and depending on how it’s justified, can lead to more extreme obvious wrong conclusions such as the presumptuous philosopher). Where as I’ve never seen any contradiction or absurdity to arise from the Halver position. I used to be a Thirder myself at one point, and for a long time I was highly uncertain about what the true answer should be. But the more and more I’ve looked at objections to the Halfer position, the more and more I’ve found it to be resilient, and the more I believe there is a true sense in which the Halfer position is correct and the Thirder position is incorrect, when you’re talking about actual belief in the coin flip and not belief in something you know will be tallied more often with more wakings (e.g. guesses-over-repeated-experiments), and the that former is more useful thing to Beauty (i.e., more actionable; not able to be money-pumped) when she’s asked unspecifically, “What do you think the coin flipped?”
Even when I died, I always felt like it was my fault, as opposed to some vague undefinable thing I couldn’t hope to fight or stop.
Absolutely this.
Even some of the best players at other games (I’m thinking Starcraft, Hearthstone) can complain about unfair randomness and opponents getting lucky. But Soulsborne bosses do a really good job of training you that everything is beatable, you just need to practice and improve. (Though there are the rare engagements that give you limited control over their outcomes—like Vyke in Elden Ring.)
Surprised not to see any From games on this list, which are the most commonly I’ve seen cited for personal growth—not in places like here in LessWrong, but on Reddit and YouTube comments in gaming sections. Though the anecdotes are often less about learning new skills and more about gaining confidence or getting back the feeling of control in one’s life. A motivation booster, essentially.
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’m afraid that’s incorrect, actually.
If you repeat the experiment, one half of the time Beauty will be in the Heads world and one half of the time Beauty will be in the Tails world. In Tails world, she’ll have twice as many wakings. If we want to measure her accuracy of credence of coins, and not measure her accuracy of credence of her own guesses, we should take the average of her answers in Tails worlds. To do so otherwise is to double count Tails answers and make them matter more when they shouldn’t.
This can seem confusing, or seem like we’re just arguing semantics. However, I think this becomes clearer by analogy:Credence is something you should be able to act on.
Let’s say after every waking, money will be deposited to Beauty’s bank account (without her knowledge). The amount that is deposited depends on the coin flip: On Heads, she will awake on Monday and receive $30; on Tails, she will awake on both Monday and Tuesday and receive $6 each day. What should Beauty believe is her expected value gain?
If Beauty answers with 1⁄3, she’ll get the wrong answer and she can be convinced to take bad bets and lose money over repetitions of the experiment.(I mad a little table for the options here—https://ramblingafter.substack.com/i/181017019/problem-seven )
I don’t like the verb “observes” since it can be confused between the notion of an observation that’s happening right now (still 1⁄2) versus tallying over all total observations over repeated experiments (therefore 1⁄3), which is why I personally prefer the noun “guesses” for distinguishing them.
“BY HALFER LOGIC, the probability that the die “landed on” each of the six sides is 1⁄6. But since she is awake, SB knows that there are only three sides that it could [have landed on/be showing] at the moment. They are still equally likely, and only one of the three is green.”
Hello, I’m not sure what you’re trying to say. There are six possibilities here (M,W,T x R,G) and yes, I do think that each of them has a 1⁄6 chance. The chance of Red is equal to the chance of Green.
Do you have any scenarios in which you can show the 1⁄2 answer being wrong?
I believe what you’ve described is just the original problem again with an extra possible waking of no question asked on Tuesday, but that doesn’t impact the credences of her waking and observing the question. The answer is still 1⁄2.