Note that the process is iterative. In the best threads, folk work together to sort through an issue. AI can be considered another collaborator in such discussions. — Banno
Write "Heads and Monday" on one notecard. Write "Tails and Monday" on another, and "Tails and Tuesday" on a third. Turn them over, and shuffle them. Then write "A," B," and "C" on the other sides.
Pick one. What is the probability that it says "Heads" on the other side? What is the probability that it says "Tails" on the other side? Call me silly, but I'd say 1/3 and 2/3, respectively.
Each morning of the experiment when SB is to be awakened, put the appropriate card on a table in her room, with the letter side up. Hold the interview at that table.
What is the probability that the card, regardless of what letter she sees, says "Heads" on the other side? Or "Tails?" This "outcome" can be defined by the letter she sees. But that does not define what an outcome is, being the description of the experiment's result, in SB's knowledge, is. If she wakes on a different day, that is a different result. Being determined by the same coin flip does not determine that.
Now, did these probabilities change somehow? For which letter(s) do they change? Or are they still 1/3 and 2/3? — JeffJo
Again, there's not much sense in this so-called "pragmatically relevant" credence. Even before being put to sleep – and even before the die is rolled – I know both that the die is most likely to not land on a 6 and that betting that it did will offer the greater expected return in the long run. So after waking up I can – and will – continue to know that the die most likely did not land on a 6 and that betting that it did will offer the greater expected return in the long run, and so I will bet against my credence.
With respect to "pragmatic relevance", Thirder reasoning is unnecessary, so if there's any sense in it it must be somewhere else. — Michael
My argument is that a rational person should not – and would not – reason this way when considering their credence, and this is most obvious when I am woken up 2^101 times if the coin lands heads 100 times in a row (or once if it doesn't).
It is true that if this experiment were to be repeated 2^101 times then we could expect 2/3 of all awakenings to occur after the coin landed heads every time, but it's also irrelevant.
Perhaps you didn't parse correctly. There is no ambiguity. If she is asked to project her state of knowledge on Wednesday, or to recall it from Sunday, of course the answer is 1/2. — JeffJo
I keep looking at the problem, and I can't find a reference to betting anywhere. The reason I don't like using betting is because anybody can re-define how and when the bet is made and/or credited, in order to justify the answer they like. One is correct, and one is wrong.
So, if a bet were to exist, and assuming she uses the same reasoning each time? She risks her $1 during the interview, and is credited her winnings then also. If she bets $1 on Heads with 2:1 odds, she gains $2 if the coin landed Heads, and loses 2*$1 if it landed on Tails. If she bets on Tails with 1:2 odds, she loses $1 if the coin landed Heads, and gains 2*$0.50=$1 if it landed Tails.
But if she bets $1 on Heads with 1:1 odds, she gains $1 if the coin landed Heads, and loses 2*$1=$2 if it landed on Tails. If she bets on Tails with 1:1 odds, she loses $1 if the coin landed Heads, and gains 2*$1=$2 if it landed Tails.
The answer, to the question that was asked and not what you want it to be, is 1/3.
And 50% and growing of public website material is produced by AI. — unenlightened
On the occasion of an awakening, what is Sleeping Beauty's expectation that when the experiment is over ...
— Pierre-Normand
This is what invalidates your variation. She is asked during the experiment, not before or after. Nobody contests what her answer should be before or after. And you have not justified why her answer inside the experiment should be the same as outside. — JeffJo
Here's one more attempt. It's really the same thing that you keep dodging by changing the timing of the question, and claiming that I have "vallid thirder logic" while ignoring that it proves the halfer logic to be inconsistent.
Get three opaque note cards.
On one side of different cards, write "Monday and Heads," "Monday and Tails," and "Tuesday and Tails.
Turn the cards over, shuffle them around, and write "A," "B," and "C" on the opposite sides.
Before waking SB on the day(s) she is to be woken, put the appropriate card in the table in her room, with the letter side face up.
Let's say she sees the letter "B." She knows, as a Mathematical fact, that there was a 1/3 probability that "B" was assigned to the card with "Heads" written on the other side. And a 2/3 chance for "Tails."
By halfer logic, while her credence that "Heads" is written on the "B" card must be 1/3, her credence that the coin landed on Heads is 1/2. This is a contradiction - these two statements represent the same path to her current state of knowledge, regardless of what day it is.
I would say she is being asked what the odds are of it being a day in which a T side vs a H side coins is flipped. — Philosophim
If she's only being asked what the percent chance of the coin ended up being at, the answer is always 50/50. The odds of the coin flip result don't change whether its 1 or 1,000,000 days. What changes is from the result of that coin flip, and that is the pertinent data that is important to get an accurate answer.
This is very similar to the old Monty Hall problem. You know the three doors, make a guess, then you get to make another guess do you stay or change?
On the first guess, its always a 1/3 shot of getting the door wrong(sic). But it can also be seen as a 2/3 chance of getting the door wrong. When given another chance, you simply look at your first set of odds and realize you were more likely than not wrong, so you change your answer. The result matches the odds.
Same with the situation here. Run this experiment 100 times and have the person guess heads 50 times, then tails 50 times. The person who guesses tails every time 50 times will be right 2/3rds of the time more than the first. Since outcomes ultimately determine if we are correct in our odds, we can be confident that 1/2 odds is incorrect.
By the way, very nice discussion! I appreciate your insight and challenging me to view things I might not have considered.
She can reason that its equally likely that the result of the coin flip is 50/50, but that doesn't mean its likely that the day she is awake is 50/50. — Philosophim
Lets flip it on its head and note how the likelihood that she would be wrong.
If she always guesses heads, she's wrong twice if its tails. If she always guesses tails, she's only wrong once. Thus, she is twice as likely to be wrong if she guesses heads on any particular day woken up, and twice as likely to guess correctly if she guesses tails. If the total odds of guessing correctly were 50/50, then she would have an equal chance of guessing correctly. She does not.
I'm not seeing the ambiguity here, but maybe I'm not communicating clearly. There are two outcomes based on context. — Philosophim
Correct. My point was that its just used as a word problem way of saying, "We have 3 outcomes we reach into a hat and pull from" — Philosophim
Because there are two different outcomes. One with one day, and one with two days. If you pick any day and have no clue if its a day that resulted from a heads or tails outcome, its a 2/3rds chance its the tails outcome. The heads and tails is also irrelevant. The math is, "Its as equally likely that we could have a series of one day or two day back to back in this week. If you pick a day and you don't know the outcome or the day, what's the odds its a tails day vs a heads day?"
The odds of whether its head or tails is irrelevant since they are the same and can be effectively removed from the problem.
The part to note is that almost all of this is a red herring. Its irrelevant if she remembers or not. Its just word play to get us out of the raw math. The odds are still the same.
Flip heads, 1 result
Flip tails, 2 results
Put the pile of results as total possible outcomes. You have 3 possible outcomes. In two of the outcomes, tails was flipped. Put it in a hat and draw one. You have a 2/3rd chance that its a tails outcome.
To be clear, it is a 50/50 shot as to whether heads or tails is picked. Meaning that both are equally like to occur. But since we have more outcomes on tails, and we're looking at the probability of what already happened based on outcomes, not prediction of what will happen, its a 2/3rds chance for tails. — Philosophim
You may have read it. You did comment on it from that aspect. But you did not address it. The points it illustrates are:
- That each "day" (where that means the coin toss and the activity that occurred during that awakening), in Mathematical fact, represents a random selection of one possible "day" from the NxN grid. If that activity appears S times in the schedule, and R times in the row, then the Mathematically correct credence for the random result corresponding to that row is R/S. This is true regardless of what the other N^2-S "days" are, even if some are "don't awaken."
- There is no connection between the "days" in a row. You call this "T-awakenings" or "the H-wakening." in the 2x2 version. They are independent. — JeffJo
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
