Ok, I read it, and I still don’t understand what you call the day 1 Objection, or the argument that you’re trying to make. What’s most puzzling is how you came to use 3⁄4 for the probability that it’s Monday. Which you attribute to thirders if I understood correctly? (Edit after finishing writing: now I think that your belief? I’m even more confused about what you argue and what you assume thirders argue). Anyway, I’ll just list what I disagree with. I’m fine with most of A “two-faced” example, except here:
You check your name tag; what was the hench-name they’d given you? Ah, right: “Ralph-wrecker”. A quick check with Claude reveals that exactly 3⁄4 of hench-names are alphabetically ahead of yours.
You reason: The chance of getting paired up with someone like Alan and seeing them get Safe was 3⁄4, meaning P(obs) = 3⁄4. On a Heads flip, both you and the other recruit would always be Safe, meaning P(obs | H) = 1. And that means your original calculation was correct:
No, P(obs | H) = 3⁄4. In other words, obs (the other applicant has a name before yours and gets a Safe uniform) and H are independent.
And that gives: P(H | obs) = P(obs | H)*P(H)/P(obs) = (3/4)*(1/2)/(3/4) = 1⁄2 which is the result you initially came up with and is correct. I think you know that the 2⁄3 reasoning is incorrect, since you disprove it correctly right in the next paragraph. But you seem to think that the mistake has a different cause (you talk about double counting...). I don’t know about that, I just know that you don’t need any argument to disprove it because the math is wrong, so you’re probably attacking an argument that no one makes.
Next paragraph is The day 1 objection.
Let’s also say that the room always has a calendar Beauty can check to learn the current day.
Well then (like I said many times at this point), the whole experiment is useless, the result is trivial. I notice that you seem to be ignoring that point (that your versions of SB problem that have 1⁄2 as a solution work exactly the same without the whole sleep and amnesia thing) every time I make it. The whole point of the sleep and amnesia is that SB doesn’t know which day it is. Of course, if she sees Monday on her calendar she has a credence of 1⁄2 for Heads. If she saw Tuesday, it would be 0! (not factorial 0, just 0)
The “Day 1” objection argues that this set of statements is impossible:
Again, I don’t understand this objection, but I disagree with P(obs Mon) = 3⁄4 (also, no idea where it comes from), so I’m pretty sure at least I am not arguing this.
If Beauty were to be asked on every waking, “What’s the chance today is Monday?” and she wanted to minimize the margin of error of her responses, then she would answer 3⁄4. This is because 3⁄4 is the proportion of time she would spend awake in Monday during many repetitions of the experiment.
Uh? The proportion of time she would spend awake on Monday during many repetitions of the experiment is 2⁄3. She should answer 2⁄3. After finishing writing this reply, I think there’s a chance that the crux of our disagreement is here.
But here’s the difference: In these scenarios, when we describe P(obs Mon) for an event “today is Monday”, the “today” refers to something specific.
In the first example, “today” refers to the average waking across repeated experiments.
In the second and third examples, “today” becomes identified as “the random day that the Riddler chose to invade”.
Yeah, SB’s answer really depends on what she believes the Riddler used as an heuristic for invading the experiment. In particular:
1) If she’s confident that the Riddler picked Monday or Tuesday at random, then was ready to cancel the invasion if SB was sleeping (in the case of Heads and Tuesday), then there’s a 2⁄3 chance that it’s Monday, and a 1⁄3 that the coin came Heads.
2) If she’s confident that the Riddler checked the result of the coin toss, then picked Monday in case of Heads and picked Monday or Tuesday with equal probability in case of Tails, then there’s a 3⁄4 chance that it’s Monday, and a 1⁄2 chance that the coin came Heads.
Anyway, sorry for not really engaging with the core points of your blog post, but I just don’t understand them. I hope this reply at least brings some insight for you.
I’m going to be sort of shitty and not give this comment the time it deserves, and I’ll only respond to parts of it. If there’s anything you’d like to focus on that I ignore, please hit me up with another comment and I’ll try to get to it when I can.
First, I’ll say that some of the confusion probably stems from the fact that the post is defending against an argument against Halferism, so I’m presenting a Halfer point and showing it to not result in any contradictions, not justifying the Halfer point directly. You being a Thirder, you wouldn’t think that P(obs Mon) = 3⁄4, since that’s not the Thirder position.
Secondly, I’d like to note my complete agreement with the statement “SB’s answer really depends on what she believes the Riddler used as an heuristic for invading the experiment”.
>> I notice that you seem to be ignoring that point (that your versions of SB problem that have 1⁄2 as a solution work exactly the same without the whole sleep and amnesia thing)
I’m not sure what you’re saying with this piece, but my guess-of-an-answer-to-what-you-might-be-saying is this: The solution does work exactly the same as asking about the coin without any sleep or amnesia, because it should be the same. That’s part and parcel to the whole Halfer position that in waking, Beauty hasn’t learned anything, and without learning anything, there’s nothing to update.
(Analogously: There’s no need to talk about “self locating” or “self indicating” in anthropics, because being-a-self gives no additional information on top of “there exists 1 of the thing that is me”.)
>> The proportion of time she would spend awake on Monday during many repetitions of the experiment is 2⁄3. She should answer 2⁄3. After finishing writing this reply, I think there’s a chance that the crux of our disagreement is here.
Damn, that’s my mistake. 2⁄3 of the time is spent awake on Monday, that’s true. I can see why that would seem to indicate the Thirder position. Probably most helpful would be to focus on my Riddler example, with the added condition that the Riddler wasn’t doing either of the actions you described. Instead, he didn’t know about Beauty at all until he happened to stumble upon her. (I’ve fixed the post up a bit.)
I’m guessing you’ll say that in this version, Beauty’s belief in Monday should not be 3/4?
Ok, well it seems that most of our disagreement has disappeared, which should allow us to focus on what remains.
That’s part and parcel to the whole Halfer position that in waking, Beauty hasn’t learned anything, and without learning anything, there’s nothing to update.
The weirdness in the thirder answer is not exactly there. If on Sunday, before the whole sleeping/amnesia thing, you asked SB what’s her credence that the coin came Head, she’d say 1⁄2. But if you asked her what would be her credence that the coin came Head when she would be woken up later, she’d answer 1⁄3. She doesn’t have to update when she wakes up, because she already has a different answer for the two questions.
And I agree that this only moves the problem: why is her answer different for the two questions? That’s where I’m still confused, but I can only say: the math checks up.
Say a crazed Riddler invades the experiment and happens upon Beauty. He hadn’t expected to find her, but he’s immediately captivated by her allure, and decides to make her the target of his games. He demands of Beauty: “You must correctly tell me the day of the week, or I will blow up Gotham!” Some fans of Gotham’s criminal underworld might know that the Riddler is usually more active on Mondays versus Tuesdays, but Beauty doesn’t know anything about him. His appearance gives her no additional nudge towards one day or another. She should reason that there was a 1⁄2 chance of Heads, therefore it’s Monday, and a 1⁄2 chance of Tails, therefore it could equally be Monday or Tuesday, and thus she should answer “Monday” with a confidence of 3⁄4.
I don’t see the difference between this version and my 1) version where the Riddler picks a day at random, and nothing happens if he picks Tuesday and the coin came Heads. Let’s say this version happens over 4 parallel worlds, 2 where the Riddler picks Monday and 2 where he picks Tuesday. We know we’re not in the one world where he picked Tuesday and the coin came Heads (Tuesday ^ Heads), so 3 worlds remain: Monday ^ Heads, Monday ^ Tails, Tuesday ^ Tails. 2 out of these 3 worlds are Monday worlds.
If he cancels the invasion when Beauty is sleeping, then that weighs the probability towards worlds where she’s not asleep, i.e. Tails. Which is why Heads goes down from 3⁄4 to 2⁄3.
(Not sure if this is identical to the $30/6/6 EV form of the questions? Not sure which is more useful)
>> 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
Ok, I read it, and I still don’t understand what you call the day 1 Objection, or the argument that you’re trying to make. What’s most puzzling is how you came to use 3⁄4 for the probability that it’s Monday. Which you attribute to thirders if I understood correctly? (Edit after finishing writing: now I think that your belief? I’m even more confused about what you argue and what you assume thirders argue). Anyway, I’ll just list what I disagree with.
I’m fine with most of A “two-faced” example, except here:
No, P(obs | H) = 3⁄4. In other words, obs (the other applicant has a name before yours and gets a Safe uniform) and H are independent.
And that gives: P(H | obs) = P(obs | H)*P(H)/P(obs) = (3/4)*(1/2)/(3/4) = 1⁄2 which is the result you initially came up with and is correct.
I think you know that the 2⁄3 reasoning is incorrect, since you disprove it correctly right in the next paragraph. But you seem to think that the mistake has a different cause (you talk about double counting...). I don’t know about that, I just know that you don’t need any argument to disprove it because the math is wrong, so you’re probably attacking an argument that no one makes.
Next paragraph is The day 1 objection.
Well then (like I said many times at this point), the whole experiment is useless, the result is trivial. I notice that you seem to be ignoring that point (that your versions of SB problem that have 1⁄2 as a solution work exactly the same without the whole sleep and amnesia thing) every time I make it. The whole point of the sleep and amnesia is that SB doesn’t know which day it is. Of course, if she sees Monday on her calendar she has a credence of 1⁄2 for Heads. If she saw Tuesday, it would be 0! (not factorial 0, just 0)
Again, I don’t understand this objection, but I disagree with P(obs Mon) = 3⁄4 (also, no idea where it comes from), so I’m pretty sure at least I am not arguing this.
Uh? The proportion of time she would spend awake on Monday during many repetitions of the experiment is 2⁄3. She should answer 2⁄3. After finishing writing this reply, I think there’s a chance that the crux of our disagreement is here.
Yeah, SB’s answer really depends on what she believes the Riddler used as an heuristic for invading the experiment. In particular:
1) If she’s confident that the Riddler picked Monday or Tuesday at random, then was ready to cancel the invasion if SB was sleeping (in the case of Heads and Tuesday), then there’s a 2⁄3 chance that it’s Monday, and a 1⁄3 that the coin came Heads.
2) If she’s confident that the Riddler checked the result of the coin toss, then picked Monday in case of Heads and picked Monday or Tuesday with equal probability in case of Tails, then there’s a 3⁄4 chance that it’s Monday, and a 1⁄2 chance that the coin came Heads.
Anyway, sorry for not really engaging with the core points of your blog post, but I just don’t understand them. I hope this reply at least brings some insight for you.
I’m going to be sort of shitty and not give this comment the time it deserves, and I’ll only respond to parts of it. If there’s anything you’d like to focus on that I ignore, please hit me up with another comment and I’ll try to get to it when I can.
First, I’ll say that some of the confusion probably stems from the fact that the post is defending against an argument against Halferism, so I’m presenting a Halfer point and showing it to not result in any contradictions, not justifying the Halfer point directly. You being a Thirder, you wouldn’t think that P(obs Mon) = 3⁄4, since that’s not the Thirder position.
Secondly, I’d like to note my complete agreement with the statement “SB’s answer really depends on what she believes the Riddler used as an heuristic for invading the experiment”.
>> I notice that you seem to be ignoring that point (that your versions of SB problem that have 1⁄2 as a solution work exactly the same without the whole sleep and amnesia thing)
I’m not sure what you’re saying with this piece, but my guess-of-an-answer-to-what-you-might-be-saying is this: The solution does work exactly the same as asking about the coin without any sleep or amnesia, because it should be the same. That’s part and parcel to the whole Halfer position that in waking, Beauty hasn’t learned anything, and without learning anything, there’s nothing to update.
(Analogously: There’s no need to talk about “self locating” or “self indicating” in anthropics, because being-a-self gives no additional information on top of “there exists 1 of the thing that is me”.)
>> The proportion of time she would spend awake on Monday during many repetitions of the experiment is 2⁄3. She should answer 2⁄3. After finishing writing this reply, I think there’s a chance that the crux of our disagreement is here.
Damn, that’s my mistake. 2⁄3 of the time is spent awake on Monday, that’s true. I can see why that would seem to indicate the Thirder position. Probably most helpful would be to focus on my Riddler example, with the added condition that the Riddler wasn’t doing either of the actions you described. Instead, he didn’t know about Beauty at all until he happened to stumble upon her. (I’ve fixed the post up a bit.)
I’m guessing you’ll say that in this version, Beauty’s belief in Monday should not be 3/4?
Ok, well it seems that most of our disagreement has disappeared, which should allow us to focus on what remains.
The weirdness in the thirder answer is not exactly there. If on Sunday, before the whole sleeping/amnesia thing, you asked SB what’s her credence that the coin came Head, she’d say 1⁄2. But if you asked her what would be her credence that the coin came Head when she would be woken up later, she’d answer 1⁄3. She doesn’t have to update when she wakes up, because she already has a different answer for the two questions.
And I agree that this only moves the problem: why is her answer different for the two questions? That’s where I’m still confused, but I can only say: the math checks up.
I don’t see the difference between this version and my 1) version where the Riddler picks a day at random, and nothing happens if he picks Tuesday and the coin came Heads.
Let’s say this version happens over 4 parallel worlds, 2 where the Riddler picks Monday and 2 where he picks Tuesday. We know we’re not in the one world where he picked Tuesday and the coin came Heads (Tuesday ^ Heads), so 3 worlds remain: Monday ^ Heads, Monday ^ Tails, Tuesday ^ Tails. 2 out of these 3 worlds are Monday worlds.
If he cancels the invasion when Beauty is sleeping, then that weighs the probability towards worlds where she’s not asleep, i.e. Tails. Which is why Heads goes down from 3⁄4 to 2⁄3.
(Not sure if this is identical to the $30/6/6 EV form of the questions? Not sure which is more useful)