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
A six is the most likely outcome, so I'm betting on it. — Pierre-Normand
Her credence remains committed to P3, else she’d calculate very different expected returns. — Michael
Those two reasonings concern the same dice but two different statements of credence in two different kinds of events/outcomes. — Pierre-Normand
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.
those credences target differently individuated events — Pierre-Normand
Indeed, and, as previously explained, that because Halfers and Thirders are typically talking past each other. They're not talking about the same events. — Pierre-Normand
Remember the flip-coin scenario where the singular H-awakenings take place in the West-Wing of the Sleeping Beauty Experimental Facility and the dual T-awakenings are taking place in the East-Wing. The West-Wing is surrounded by a moat with crocodiles and the East-Wing is surrounded by a jungle with lions. On the occasion of her awakening Sleeping Beauty finds a rare opportunity to escape and can either choose to bring a torch (that she can use to scare off lions) or a wooden plank (that she can use to safely cross the moat). A Thirder analysis of the situation is natural in that case since it tracks singular escape opportunities. Her credence that she will encounter crocodiles is 2/3 (as is her credence that the coin landed Tails). Taking the plank is the safest bet and, indeed, two thirds of Sleeping Beauties who make this bet on the rare occasions where this opportunity presents itself to them survive. — Pierre-Normand
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
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
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.
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
According to the often-misrepresented, original Thirder analysis by Adam Elga, there are two independent random elements: the coin toss, and the day. They combine in four (not three) ways. But I suppose Elga suspected how obtuse halfers would be about them, so he only considered the two overlapping pairs of two that you think constitute the entire sample space.According to a standard Thirder analysis, prior to being put to sleep, SB deems the two possible coin toss outcomes to be equally likely.
That's not the reasoning.When she awakens, she could be in either one of three equiprobable situations: Monday&Tails, Monday&Heads and Tuesday&Tails (according to Elga's sensible argument)
.SB's credence in the truth of the statement "Today is Tuesday" is 1/3.
As far as I know, SB can tell that this day is just one of those three possibilities. Please explain if you think otherwise. And before the experiment begins (this is called the "prior" to those who understand probability), she also knows that the experiment exists on Tuesday&Heads, even though she will not be awake to observe it. So when she is awake, she eliminates that possibility.Before the experiment began, SB could (correctly) reason that is was equally likely that she would be awakened once when the coin toss result is Heads and twice when the coin toss result is Tails.
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
it's rational to bet on the least likely outcome (namely, a non-six result, which occurs only 5/11th of the times) since this is the betting behavior that maximizes the expected return. In fact, it could be argued that this arbitrary payoff structure is misleading in the present context since it is being designed precisely to incentivise the bettor to bet on the least likely outcome according to their own credence. — Pierre-Normand
It's the (well defined) credence in combination with the payoff structure that jointly govern the rational betting behavior. — Pierre-Normand
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. None of which can answer the questions I raised. Using this tableI didn't ignore your post. — Pierre-Normand
The SB setup is a very close analogy to this. Coins landing Tails play a similar causal role. Just replace "increased proclivity to being noticed by a passerby" with "increased proclivity to awaken a random test subject in the Sleeping Beauty Experimental Facility". — Pierre-Normand
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
Of course, one salient disanalogy between this penny drop analogy and the SB problem is that, in the standard SB problem, each coin is being tracked separately and noticed at least once, on Monday. But I don't think this disanalogy undermines the main point. It's because tail-outcomes causally increase the proportion of awakening episodes at which SB would encounter them that, on each occasion where she encounters them, SB can update her credence that the coin landed Tails. That this rational ground for Bayesian updating remains valid even in cases of singular experimental runs with amnesia (as in the original SB problem) is something that I had illustrated by means of a Christmas gift analogy (see the second half of the post). — Pierre-Normand
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
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
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 didn't provide a detailed response to your post because you didn't address it to me or mention me. — Pierre-Normand
I use "single day" because each day is an independent outcome to SB. — JeffJo
Thank you for that. But you ignored the third question:the variation that you actually propose, when only one activity is being experienced on any given day, yields a very straightforward Bayesian updating procedure that both Halfers and Thirders will agree on. — 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.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
I don't see how it bears on the original problem — Pierre-Normand
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
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.
Halfers don't condition on the propostion "I am experiencing an awakening". — Pierre-Normand
They contend that for SB to be awakened several times, rather than once, in the same experimental run
Halfers, however, interpret SB's credence, as expressed by the phrase "the probability that the coin landed Tails" to be the expression of her expectation that the current experimental run,
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