Yep. What makes it an independent outcome, is not knowing how the actual progress of the experiment is related to her current situation. This is really basic probability. If you want to see it for yourself, simply address the Camp Sleeping Beauty version. — JeffJo
I spent the last hour composing a post responding to all my mentions, and had it nearly finished only to have it disappear leaving only the single letter "s" when I hit some key. I don't have the will to start over now, so I'll come back to it later. — Janus
It's s different probability problem based on the same coin toss. SB has no knowledge of the other possible days, while this answer requires it. — JeffJo
This experiment is now becoming "beyond the pale" and "incorrigable" to me... — ProtagoranSocratist
Sleeping beauty is a mythical character who always sleeps until she is woken up for whatever reason. However, there's not part of her story dictating what she remembers and doesn't, so if amnesia drugs are involved, then the experimentors are free to then craft the percentage that the outcome shows up... — ProtagoranSocratist
assuming there is nothing mysterious or "spooky" influencing a coin flip, then the answer is always is always 50/50 heads or tails. Maybe I misunderstand. — ProtagoranSocratist
As I understand it, the insight is what you’re supposed to provide in your post. I don’t really care where you get it from, but the insight should be in your own words based on your own understanding and experience and expressed in a defensible way. The documentation you get from the AI response can be used to document what you have to say, but then you’re still responsible for verifying it and understanding it yourself. — T Clark
I guess my question is whether the user’s understanding is genuine, authentic, and owned by them. — T Clark
What are we supposed to do about it? There's zero chance the world will decide to collectively ban ai ala Dune's thinking machines, so would you ban American development of it and cede the ai race to China? — RogueAI
Isn't the best policy simply to treat AI as if it were a stranger? So, for instance, let's say I've written something and I want someone else to read it to check for grammar, make comments, etc. Well, I don't really see that it is any more problematic me giving it to an AI to do that for me than it is me giving it to a stranger to do that for me. — Clarendon
I don’t disagree, but I still think it can be helpful personally in getting my thoughts together. — T Clark
Yes, that makes the answer 1/2 BECAUSE IT IS A DIFFERENT PROBLEM. — JeffJo
There are three Michelin three-star restaurants in San Francisco, where I'll assume the experiment takes place. They are Atelier Crenn, Benu, and Quince. Before the coin is tossed, a different restaurant is randomly assigned to each of Heads&Mon, Tails&Mon, and Tails&Tue. When she is awoken, SB is taken to the assigned restaurant for her interview. Since she has no idea which restaurant was assigned to which day, as she gets in the car to go there each has a 1/3 probability. (Note that this is Elga's solution.) Once she gets to, say, Benu, she can reason that it had a 1/3 chance to be assigned to Heads&Mon. — JeffJo
You appear to be affirming the consequent. In this case, Tails is noticed twice as often because Tails is twice as likely to be noticed. It doesn't then follow that Tail awakenings happen twice as often because Tails awakenings are twice as likely to happen. — Michael
"1) Per run: most runs are 'non-six', so the per-run credence is P(6)=1/6 (the Halfer number).
2) Per awakening/observation: a 'six-run' spawns six observation-cases, a 'non-six' run spawns one. So among the observation-cases, 'six' shows up in a 6/5 ratio, giving P('six'|Awake)=6/11 (the Thirder number)."
— Pierre-Normand
This doesn't make sense.
She is in a Tails awakening if and only if she is in a Tails run.
Therefore, she believes that she is most likely in a Tails awakening if and only if she believes that she is most likely in a Tails run.
Therefore, her credence that she is in a Tails awakening equals her credence that she is in a Tails run.
You can't have it both ways.
If it helps, it's not a bet but a holiday destination. The die is a magical die that determines the weather. If it lands on a 6 then it will rain in Paris, otherwise it will rain in Tokyo. Both Prince Charming and Sleeping Beauty initially decide to go to Paris. If after being woken up Sleeping Beauty genuinely believes that the die most likely landed on a 6 then she genuinely believes that it is most likely to rain in Paris, and so will decide instead to go to Tokyo.
Still: the effects of one flip never effect the outcome of the other FLIPS, unless that is baked into the experiment, so it is a misleading hypothetical question (but interesting to me for whatever reason). The likelihood of the flips themselves are still 50/50, not accounting for other spooky phenomenon that we just don't know about. So, i'll think about it some more, as it has a "gamey" vibe to it... — ProtagoranSocratist
Why? How does something that is not happening, on not doing so on a different day, change her state of credence now? How does non-sleeping activity not happening, and not doing so on a different day, change her experience on this single day, from an observation of this single day, to an "experimental run?"
You are giving indefensible excuses to re-interpret the experiment in the only way it produces the answer you want. — JeffJo
Right. And this is they get the wrong answer, and have to come up with contradictory explanations for the probabilities of the days. See "double halfers." — JeffJo
I understand the 1/3rd logic, but it simply doesn't apply here: the third flip, given the first two were heads (less likely than one tail and a head, but still very likely), is also unaffected by the other flips. — ProtagoranSocratist
This is useful information. I had it in my mind that it didn't use the spaces, so I started using spaces to distinguish myself. I guess I'll go back to spaceless em dashes. — Jamal
I would think handing your half-formed prose to a bot for it to improve it is plagiarism, regardless of the number of words changed or inserted. It's a different thing from you deliberately searching for a synonym. No? — bongo fury
Getting the thesaurus to suggest whole phrases and sentences is obviously plagiarism. — bongo fury
Then try this schedule:
. M T W H F S
1 A E E E E E
2 A A E E E E
3 A A A E E E
4 A A A A E E
5 A A A A A E
6 A A A A A A
Here, A is "awake and interview."
If E is "Extended Sleep," the Halfer logic says Pr(d|A)=1/6 for every possible roll, but I'm not sure what Pr(Y|A) is. Halfers aren't very clear on that. — JeffJo
But if E is anything where SB is awoken but not interviewed, then the straightforward Bayesian updating procedure you agreed to says Pr(d|A)=d/21, and if Y is an index for the day, Pr(Y|A)=Y/21.
My issue is that, if A is what SB sees, these two cannot be different.
Thank you for that. But you ignored the third question:
Does it matter if E is "Extended sleep"? That is, the same as Tuesday&Heads. in the popular version?
"I don't see how it bears on the original problem where the new evidence being appealed to for purposes of Bayesian updating isn't straightforwardly given"
— Pierre-Normand
Then you don't want to see it as straightforward. Tuesday still exists if the coin lands Heads. It is still a single day, with a distinct activity, in the experiment. Just like the others in what you just called straightforward. — JeffJo
I use "single day" because each day is an independent outcome to SB. — JeffJo
This, I think, shows the fallacy. You're equivocating, or at least begging the question. It's not that there is an increased proclivity to awaken in this scenario but that waking up in this scenario is more frequent.
In any normal situation an increased frequency is often explained by an increased proclivity, but it does not then follow that they are the same or that the latter always explains the former – and this is no normal situation; it is explicitly set up in such a way that the frequency of us waking up Sleeping Beauty does not mirror the probability of the coin toss (or die roll). — Michael
If you are allowed to place 6 bets if the die lands on a 6 but only 1 if it doesn't then it is both the case that winning bets are more frequently bets that the die landed on a 6 and the case that the die is most likely to not land on a 6.
I think your comment sidestepped the issue I was raising (or at least misunderstood it, unless I'm misunderstanding you), but this reference to Bayesian probability will make it clearer.
[...]
it cannot be that both Halfers and Thirders are right. One may be "right" in isolation, but if used in the context of this paradox they are equivocating, and so are wrong in the context of this paradox. — Michael
Yes, so consider the previous argument:
P1. If I keep my bet and the die didn't land on a 6 then I will win £100 at the end of the experiment
P2. If I change my bet and the die did land on a 6 then I will win £100 at the end of the experiment
P3. My credence that the die landed on a 6 is 6/11
C1. Therefore, the expected return at the end of the experiment if I keep my bet is £
C1(sic). Therefore, the expected return at the end of the experiment if I change my bet is £
What values does she calculate for and ?
She multiplies her credence in the event by the reward. Her calculation is:
C1. Therefore, the expected return at the end of the experiment if I keep my bet is £45.45
C2. Therefore, the expected return at the end of the experiment if I change my bet is £54.55
This is exactly what Prince Charming does given his genuine commitment to P3 and is why he changes his bet.
So why doesn’t she change her bet? Your position requires her to calculate that > but that’s impossible given P1, P2, and P3. She can only calculate that > if she rejects P3 in favour of “my credence that the die landed on a 6 is 1/6”. — Michael
You didn't respond to a single point in it. You only acknowledged its existence, while you continued your invalid analysis about changing bets and expected runs. — JeffJo
This is a trivial conditional probability problem. The reason I posed the "Camp Sleeping Beauty" version, is that it exposes the red herrings. And I assume that is the reason you ignore it, and how the red herrings are exposed. — JeffJo
This is where I believe the mistake is made. The question she is asked after being woken up is the same question she is asked before being put to sleep. There is no ambiguity in that first question, and so there is no ambiguity in any subsequent question. There is a single event that is the target of the question before being put to sleep and we are asking if being put to sleep and woken up gives Sleeping Beauty reason to re-consider her credence in that event, much like Prince Charming re-considers his credence in that event after being told that his coin is loaded. Neither Sleeping Beauty nor Prince Charming is being asked to consider their credence in one of two different events of their own choosing. — Michael
You seem to continue to conflate an outcome's expected return with its probability and assert that one's behaviour is only governed by one's credence in the outcome. — Michael
Neither of these things is true. I've shown several times that the least likely outcome can have the greater expected return and so that this assessment alone is sufficient to guide one's decisions.
No number of analogies is going to make either "she wins two thirds of the time if she acts as if A happened, therefore she believes (or ought to believe) that A most likely happened" or "she believes that A most likely happened, therefore she acts (or ought to act) as if A happened" valid inferences.
But the most important part of my previous comment were the first two paragraphs, especially when considering the standard problem.
SB has no unusual "epistemic relationship to the coin," which is what the point of my new construction was trying to point out. That fallacy is based on the misconception that Tuesday somehow ceases to exist, in her world, if the coin lands on Heads. It still exists, and she knows it exists when she addresses the question. — JeffJo
That you're more likely to escape if you assume that the coin landed tails isn't that the coin most likely landed tails. You just get two opportunities to escape if the coin landed tails. — Michael
This makes no sense. There is only one kind of event; being woken up after a die roll. Her credence in the outcome of that die roll cannot be and is not determined by any betting rules. Maybe she's not allowed to place a bet at all. — Michael
After waking up, either she continues to believe that the probability that the die landed on a 6 is 1/6, as Halfers say, or she now believes that it is 6/11, as Thirders say.
Only then, if allowed, can she use her credence to calculate the expected returns of placing or changing a bet, accounting for the particular betting rules. And as I believe I showed above, only a credence of 1/6 provides a consistent and sensible approach to both betting scenarios.
Her credence remains committed to P3, else she’d calculate very different expected returns. — Michael
I don't even have to be put to sleep and woken up to do this. I can just say before the experiment starts that I choose to place 6 bets that the die will land on a 6 instead of 1 bet that it won't. — Michael
So you need to first specify the mechanism by which one has "encountered" a door, and this mechanism must be comparable to the Sleeping Beauty scenario for it to be an apt analogy. — Michael
Sorry, I deleted that post because it's late and I'm tired and I may have messed up the specific numbers. The general gist is what I said before. Your argument is that her reasoning after being woken up is:
A1. If I keep my bet and the die didn't land on a 6 then I will win £100
A2. If I change my bet and the die did land on a 6 then I will win £100
A3. My credence that the die landed on a 6 is 6/11
A4. Therefore, the expected return if I keep my bet is £83.33
A5. Therefore, the expected return if I change my bet is £16.67
But A3, A4, and A5 are inconsistent. If A3 really was true then she would calculate different values for A4 and A5, concluding that it is profitable to change her bet. But she doesn't do this. — Michael
Thirders then claim that:
P(6|Monday)=6/11
P(¬6|Monday)=5/11 — Michael