But the time period after the coin is flipped still exists, and the coin can be Heads during that time. — JeffJo
While in each case the biconditionals "I am now in an H-awakening iff I am now (and will be) in an H-run" or (on Wednesday) "I was in an H-awakening iff I am now in an H-run" hold, the probabilities don't necessarily match due to the two-to-one mapping between T-awakenings and T-runs. — Pierre-Normand
The specific days she’s woken or kept asleep are irrelevant. — Michael
The only things that matter are that she has either one or two interviews – determined by a fair coin toss – and that she doesn’t know if she’s already had one. Everything else is a red herring. — Michael
So tell me what prior probability is ruled out in my experiment above. — Michael
The answer follows trivially from what I have said before - have you read it? — JeffJo
I agree, with a caveat. The specific details of whether she is woken at a specific point ("day") in the experiment do matter. You can test this in Elga's solution. If she is told that the current day is Tuesday, then she knows Pr(Heads) must decrease to 0. If she is told that it is not Tuesday, the Law of Total Probability actually requires it to go up. It goes from 1/3 to 1/2 in the Thirder soluton, and from 1/2 to 2/3 in the Halfer solution. This is quite relevant, even if you think it is wrong AND CAN PROVIDE A VALID REASON
But it isn't the day name that matters, it is the details associated with that day name. And since these details are different on different "days," we can track them by naming the "days."
Now, you could argue about why those details might matter; but so far you have refused to. You have just asserted they don't (in spite of evidence like I just presented). But naming them cannot affect the correct answer, no matter how those details affect it. SO THERE IS NO REASON TO NOT NAME THE "DAYS." And even the possibility that they might have an effect makes them relevant.
In other words, you are proffering the red herring here. You are insisting that we must ignore a piece of potential information, because you think it has no affect. If so, there is no harm in including it. — JeffJo
This procedure creates what, in your words, are "the only things that matters." She has either one, or two, interviews and does not know if another will/did happen. The only thing that is different, is that your possible two interviews occur under different circumstances; one is mandatory, and one is optional. What we disagree about is whether the part that is missing - the non-interview when the option is not taken - matters.
Here they occur under identical circumstances. That is, either steps 2.1 thru 2.4, or steps 3.1 thru 3.4. And those circumstances can be used to answer the question. I did "name" the details by calling them state S1 or S2, but since they are identical to SB she can call them state S.
There are three possible combinations of the two coins in state S, and they are equally likely. Her credence in state S=(H,T) is 1/3.
This has nothing to do with what may or may not get "ruled out" in a solution to your version of the experiment. That difference is the red herring in the "most frequent" presentation of the problem. This is a self-contained experiment with a trivial answer.
But it's an answer you don't like. So you will either ignore it, or repeat the non sequitur that it includes the "ruling out of a 1/4 probability" that we are debating about above, which is circular logic. — JeffJo
I accept 6 and reject 5. My credence that my current interview is a heads interview isn't equal to the fraction of interviews which are heads interviews.
My argument is:
P1. My credence is the degree to which I believe that a proposition is true
P2. My current interview is a heads interview iff I have been assigned one heads interview
C1. Therefore my credence that my current interview is a heads interview is equal to my credence that I have been assigned one heads interview (from P1 and P2)
P3. If I have been assigned at random by a fair coin toss either one heads interview or two tails interviews then the probability that I have been assigned one heads interview is 1/2
P4. I have been assigned at random by a fair coin toss either one heads interview or two tails interviews
C2. Therefore the probability that I have been assigned one heads interview is 1/2
(from P3 and P4)
P5. My credence that I have been assigned one heads interview is equal to the probability that I have been assigned one heads interview
C3. Therefore my credence that I have been assigned one heads interview is 1/2
(from C2 and P5)
C4. Therefore my credence that my current interview is a heads interview is 1/2
(from C1 and C3) — Michael
Your propositions P1 through P4 and C1 though C4 above frequently shift between those two perspectives, which vitiates the validity of some inferences. — Pierre-Normand
Therefore, when shifting to the episodic perspective, it would be a mistake is to divide the probability of the T-timeline (1/2) between the two T-awakenings, suggesting each has a probability of 1/4. This line of thinking presumes these awakenings to be exclusive events within the T-timeline — Pierre-Normand
I know I referred you to one of my previous posts, but I’ll respond to this directly too.
We’re discussing credence.
If I am certain that A is true if and only if B is true and if I am pretty sure that A is true then ipso facto I am pretty sure that B is true. — Michael
When I previously addressed this inference of yours, I conceded that it is generally valid, but I also pointed out that it involved a possible conflation of two meanings of the predicate P(). The problem I identified wasn't with the validity of the inference (within the context of probability calculus), but rather with the conflation that could occur when the expression P(A) appears twice in your demonstration. — Pierre-Normand
What makes you "pretty sure" that A is true is the expectation that A is much more likely to occur than not-A. As such, this probabilistic judgment is implicitly comparative. It is therefore dependent on how you individuate and count not only A events but also not-A events. As I've argued elsewhere, a shift in epistemic perspective can alter the way you count not-Heads events (i.e., Tails events), transforming them from non-exclusive to exclusive. For example, when you move from considering possible world timelines to specific awakening episodes, what were concurrent alternatives (not-H events) become exclusive possibilities. This change in perspective modifies the content of your comparative judgment "H is much more likely to occur than not-H," and consequently affects your credence. — Pierre-Normand
There is only one meaning I'm using: "the degree to which I believe that the proposition is true".
If I am certain that A is true if and only if B is true then the degree to which I believe that A is true is equal to the degree to which I believe that B is true. This is true for all As and Bs. — Michael
Given the above, as I said before, these cannot all be true:
1. My current interview is a heads interview iff I have been assigned one heads interview
2. The fraction of interviews which are heads interviews is 1/3
3. The fraction of experiments which have one heads interview is 1/2
4. My credence that my current interview is a heads interview is equal to the fraction of interviews which are heads interviews
5. My credence that I have been assigned one heads interview is equal to the fraction of experiments which have one heads interview
You seem to assert that 4 and 5 are true by definition, but they're not. Given the definition of the term "credence", and given the truth of 1, 2, and 3, it must be that one or both of 4 and 5 are false. — Michael
So simply asserting that "the fraction of interviews which are heads interviews is 1/3, therefore my credence that my current interview is a heads interview" is a non sequitur.
This overlooks the issue that your credence can change over time when your epistemic perspective changes. — Pierre-Normand
4 and 5 aren't true by definition; rather, they are definitions. — Pierre-Normand
My current interview being the first or the second T-awakening are exclusive events. — Michael
But I brought up something like this here: — Michael
These mean two different things:
1. My credence favours tails awakenings
2. There are more tails awakenings than heads awakenings
I don’t think we can move forward if you insist that they mean the same thing. — Michael
It's precisely because they mean different things that I've provided detailed arguments for deducing 1 from 2 (alongside with other premises). However, the truth of 2 certainly is relevant to the deduction of 1. Nobody would be a Thirder in a scenario where coins lading tails would generate as many awakenings as coins landing heads. — Pierre-Normand
And the only point of mentioning new information, was to show that the information you ignore has meaning. Not to solve the problem or alter the problem. But you knew that.When I said that the only things that matter are:
1. She has either one or two interviews determined by a fair coin toss and
2. She doesn’t know if she’s already had one
I was referring to her just waking up, not being told any further information. — Michael
And again, you keep using circular logic. You deny that events with non-zero prior probability are "ruled out" in your solution. So you claim that my solution, which does "rule out," must be wrong. This is a fallacy; your presumption that you are right is your only defense. You have never argued for why you think they aren't events.No prior probability is ruled out here when woken so your example isn't equivalent. — Michael
And again, you keep using circular logic. You deny that events with non-zero prior probability are "ruled out" in your solution. So you claim that my solution, which does "rule out," must be wrong. This is a fallacy; your presumption that you are right is your only defense. You have never argued for why you think they aren't events. — JeffJo
1. Sleeping Beauty is given amnesia
2. She is asked her credence that a coin has been tossed
3. A coin is tossed
4. If the coin lands tails then:
4A. She is given amnesia
4B. She is asked her credence that a coin has been tossed
Thirder reasoning is that because step 2 is twice as likely to occur as step 4B then I am twice as likely to be in step 2 as step 4B.
Halfer reasoning is that because step 2 is twice as likely to occur as step 4B and that because if 4B will occur then I am equally likely to be in step 2 as step 4B then I am three times as likely to be in step 2 as step 4B. — Michael
whenever she awakens, the coin landed (or will land) tails two times out of three — Pierre-Normand
This is not true. There are three possible awakenings, Monday-Heads, Tuesday-Heads, Tuesday-Tails, and SB's job on awakening is determine the probability that she is experiencing each of these. The coin has a 50% chance of landing heads, and if it does, the awakening will be on Monday 100% of the time. Therefore, P(Monday-Heads) = 50%. The coin has a 50% chance of landing tails, and if it does, the awakening will be on Monday 50% of the time, and Tuesday 50% of the time. Therefore, P(Tuesday-Heads) = P(Tuesday-Tails) = 25%. If this is true, and I don't see how it can be reasonably argued against, on each awakening the coin is equally likely to be heads and tails. — hypericin
Now, let's look at a particular moment of Leonard's visit. As he walks, before reaching a new enclosure, he might reason this way: "Since each fork in the path gives an equal chance of leading to a T-path or an H-path, there is a 50% chance that the next enclosure I'll see will have a hippo." Thus, when he approaches an enclosure, he might conclude there is a 25% chance of it being a tiger enclosure, and a 25% chance of it being a toucan enclosure.
Is this reasoning accurate? — Pierre-Normand
From the episodic perspective, Sleeping Beauty knows that conditionally on her present awakening being the first, it is equally probable that it is a H-awakening (and that the coin will land heads) or that it is a T-first-awakening (and that the coin will land tails). She also knows that in the event the coin will land (or has landed) tails, it is equiprobable that she is experiencing a T-first-awakening or a T-second awakening. Since the three possible outcomes are exclusive from her episodic perspective, their probabilities must sum up to 1 and since P(H-awakening) = P(T-first-awakening) and P(T-first-awakening) = P(T-second awakening), all three possible outcomes must have probability 1/3. — Pierre-Normand
Yet that is the basis of your argument. You even reiterate it here. And it is part of your circular argument, which you used this non sequitur to divert attention from:I can't prove a negative. — Michael
I have. You ignore it. But this is a fallacious argument. Claiming I did something different does not prove the way you handles the different thing is right and mine was wrong.If there is some prior probability that is ruled out when woken then tell me what it is.
Quite an ultimatum, from one who never answers questions and ignores answers he can't refute. Since you haven't proven why the event "Heads&Tuesday" doesn't exist - and in fact can't, by your ":can't prove a negative" assertion, I have every reason to accept that it does exist.If you can’t then I have every reason to accept that there isn’t one.
I just need you to agree that it is equivalent to your procedure first. — JeffJo
Something is indeed is ruled out when she wakes. — JeffJo
The next enclosure is the toucon enclosure iff I first turned right at the fork (P = 1/2) and then passed the tiger enclosure.
2. My credence that the next enclosure is the toucon enclosure is equal to the probability that the first event happened multiplied by the probability that the second (dependent) event happened. — Michael
1. Conditionally on its being a first encounter on a path segment, P(Tiger) = P(Hippo)
2. Conditionally on Leonard being on a T-path segment, P(Tiger) = P(Toucan)
3. The three possible outcomes are exhaustive and mutually exclusive
4. Therefore, P(Tiger) = P(Hippo) = P(Toucan) = 1/3 — Pierre-Normand
You need to prove this inference:
P(Hippo|Hippo or Tiger) = P(Tiger|Hippo or Tiger)
Therefore P(Hippo) = P(Tiger) — Michael
I’m only considering one fork as only that is comparable to the Sleeping Beauty problem. What’s true of multiple forks isn’t true of one fork, as evidenced by (1);
1. The next enclosure is the toucon [sic] enclosure iff I first turned right at the fork (P = 1/2) and then passed the tiger enclosure.
This isn’t true if there are two forks.
So what is wrong about my analysis of one fork? — Michael
Consider first the two possible outcomes conditional on today being Monday. Since Sleeping Beauty always is awakened on Monday regardless of the coin toss result, P(Monday-Heads) = P(Monday-Tails). Consider next the two possible outcomes conditional on the coin having landed tails. Since in that case Sleeping Beauty is awakened once on Monday and once on Tuesday, P(Monday-Tails) = P(Tuesday-Tails), which is something that the Thirders, Halfers and Double-halfers all agree on. We therefore have that P(Monday-Heads) = P(Monday-Tails) = P(Tuesday-Tails). Lastly, since Sleeping Beauty isn't inquiring about the probabilities that any of those three outcomes will occur at least once during her current experimental run, but rather about the probability that her current awakening episode is the realization of one of those three outcomes, the three possibilities are exclusive and exhaustive, and their probabilities must therefore sum up to 1. They therefore all three are 1/3, and P(Tails) = P(Monday-Tails) + P(Tuesday-Tails) = 2/3. — Pierre-Normand
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