Just Use Bayes: Sleeping Beauty and Monty Hall

Steff

Just Use Bayes: Sleeping Beauty and Monty Hall — LessWrong

12 Just Use Bayes: Sleeping Beauty and Monty Hall

28th Mar 2026

14 min read

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There's a question that's held my fascination for months. At times, it's had me spinning in circles, caught up in seemingly impossible contradictions. And it's not the famous Sleeping Beauty problem... or at least, not exactly.

There is a certain Vincent Conitzer, of Duke University, who presents what he calls a "Devastating" example supposedly disproving the logic behind the "1/2" answer to the original problem.

His argument is valid, but based on a faulty premise. He assumes that Halfers, to get their answer, always look at remaining possibilities and renormalize equally between them. (At no point does he justify this assumption.)

So Conitzer is making a mistake in his critique of the Halfer position. I would say that anyone who still believes in the "1/3" answer to the original SB problem is making a mistake. But I'm not any less error prone myself. Holy hells, have I made mistakes. Conitzer's example is a variant of SB with an easier answer, because it's a question where the correct answer and the intuitive answer are the same. Yet at one point I got so confused, I believed in an answer that really should be self-evidently wrong to anyone.

The question that's fascinated me for months isn't any particular object-level SB variant, but rather the meta-level question: What makes Sleeping Beauty so difficult?

I think the typical answer here is, in a word, "anthropics".

Take, for instance, Eliezer Yudkowsky, who in Project Lawful writes (via a character he uses to serve as a mouthpiece for lecturing on rationality):

This however would get us into ‘anthropics’, and we are not getting into ‘anthropics’. That, by the way, is a general slogan of dath ilani classes on probability theory: We are not getting into ‘anthropics’. I’m not even going to translate the word. As long as you don’t make any copies of people, you can stay out of that kind of trouble. That trouble-free life should be our ambition for at least the next several weeks.

I’m not satisfied with that. “Anthropics” doesn’t get a free pass to be inexplicably mysterious. It should bend to the laws of math, like everything else.

This is a post with a two-part thesis:

The Equiprobability bias is really good at leading humans astray.
If you want the right answer, you don't need the "SIA" or "SSA" or the like. Just use Bayes' Theorem.

But before we dive into the specifics of Sleeping Beauty and its variants, first I'd like to start with a very different example with surprisingly similar underlying math...

Monty Hall

“Well, a remarkable thing about this problem, simple as it is, is that it has sparked just endless debate”

In the Monty Hall problem, when you first pick a door, the chance of it being the right one is 1/3. When Monty opens a door and then you choose to switch, you’ve now narrowed down a possibility space from 3 options to 2: It can either be behind door 1 or door 3. A 50/50 chance! Yay! That’s an improvement.

Well, actually, that’s wrong. It’s not 50/50. If you switch, your chances go up to ⅔.

According to Conitzer, the “Halfer rule” says that whenever possibilities are eliminated, you should renormalize between all remaining possibilities equally. Here, that gets the wrong answer of 2/3.

I should hope no one uses Conitzer's inaptly-named "Halfer rule". Let’s see instead how my proposed solution of relying on Bayes handles this problem:

Say that you guess door 1, Monty opens door 2 to reveal a zonk, and you decide to switch to door 3.

Before any choices have been made, our prior probabilities for each door hiding the sports car are as follows:

P(1) = P(2) = P(3) = 1/3

The prior probability it’s not behind door 2?

P(~2) = ⅔

We want to know the chance it’s behind door 3, given that it’s not behind door 2, AKA:

P(3 | ~2) = ?

We also know that if it’s behind door 3, it cannot be behind door 2:

P(~2 | 3) = 1

All that’s left is Bayes:

P(3 | ~2) = P(~2 | 3)*P(3)/P(~2) = (1)*(1/3)/(2/3) = 1/2

Somehow, we got the wrong answer of 1/2 again.

But that's alright. We'll come back to this.

The “devastating” problem

Conitzer’s variant begins like this: Sleeping Beauty gets put to sleep, wakes up Monday morning, and is shown a coin flip toss. Then regardless of the coin flip’s result, her memory of Monday will be erased and the procedure repeated: She gets put to sleep, wakes up Tuesday, is shown a second coin toss, and has her memory again wiped.

Conitzer’s question: After waking and observing a coin flip to Heads, what should Beauty believe is the chance that the two coin flips will have had different results? I.e., what’s the chance that either today is Monday and the coins will be Heads-Tails (HT) OR today is Tuesday and the coins were Tails-Heads (TH)?

If you believe that, after observing Heads, you should update to P(coins are different) = 2/3, then consider that the situation is symmetrical with Tails. Upon observing Tails, you should also update to 2/3. But that means no matter what you observe, you will update to 2/3—and if you already know this update will happen, you should just go ahead and update now. But that would mean believing two fair coins have a 2/3 likelihood of flipping to different results, even before having observed anything.

(For a period of time, this is what I actually believed! The reason for my confusion wasn't necessary to the main points of this post, but if you're interested, I explain in the bonus section at the end.)

Again, the “Halfer rule” (of renormalizing equally among remaining possibilities) gets the wrong answer of ⅔. That logic goes like this: When you observe a Heads, you learn that Tails-Tails (TT) has not happened. Because you don’t know which day it is, however, you haven’t learned anything else. That means all of HH, HT and TH remain as possibilities…

…with equal likelihood. HT and TH are two of the three options, so the chance the coin flips are different must be two out of three.

What about Bayes?

P(Coins are different | observe ~TT) = ?
P(Different) = 1/2
P(~TT) = 3/4
P(~TT | Different) = 1
P(Different | ~TT) = P(~TT | Different) * P(Different) / P(~TT) = (1)*(1/2)/(3/4) = 2/3

These get the same results because they mean the same thing. Eliminate TT, and this is what happens.

This is a fun problem! It’s weird. Unless you already understand it perfectly, you should be surprised by this. If you feel confused, that’s a good thing! (And if you're not confused, then that's also a good thing, and kudos.)

We woke up, we observed a Heads, we logically eliminated TT from the possibilities, and of course could not eliminate any of HT, TH, or HH. So where’s the flaw in the above reasoning?

The Monty Hall solution

There’s a different Bayes calculation we can apply to the Monty Hall problem that will give us the correct answer:

P(observe ~2) = 1/2
P(~2 | 3) = 1
P(3 | ~2) = P(~2 | 3)*P(3)/P(~2) = (1)*(1/3)/(1/2) = ⅔

Except this only works because I redefined the meaning of “~2”. Before, it referred to the initial chance that Door 2 would be hiding the car. Now, it means: Once you’ve chosen to open Door 1, what’s the chance that Monty will reveal a zonk by opening Door 2?

He has to choose either Door 2 or Door 3 to open, and they’re equally probable from our perspective, hence: P(~2) = 1/2. If the car is behind Door 3, then we know Door 2 must be opened, and thus: P(~2 | 3) = 1. Then the rest is just plug-and-play with Bayes.

Our earlier calculation wasn’t wrong. It just didn’t capture as much information about the situation. True and accurate statements can still be incomplete.

But…

…you might have noticed…

…this doesn’t actually resolve anything. If both mathematical calculations are accurate, how do we choose which one to use over the other?

The answer: You don’t have to choose one over the other. You can just use both! If you’ve got two pieces of evidence, A and B, then Bayes lets you update on A first, then followed by B. Or B followed by A. That’s true even if A happens to be a superset of the information with B. No matter how you slice it, the math will work out.

The math

Let “2” mean the event that Door 2 is hiding the car.

Let “O2” mean the event that Monty will Open 2.

Our initial calculation showed:

P(3 | ~2) = 1/2

Now we want to update on O2, in addition to just ~2:

P(3 | observe O2 ^ ~2) = ?

If door 2 doesn’t have the car but door 3 does, Monty will be forced to open door 2:

P(O2 | 3 ^ ~2) = 1

If you only know that the car’s not behind door 2, then there’s a 1/2 chance it’s behind door 3 and 1/2 it’s behind door 1:

P(O2 | ~2) = (1/2)(1) + (1/2)(1/2) = 3/4

Then applying Bayes:

P(3 | O2 ^ ~2) = P(O2 | 3 ^ ~2) * P(3 | ~2) / P(O2 | ~2) = (1)(1/2)/(3/4) = 2/3

Now let’s try going in the opposite direction, starting with the update we calculated using event O2, and seeing what happens when add in “2”:

P(3 | O2) = 2/3
P(3 | observe ~2 ^ O2) = ?

This one’s easy:

P(~2 | O2) = 1 (if Monty opens door 2, door 2 can’t have the car)
P(~2 | 3 ^ O2) = 1 (ditto)
P(3 | ~2 ^ O2) = P(~2 | 3 ^ O2) * P(3 | O2) / P(~2 | O2) = (1)(⅔)/(1) = 2/3

O2 implies ~2, so we get no change when considering ~2 as a subsequent, standalone update.

Sleeping Beauty works the same way

Earlier, I considered the event “TT” and described:

P(~TT) = 3/4
P(~TT | Different) = 1

Let’s instead consider the observation of not whether you will, on either Monday or Tuesday, eventually observe a Heads, but consider the event of observing a Heads right now, in this particular moment of observation, and let’s call that event “H”.

P(H) = 1/2

Because, you know… fair coin. But here’s a diagram anyway:

P(H | Different) = 1/2

Because if there’s one Heads and one Tails, but you don’t know which will be on which day, and you don’t know which day it is, then yeah… it’s still an even chance.

And now the calculation:

P(different | H) = P(H | different) * P(different) / P(H) = (1/2)*(1/2)/(1/2) = 1/2

Like with the Monty Hall problem, we have two ways of looking at the situation and two similarly defined events, one which gets the right answer (in this case, “H”) and one that does not (“~TT”). It can seem like a paradox: Without already knowing the right answer, how do we know to use “H” instead of “~TT”? And as before, if you’re not sure about which event to update on, you can simply update on both.

Let’s confirm.

If we wake up and see Heads, then we know Tails-Tails isn’t happening. Therefore H → ~TT, and we should expect that if we update on H first, a subsequent update on ~TT should effect no change:

P(Different coins | observe ~TT ^ H) = ?
P(~TT | different ^ H) = 1
P(~TT | H) = 1
P(different | H) = 1/2 from Step 1
P(different | ~TT ^ H) = P(~TT | different ^ H) * P(different | H) / P(~TT | H) = (1)*(1/2)/(1) = 1/2

And in the other direction, if we update on ~TT first and then H?

Calculated previously:

P(different | ~TT) = ⅔

Also, if we know Tails-Tails didn’t happen, then we know the chance of experiencing a Heads waking is higher:

P(H | ~TT) = ⅔

Also, if we know the coins are different, then we already know ~TT. Since we already established that P(H | different) = 1/2, that means this is the same:

P(H | different ^ ~TT) = 1/2

Putting it all together:

P(Different | H ^ ~TT) = P(H | different ^ ~TT) * P(different | ~TT) / P(H | ~TT) = (1/2)*(2/3)/(2/3) = 1/2

I love this kind of thing, math working out the way you expect it to.

So the general guidance I’d like to give, if you’re seeing Bayes spit out different answers and you’re not sure which to go with: Just go with both! Add as much information as you can assemble, then see where the math takes you.

Furthermore, I think there will always be a clue you can find in retrospect which will act to confirm your findings. For instance:

Noticing that H → ~TT, but ~TT → H doesn’t make sense
Noticing that the event “not Door 2” was being defined in different ways
Noticing that when Sleeping Beauty wakes up in her original problem, she has learned no new information (and therefore has nothing to update on)
From my bonus section: Noticing that the conditions of the bettor’s approach represents new information not yet considered

Why do people answer 1/3 to Sleeping Beauty?

“The Equiprobability Bias: [Possible] Events tend to be viewed as equally likely” — Joan B. Garfield

Roll two dice, and you can get any sum between 2 and 12. These sums are not all equally likely; there are more ways to sum to 7, for example, than to get snake eyes. Yet “people show a strong tendency to believe that 11 and 12 are equally likely” (Nicolas Gauvrit and Kinga Morsanyi).

A better version of this dice problem was made famous by the Grand Duke of Tuscany and Galileo, around the year 1620. The Grand Duke considered three dice rolls, then asked Galileo if the sum of “10” was any more likely than the sum of “9”.

As Florent Buisson explains here, both sums can be achieved in six ways, via the following permutations:

9 = 6 + 2 + 1 = 5 + 2 + 2 = 5 + 3 + 1 = 4 + 3 + 2 = 4 + 4 + 1 = 3 + 3 + 3

10 = 6 + 2 + 2 = 6 + 3 + 1 = 5 + 3 + 2 = 5 + 4 + 1 = 4 + 4 + 2 = 4 + 3 + 3

It’s easy to think that “9” and “10” must be equally probable, and yet: “10” is more likely than “9”, because though it has the same number of summing permutations, “10” has more combinations. (4+3+3 is three times as likely to result as 3+3+3, which can only result if all dice rolled a 3.)

I believe the Equiprobability Bias is why so many people get confused by the SB problem.

We saw the exact same dynamic at play with Monty Hall, thinking that the two doors that remain after one is eliminated must each be 1/2.

The same thing happens again with Bertrand’s box paradox.

Are you surprised by Benford’s Law? If so, it counts towards the Equiprobability Bias as well.

It should make sense: When we first learn probability as kids, we’re trained on examples involving coins and dice and drawing cards from decks, all equiprobable events. It’s like the $1.10 ball-and-bat problem: a failure of over-generalizing simple rules and not working through the necessary steps.

With SB, we’re being presented three equal-seeming options. Then we over-generalize. I’m no exception to this—I used to be a Thirder myself, until I thought through the problem more. And the gods know I’ve made all sorts of reasoning and mathematical errors in my life.

But I also think there are some very smart people getting unnecessarily tripped up by SB due to a tendency to get tangled up into webs as they contend with the aforementioned topic of anthropics.

You don’t need the SIA, SSA, self-locality, or anthropics

There’s another problem I’ve yet to mention in this post, despite the fact that it is completely isomorphic to Conitzer’s SB variant.

And by “isomorphic” I mean the math is EXACTLY the same.

It’s called the “Boy or Girl paradox”, and I don’t think it should be any more or less difficult to understand than Conitzer’s problem.

Except it is.

I don’t think Bentham’s Bulldog and other Thirders would trip up on the Boy or Girl paradox in the same way. Where the Boy or Girl problem merely involves the meeting of kids or dogs in different contexts, Sleeping Beauty uses amnesia to create experientially identical “observer moments”. Thinking about observer moments, or anthropics, can lead you down twisty chains of thought.

For instance: In the Conitzer variant, some would argue that the difference between events “H” and “~TT” is one of “self-locality”: ~TT is a description of the world from a more removed, objective perspective. H is a description of you during a particular observer moment. This is the fundamental difference, they’d argue. But this sort of thinking gives rise to paradoxes, which then require all sorts of convoluted concepts like the “SIA” or “SSA” to resolve, and can lead to EXTREMELY absurd conclusions such as the presumptuous philosopher.

As I argue here and Ape in the Coat argues here, neither the SIA nor the SSA is correct because both rely on the Doomsday argument’s faulty anthropic assumptions.

We don’t need “self-locality” when we can simply recognize that “H” holds more informational content than “~TT”.

Except I’m not satisfied leaving it at that.

How is it I was once a Thirder myself, and yet I’ve never believed that the 1/2 Boy or Girl problem could have any other answer than 1/2? If the math is the same, where does the extra confusion come from?

I believe I’ve found my answer, and it’s the same answer I’ve already presented: The Equiprobability Bias.

With Boy or Girl, there are two kids or dogs with two possible genders each, so your brain naturally thinks of four options. You’re more likely to think of three options (and therefore probabilities of 1/3 and 2/3) after seeing a diagram that, say, eliminates “GG” as an option.
Whereas with SB, the wakings form a viscerally graphic and intuitive set of three.

That’s really it, I think: the seed from which all these other tangled roots have grown.

If you have any other SB variants you’ve struggled with, or that you’ve created, please share them! I bet that with Bayes, we can find their answers.

BONUS: My added confusion

Let’s say that during Conitzer’s setup, after Beauty awakes and sees a Heads, she meets a bettor who explains:

I decided to pick a random “Heads” day to approach you.
(This means if TT happened, I would not have approached.)
Do you want to take the bet that both coins flipped Heads?
If you’re right, you’ll gain $3. Wrong and you’ll lose $2.

Previously, I showed that P(Different | observation) = 1/2. If that’s the case, then the expected payoff for accepting the bet would be (1/2)(+$3) + (1/2)(-$2), which is positive. You should accept the bet!

Right?

Well, actually…

The expected payoff is (1/3)(+$3) + (2/3)(-$2), a negative value, which makes this a bad bet.

It seems as if there’s a sort of logic that’s needed to get the right answer to Conitzer’s problem, which leads to 1/2, and there’s another sort of logic needed to get the right answer here, which leads to 2/3. We distinguished them with the descriptors of “Beauty will at some point see Heads” versus “Beauty is currently seeing Heads”—a more general statement versus a “self-locating” one.

That’s what a particular r/slatestarcodex redditor argued with me, and it tripped me up bad. We went in circles and circles on this, yet at no point did either one of us realize that those 2/3 values represent different things.

This becomes easy to see if we consider one last variant, wherein the bettor approaches Beauty but changes the conditions of his approach:

I chose a random day (Heads or Tails) to approach you.
Do you want to take the bet (same as before)?

In this case: Yes! Take the bet! The chance the coins are different or the same really is 1/2 and you stand to gain in expected value.

The key is this: “I decided to pick a random “Heads” day to approach you” is new information. “Beauty is currently seeing Heads” represented the most up-to-date information before the bet, with its answer of 1/2. Learning about the approach creates an update towards 2/3…

…and makes this yet another example of the guideline: When in doubt, just try Bayes again.

Rationality

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[-]justinpombrio3mo112

Why do people answer 1⁄3 to Sleeping Beauty?

There's a simpler explanation. You say yourself that the answer to this original question is 1/2:

When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?

and that the answer to this slightly different question is 1/3:

When you are first awakened, to what degree ought you believe that your guess will be correct if you guess the coin toss was Heads?

People disagree about the answer to the Sleeping Beauty problem because those questions are nearly identical and easy to confuse. And because in almost all circumstances the answer to questions that vary this way is the same, so you don't need to distinguish them.

Bit of a side note, but personally the distinction I like to draw is between the probability that the coin landed heads or tails (which happens on Sunday, exactly once, and has two possible events, giving the answer 1/2), and the probability that Beauty observes heads (which happens on Monday or maybe Tuesday, and has three possible events, giving the answer 1/3).

[-]Vanilla_cabs2mo20

I agree with your point that the answer depends on the exact formulation of the question. But I think that the two questions that you submitted don't work.

First, there's no difference between them: the degree 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 (implied: knowing that I am in a wake-up event).

Second, if you preface your question with "when you are first awakened", then the answer will always be 1/2 whatever comes next. The reason is that you ask to consider only Monday, and in that case the existence of Tuesday is irrelevant so it's just Monday^heads VS Monday^tails.

It is hard finding a formulation that unambiguously yields 1/2 as an answer without being equally trivial. The reason is that the answer is always 1/3 as long as you consider the context of being in a wake-up event and take every answer into account separately. If you somehow group the answers into a single one or ignore all answers but one (by asking "when you are first awakened", or by discarding all answers except one on a random day, etc.) then you get 1/2, but again trivially, making the existence of Tuesday and the induced amnesia completely pointless.

[-]justinpombrio2mo20

I agree. I didn't notice the significance of the "when you are first awakened" clause, and agree that makes the answer 1/2. I interpreted it as "right after site is woken up, be it in Monday or Tuesday", though that's in retrospect not what it's saying, it's saying "on Monday".

You know, Wikipedia shows two versions of the problem, one of which uses that clause and one of which doesn't, and doesn't note the difference (that I've seen). This problem really is a mess of hard to notice ambiguous wordings, with a little bit of philosophy problem underneath.

[-]Vanilla_cabs2mo10

I went to check Adam Elga's paper. It's very short, only 5 pages written with a big font. He does ask

When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?

But he doesn't mean that you (the person experimented upon) know that it's your first awakening. Rather, it's a way to single out Monday's awakening for the sake of the reader. It's obvious later in the paper:

If (upon first awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either T1 or T2.

So the 2 versions of the problem on wikipedia are consistent. Still, if I was confused by the formulation of the first, I bet I'm not the only one.

[-]Steff21d10

>> 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

[-]Vanilla_cabs19d10

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:
P(H | obs) = P(obs | H)*P(H)/P(obs) = (1)*(1/2)/(3/4) = 2/3

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:
P(obs Mon) = 3/4
P(obs Mon | H) = 1
P(H | obs Mon) = P(obs Mon | H)*P(H)/(P(obs Mon) = (1)*(1/2)/(3/4) = 2/3

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.

[-]Steff19d10

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?

[-]Vanilla_cabs19d10

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.

[-]Steff19d10

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)

[-]Steff2mo10

Do you have any scenarios in which you can show the 1/2 answer being wrong?

[-]JeffJo2mo20

Certainly, since it is wrong.

There is much obfuscation surrounding the SB problem, that appear in this thread. One is that when you ask about a probability in the present tense about an occurrence described in the past tense, you are asking about the conditional probability of that event given the present information. What this means is that there is only one question that is relevant: "Given SB's knowledge when she is awake, what is the probability that the coin result was Heads?" Not "What was the probability, ignoring any information obtained since, that a coin flip result is Heads?" (For comparison, say I draw a card at random from a standard 52-card deck, and then tell you that it is a spade. The probability, that I drew the Ace of Spades, is 1/13. Not 1/52, the probability when I drew it.)

Then, and this will become important later, the question Adam Elga asked is equivalent to, but not identical to, the one he answered. In the one he asked, SB is woken once, or twice, based on a coin flip. In the one he answered, SB is always woken on Monday, and then again on Tuesday, if the coin result was Tails.

But in the one he asked, he added two details that presaged his solution. Neither detail has any impact on what was asked. The first is that he mentioned that the experiment occurs over two days. But days are irrelevant in what was asked. The only significance of these two days are in his solution. And he asked "when you are first awakened...". His solution supposed that SB could be told that it is Monday, or that the coin landed Tails, and evaluated that probability after this information was revealed. The point of "when you are first awakened..." was simply to evaluate the probability before either revelation.

+++++

My point is that these timing details are completely irrelevant, and are used to confuse the probability by establishing different kinds of conditions for each waking. A trivial solution can be obtained by making these conditions equivalent for each. Here is one way:

The experiment proceeds over three days, not two. Monday, Tuesday, and Wednesday.
Only one set of conditions produce a waking on each day.
Replace the coin with a six-sided die. Three faces will be green, representing "Heads," and three will be red, representing "Tails."
The words "Monday," "Tuesday" and "Wednesday" are each written on two sides, once in each color.
The die is rolled on Sunday Night, and preserved throughout the experiment. If it [landed on/is currently showing] a green side, SB is woken only on the day indicated on that side. If red, SB is woken twice, on the two days not indicated on that side.
SB is asked for the probability that the die [landed on/is showing] a green side.

BY HALFER LOGIC, the probability that the die "landed on" each of the six sides is 1/6. But since she is awake, SB knows that there are only three sides that it could [have landed on/be showing] at the moment. They are still equally likely, and only one of the three is green.

"New information" does not mean learning something happened that was not guaranteed to happen. It means that the information eliminates at least one result that was possible without the information. My version does this transparently, by ruling out three die results.

In the classic version (the one in Elga's solution), the information is that the combination of coin and day allows SB to be awake. Before that information was obtained, Heads+Tuesday is possible. By being awake, SB can eliminate one of the four equally-likely possibilities, leaving three. The obfuscation in that version happens because the two combinations that can occur on Monday are treated as equivalent to only one that can occur on Tuesday.

[-]Steff2mo10

"BY HALFER LOGIC, the probability that the die "landed on" each of the six sides is 1/6. But since she is awake, SB knows that there are only three sides that it could [have landed on/be showing] at the moment. They are still equally likely, and only one of the three is green."

Hello, I'm not sure what you're trying to say. There are six possibilities here (M,W,T x R,G) and yes, I do think that each of them has a 1/6 chance. The chance of Red is equal to the chance of Green.

[-]JeffJo2mo30

I’m not sure what you’re trying to say.

With all due respect, I don't sense that you are trying to understand it. Your position seems to be firmly that SB has gained no information, so she can't use conditional probability. Mine is that she does, and all I'm trying to do is make it hard to ignore why it is.

As you point out, there are six possible die rolls. But on any one day, only three remain possible. This is what is known, in probability theory, as a "condition." It is what allows one to "update" a prior probability, which is what you calculated, to a "posterior" or "conditional" probability, which is what the Sleeping Beauty Problem asks for. This is true even if she gains no information. The condition will be the entire sample space, and the "updated" probability will be the same as the prior probability. So there is no reason to not explore what the condition is.

It's easier if we number the days 0 (for "Monday"), 1 and 2. The set of possible die rolls - can we agree that this is the sample space? - is {G_0, G_1, G_2, R_0, R_1, R_2}. When she is awake, she does not know which day it is, but she does know that it is only one day. Call it day D. The condition is the event {G_D, R_MOD(D+1,3), R_MOD(D+2,3)}. The condition is the same regardless of what D is. Each outcome in this set has a prior probability of 1/6. The posterior (or conditional) probability is of G_D is (1/6)/[(1/6)+(1/6)+(1/6)]=1/3.

[-]Steff2mo10

What observation are you trying to update on?

P(G) = 1/2
P(waking) = 1
P(waking | G) = 1
P(G | waking) = P(waking | G)*P(G)/P(waking) = 1/2

Let's say she learns it's Monday. Half the time, she'll be awake on a random day. Half the time, she'll be awake on 2/3 of days.
P(M) = P(T) = P(W) = (1/2)(1/3) + (1/2)(2/3) = 1/2
(Note that this is distinct from the probability that a randomly chosen waking is M/T/W - more on that in a second)

P(M | G) = 1/3
P(G | M) = P(M | G)*P(G)/P(M) = (1/3)(1/2)/(1/2) = 1/3

Again, no updates are occurring.

What did I mean by "a randomly chosen waking is M/T/W"? An example would be the chance of Beauty receiving a knock on the door, if an outside experimenter flips a coin to decide whether to knock on her door on her first red waking or the second (or the only choice of green waking, if it was green).

The chance that she gets a knock on a Monday would be (1/2)(1/3) + (1/2)(2/3)(1/2) = 1/3.

The Thirder mistake is to treat different wakings as a random draw, rather than as sequential events that we know will both happen. (Also the same mistake made on the Doomsday Argument.) This is exactly analogous to the "Day 1 objection" to the Halfer position of the original problem, wherein Beauty learns that it's Monday - similar to the scenario you're describing with green and red sides of a dice, but simpler to reason about, so I think it would be better to discuss.

And I break down exactly why that objection fails here:
https://ramblingafter.substack.com/p/sleeping-beauty-meets-two-face

--

Meta-note for anyone else reading this: As far as I've seen, the Thirder logic results in credences of expected value that cannot be used without leading to losing bets (and depending on how it's justified, can lead to more extreme obvious wrong conclusions such as the presumptuous philosopher). Where as I've never seen any contradiction or absurdity to arise from the Halver position. I used to be a Thirder myself at one point, and for a long time I was highly uncertain about what the true answer should be. But the more and more I've looked at objections to the Halfer position, the more and more I've found it to be resilient, and the more I believe there is a true sense in which the Halfer position is correct and the Thirder position is incorrect, when you're talking about actual belief in the coin flip and not belief in something you know will be tallied more often with more wakings (e.g. guesses-over-repeated-experiments), and the that former is more useful thing to Beauty (i.e., more actionable; not able to be money-pumped) when she's asked unspecifically, "What do you think the coin flipped?"

[-]JeffJo2mo21

What observation are you trying to update on?

There are three die rolls where SB would be awake at thevmoment.

There are three where she would be asleep.

She is awake.

What is so difficult to see about this?

the Thirder logic results in credences of expected value that cannot be used without leading to losing bets

That depends entirely on how you evaluate the"bet." If it is evaluated once, over the two days of the experiment, "1/2" produces balanced bets and "1/3" produces unbalanced bets. If it is evaluated individually each time SB is asked the question, "1/2" produces unbalances bets and "1/3" produces balanced bets.

Please tell me why you think the question is not about the moment the question is asked, but instead includes another moment that may, or may not, happen.

[-]Steff2mo1-1

That depends entirely on how you evaluate the"bet."

I pick a random day to approach Beauty with an offer of a bet. If she guesses right, on Wednesday I'll give her $1.5, otherwise she'll give me $2. If she truly believes that the coin flipped Tails with a probability of 2/3, then she'll take this bet.

What is so difficult to see about this?

You're not making yourself clear. Yes, from the outside perspective of a randomly chosen day, there are different die rolls associated with whether Beauty is asleep or not on that day. How does that contradict the math I presented? Would you like to share some Bayesian equations of your own? Did you read my post about the Day 1 Objection, which I strongly suspect is very relevant?

Apologies if that comes off as rude, I'm just finding it difficult to respond usefully to this latest comment. ("What is so difficult to see about this?" isn't a helpful question - I could just as easily pose the same question back to you!)

[-]JeffJo2mo31

I pick a random day ...

You say you pick a random day, but you do not specify the set from which you pick. Is it {Mon, Tue}, {H+Mon, T+Mon, T+Tue}, or {H+Mon, T+Mon, H+Tue, T+Tue}? In each case, it is possible that the bet will not be offered, and possible that SB is asked for a credence that is never "tested" with a bet. (And please, don't say "Monday if Heads, and Monday or Tuesday if Tails." That's biasing the method to what the method is supposed to test.)

That is not how the problem works. You are manipulating the odds to make your answer look correct. This is why it does not represent the canonical version of the problem, where she is asked on each waking day, not a random one.

The bet must be offered on each waking day, and evaluated each time SB is asked fro a credence, to correctly represent her credence. If you do this, the experiment is a zero-sum game if and only if 2:1 odds are used. But I suspect you will disagree with me. Which is why I do not use betting arguments. Even though I know yours can't be right based on the issues mentioned above, I know I won't convince you with my betting scheme, either

You’re not making yourself clear.

I am. You are doing everything you can to avoid seeing it.

Would you like to share some Bayesian equations of your own?

I have described it well enough. But, since you insist:

In the canonical version, on any single day (and this includes Tuesday, after Heads), the sample space for that day includes four outcomes: {H+Mon, T+Mon, H+Tue, T+Tue}. Each has a prior probability - in Bayesian probability, this means a result of the experiment prior to evidence being considered - of 1/4. SB's evidence is that it can't be H+Tue, so the "new data" is the event {H+Mon, T+Mon, T+Tue}.

Bayesian probability says:

Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = Pr({H+Mon, T+Mon, T+Tue}|{H+Mon}) * Pr({H+Mon}) / Pr({H+Mon, T+Mon, T+Tue}}

Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = (1)*(1/4)/(3/4) Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = 1/3

And again, the mistake you make in your equations is ignoring a valid outcome of the experiment because it goes unobserved by SB.

You have ignored how I point out that the need to consider it is demonstrated by letting SB observe it in a way that distinguishes it from the others: Wake her on Tuesday, after Heads, but take her shopping instead of asking about the coin. The above equation clearly gives her credence, in the situations where she is asked.

And the point is that the correct Bayesian solution does not depend, in any way, on what would happen on Tuesday, after Heads. Only that she makes he same observation for any outcome in {H+Mon, T+Mon, T+Tue}, and uses Bayesian probability based on that observation.

Did you read my post about the Day 1 Objection, which I strongly suspect is very relevant?

Did you note that the problem asks about her credence based on her current circumstance, and that there are circumstances that differ? Circumstances that can exist regardless of whether SB is awake.

So Pr(Awake) is not 1. "Awake" will not happen on Tuesday, after Heads, which is a possible result.

Please, please, please try to see that I am not saying she might not be awakened over the course of the experiment. I am saying that there are circumstances, consistent with how the question is posed on a single day that is made independent of the other day by amnesia, where she will not be wakened.

And the point of the six-sided die version was to isolate the problem to just the circumstance where she is awake. With no ambiguity about one or two wakings, and how to handle them. That when SB is asked the question, there are exactly three possible faces that could be showing, and one of them is green.

[-]Steff2mo1-1

The bet must be offered on each waking day, and evaluated each time SB is asked fro a credence, to correctly represent her credence. If you do this, the experiment is a zero-sum game if and only if 2:1 odds are used. But I suspect you will disagree with me. Which is why I do not use betting arguments.

I actually think this might be useful to drill into further. Why couldn't SB use her credence of the coin for the bet I described? Are you thinking that SB actually has two different credences about the coin, and which one she applies will depend on the exact question asked?

You have ignored how I point out that the need to consider it is demonstrated by letting SB observe it in a way that distinguishes it from the others: Wake her on Tuesday, after Heads, but take her shopping instead of asking about the coin.

Hoping for a clarification with this one: Do you mean that she goes shopping on Tuesday and is told on Tuesday? Or that a random day was chosen for her to go shopping, and she knew that it'd be a random day? I think I would agree with your presented math depending on exactly what she's learning.

Okay I'm getting tired of fighting the editor, going to quote differently:

>> That when SB is asked the question, there are exactly three possible faces that could be showing, and one of them is green.

I'm not denying that those are the possibilities, but that they share equal probabilities.

>> Did you note that the problem asks about her credence based on her current circumstance, and that there are circumstances that differ? Circumstances that can exist regardless of whether SB is awake.

>> So Pr(Awake) is not 1. "Awake" will not happen on Tuesday, after Heads, which is a possible result.

Unfortunately I'm not following this part. P(Awake on Tues) is not 1, but that's not what she's observing, she's observing being awake, which she already knew would happen.

[-]JeffJo2mo*21

Why couldn’t SB use her credence of the coin for the bet I described?

First, I'm still waiting for you to fully describe the random day. Is it chosen from a set of two, three, or four?

But to answer your question, she can. On the inside of the experiment, where there are three equivalently-placed choices. It's a common justification for 1/3. You can't, on the outside, where there is no way to make a random choice that is equivalent. It's a different experiment than the one where she is asked about her credence.

Do you mean that she goes shopping on Tuesday and is told on Tuesday?

On Monday, and on Tuesday after Tails, she is asked about her credence. On Tuesday after Heads, she goes shopping. She doesn't have to be told anything in this case, since her credences that it is Tuesday, and that the coin landed Heads, will both be 100%. This takes what some want to call "probability mass" away from her Sunday-Night credence in Heads. That was 1/2, and removing probability mass is what reduces it to 1/3.

And the point is that IT DOES NOT MATTER what would happen on H+Tue. That is still a day in the experiment, just as valid a result as T+Mon. If you do this, the proper odds for the bet if she is offered one are 2:1.

I’m not denying that those are the possibilities, but that they share equal probabilities.

Please justify how, without using the circular argument that it is the only way to get the answer you want.

But you gave me an idea, to alter the "shopping" version to the red/green die version.

Wake her on all three days. If the result was a green number matching the current day, or a red number not matching it, ask her for her credence in green. If it was a red number matching the current day, or a green one not matching it, ask her for her credence in red.

Before she is asked, there are six-equally likely faces (which contradicts your claim about them being unequal). If she is asked about green, it is reduced to three (still equally likely) faces, where one is green. There is absolutely no difference between these "ask about green" cases and the "awake" cases from before.

I’m not following this part.

And it is what I keep trying to describe. You, on the outside, see a two-day experiment where SB is wakened once, or twice. To SB, it is a one-day sample of that experiment, but limited to the three equally-likely situations where she will be awake. When the sun rose this morning (remember that it is a one-day sample), there were four possible combinations. Three of them, still equally likely, result in waking her. Her evidence is that it is one of these three, and not H+Tue.

Pr(Awake) vs. Pr(Awake on Tues) is awkward because you are confusing two-day and one-day probabilities. So I'll use the shopping variant. Instead of "Awake" I'll use "Ask" and "Shop."

In the two-day experiment - that is, before the coin flip on Sunday and anticipating the next two days - Pr(Ask)=1 and Pr(Shop)=1/2. Note that these are not independent events, so these don't add up to 1. When SB awakes, but before she learns what will happen today, Pr(Ask)=3/4 and Pr(Shop)=1/4. These are independent events that represent all possibilities, so they do add up to 1.

BOTH OF THESE ARE "PRIOR" PROBABILITIES, meaning they describe the possibilities with no evidence. What you are "not following," is that the probability space for SB's credence is the second one, not the first.

Your Bayesian analysis is wrong because it is based on the first. Pr(Awake)=1 means she will be awake at some time over two days. This is why you think you can choose one day at random from what I suspect you mean the two "days" Monday and Tuesday_If_Tails.

SB's credence is about one day only. Its prior sample space (remember, no evidence) is {H+Mon,T+Mon, H+Tue,T+Tue} with a 1/4 probability for each. Her evidence is that H+Tue is eliminated, either because she is awake of because she didn't go shopping.

Your Bayesian analysis "folds" H+Tue into H+Monday because that is the only way she is awake after Heads.

[-]Steff2mo10

I'm hoping to find the time for a full response later, but right now:

>> First, I'm still waiting for you to fully describe the random day. Is it chosen from a set of two, three, or four?

It's a random choice among the days she'll be awoken. If she's to awake just Mon, then she'll be offered on Mon. If she's to awake on Mon and Tues, then a coin is flipped.

[-]JeffJo2mo10

You are right, I didn't look closely at your betting argument, because (as I have said) betting arguments are invalid. They can be manipulated - as you do - to get the answer you want.

I pick a random day to approach Beauty with an offer of a bet. If she guesses right, on Wednesday I’ll give her $1.5, otherwise she’ll give me $2. If she truly believes that the coin flipped Tails with a probability of 2⁄3, then she’ll take this bet.

You are saying the odds (easier to work with for betting) you offer are 1.5:2 for Tails (representing probability 4/7), and so should be 2.67:2 (representing 3/7) for Heads. If Beauty thinks they are 1:2 and 4:2 (2/3 and 1/3), Beauty should indeed take this bet, on Tails.

But if this is what you mean, you did not do what claimed to do. You did not puck a random day, you choose a random timeline and inserted a bet into a waking moment in that timeline. She may as well have chosen the bet on Sunday.

By flipping the coin on Sunday and resolving the bet, once, on Wednesday, you have manipulated the wager. What happens in between - all the sleeping, waking, and amnesia - is completely irrelevant. So what is the point of what happens in between? Isn't the entire point of the thought experiment to evaluate how, and yes, if, it matters? Yet you designed your question to eliminate any issue, not to resolve it.

Try this: actually choose a random day for the offer, but use her "thirder" odds. Note that now there is a chance that no bet will be offered, if the random day is Tuesday and the coin landed on Heads.

If the coin landed Tails, Beauty will be offered a wager. She will win $1 if she bet on Tails (at 1:2), and lose $2 if she bet on Heads. But if the coin landed Heads, she will lose an expected $1 if she chooses Tails when the bet is offered, and win an expected $2 (at 4:2) if she chooses Heads when the bet is offered.

In other words, if we remove the bias you built into your challenge, Beauty expects to wager $2 and a either choice wins $1 if correct.

What if you have Beauty put $2 into each of two identical envelopes? Each time she is awake, we give her one and she chooses either Tails (at 1:2 odds) or Heads (at 4:2). We again evaluate on Wednesday, but this time the number of bets evaluated can vary.

If the coin lands Tails, she is up $2 if she chooses Tails (two wins of $1 since we assume she is consistent over the two days) and down $2 if she chooses Heads. If it lands Heads, she is down $4 if she chooses Tails (two losses of $4) and up $4 if she chooses Heads. This is a balanced wager even though the balance is differs depending on the result.

But it is the differing balance that halfers object to. It is an invalid objection, but again it is why I try to avoid betting arguments altogether.

+++++

It’s a random choice among the days she’ll be awoken. If she’s to awake just Mon, then she’ll be offered on Mon. If she’s to awake on Mon and Tues, then a coin is flipped.

Sorry, I somehow confused your similar response to another comment with this one. This models a very different experiment. One where Beauty is woken exactly once regardless of the coin. It happens on Monday if the coin landed Heads. Or on Monday or Tuesday, according to the second coin, if the first coin landed on Tails.

[-]Vanilla_cabs2mo2-1

I have read your blog post. You are aware of the traditional arguments for both sides. You also agree that halfers and thirders answer slightly different questions.

If you want to maximize the number of times you answer correctly, go Thirder.
If you want to maximize the number of flips you guess correctly, go Halver.

You even say yourself that the halfers answer a version of the problem where Tuesday doesn't matter:

To test Beauty’s accuracy at guessing coin flips, we should see what she says for every Heads, and see what she says for every Tails. It doesn’t make sense to ask her the same question again for every Tails waking, because [...] we already know her answer.

And since Tuesday doesn't matter, the whole protocol (sleeping, waking up, the amnesia) is unnecessary: it comes down to guessing the result of a fair coin toss. Which is my point: halfers answer a trivial version of the problem. Same for your "unfair coin" variant.

I'll use what I read from you post to speculate a little, so feel free to correct me. If I had to guess, I think our disagreement might come from 2 points:

- the version of the SB problem that you favor is Adam Elga's, which asks "When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?" Which indeed has 1/2 as its unique answer, but is trivial.

- maybe you view the fair coin has having an intrinsic 1/2 chance of landing heads. I view that all probabilities are in a person's head (nothing original, I just absorbed the standard Bayesian view as taught in the Sequences on LW, see https://www.lesswrong.com/posts/f6ZLxEWaankRZ2Crv/probability-is-in-the-mind for example). A probability expresses a state of knowledge at a given moment, not an intrinsic property of a physical object.

You said it yourself in your blog post when examining the thirder version:

The questions ask about degrees of belief.

I am guessing that to you, it's an exception. To me, there is nothing else: all probabilities are degrees of belief.

I discovered the SB problem through the version deemed canonical by wikipedia, which is different from Elga's:

Any time Sleeping Beauty is awakened and interviewed she will not be able to tell which day it is or whether she has been awakened before. During the interview Sleeping Beauty is asked: "What is your credence now for the proposition that the coin landed heads?"

In that version, each day is treated equally.

And to that version, my answer is: when SB wakes up, the coin has a 1/3 chance to have landed up heads. You can verify it simply (and you did in your post) by remarking that if you iterate the experiment a large number of times, 1/3 of the wake-up events happen after a heads toss.

I might be unfair, but I have noticed that, faced with such a version that does not trivially reduce to a coin toss, you and other halfers make this weird dance where you say that SB (waking up) ought to believe that the coin has a 1/3 chance to have come up heads, but really, really, in the real world the coin has a 1/2 chance to have come up heads. As if the point of forming beliefs wasn't to accurately model the real world.

Adding a betting structure, at best, doesn't change anything (at worst it can give wrong intuitions): the reason why betting 1/3 is the best (under a well-formed betting scheme) is because it is the actual probability.

You already know a lot about the arguments of the thirders. This here is just my point of view, but hopefully, it helps fill some of the gaps.

[-]Steff2mo10

In that version, each day is treated equally.
And to that version, my answer is: when SB wakes up, the coin has a 1/3 chance to have landed up heads. You can verify it simply (and you did in your post) by remarking that if you iterate the experiment a large number of times, 1/3 of the wake-up events happen after a heads toss.

I'm afraid that's incorrect, actually.

If you repeat the experiment, one half of the time Beauty will be in the Heads world and one half of the time Beauty will be in the Tails world. In Tails world, she'll have twice as many wakings. If we want to measure her accuracy of credence of coins, and not measure her accuracy of credence of her own guesses, we should take the average of her answers in Tails worlds. To do so otherwise is to double count Tails answers and make them matter more when they shouldn't.

This can seem confusing, or seem like we're just arguing semantics. However, I think this becomes clearer by analogy:

Credence is something you should be able to act on.

Let's say after every waking, money will be deposited to Beauty’s bank account (without her knowledge). The amount that is deposited depends on the coin flip: On Heads, she will awake on Monday and receive $30; on Tails, she will awake on both Monday and Tuesday and receive $6 each day. What should Beauty believe is her expected value gain?

If Beauty answers with 1/3, she'll get the wrong answer and she can be convinced to take bad bets and lose money over repetitions of the experiment.

(I mad a little table for the options here - https://ramblingafter.substack.com/i/181017019/problem-seven )

[-]Vanilla_cabs2mo20

I took a look at your table, and I have to say, it was a great variant. It took me a while to de-confuse my thoughts about it, and made me realize I wasn't as comfortable with the SB problem as I thought.

Initially I was a bit uncertain what you meant by "round", whether it was a wake-up event of a full iteration of the experiment. After rereading a few times, I became more certain that it was the latter.

So, you gain $30 if the coin lands heads and $6 + $6 = $12 if it lands tails. So your expectation of gain per iteration of the experiment is 1/2 * $30 + 1/2 * $12 = $21

There are 3 equally likely wake-up events, so the expectation of gain per wake-up event is 1/3 * $30 + 1/3 * $6 + 1/3 * $6 = $14

We can notice that there are on average 1.5 wake-up events per iteration, and $14 * 1.5 = $21, so that checks out.

Now, let's say I experience a wake-up event. I believe that there is 1/3 chance that the coin landed heads. You think that my calculation for the expected gain per iteration will be 1/3 * $30 + 2/3 * $6 = $14

However, I reason: if the coin landed tails, I won't gain $6, I will gain $6 * 2, once for Monday and once for Tuesday. So my expected gain per iteration is 1/3 * $30 + 2/3 * $6 * 2 = $18

That's the point where I was confused. What did that $18 mean? What actually helped me figure it out was the betting scheme that you added in your post.

Sleeping Beauty is put to sleep like normal with only the usual information supplied to her about the original SB problem. When she awakes, there is a surprise message left for her:
You may agree to play the following game: If the coin had flipped Heads, your bank account will receive $30 right now. If the coin had flipped Tails, your bank account will receive $6 right now. However, once this round of experiment is over and you awake, you will pay $18 for having played this game. (If you are inconsistent between wakings about whether you agree to play the game, the game is considered defunct and all bank transfers will be reverted.)

This betting scheme is a based on a version of the SB problem where the Monday guess and the Tuesday guess are counted as a single one. So, Tuesday is unnecessary, so (see previous comment) it boils down to the trivial prediction of the result of a fair coin toss. $21 - $18 = $3 gain expectation.

Next I tried a few variations:

if the bets are per wake-up (SB can bet once on Monday and another time on Tuesday), but SB is only paid for her current situation (if she bets on Monday she gains $6 and pay $18, same for Tuesday), then her gain expectation is 1/3($30 - $$6 - $18) = -$4. In other words, it's $14 (average gain expectation per wake-up event) - $18 (cost to bet per wake-up event).
if the bets are per wake-up, but SB is paid for both Monday and Tuesday when she bets on either Monday or Tuesday, then her gain expectation is 1/3($30 - $$12 - $18) = $0. So that's what the $18 found with the thirder calculation meant: it's the gain expectation, over the whole iteration, but knowing that she's in a wake-up event. In hindsight, I should have guessed, because it naturally corresponds to the formulations where the thirder answer is correct. It is not meaningless: if you repeat the experiment a large number of times, and then tag each wake-up event with the profit that SB made during the iteration that included that event, then the average of tags will be $18. In other words, when SB's Monday and Tuesday answers are not conflated in a single one, it does answer the question (to SB during a wake-up event): how much money do you expect to gain during this iteration? It's not intuitive, but it checks out.

So, cool variation, but ultimately each interpretation finds the result that it expected.

[-]Steff21d20

Sorry for taking so long!

$14 was a mistake, that's my bad and I've updated the post.

>> This betting scheme is a based on a version of the SB problem where the Monday guess and the Tuesday guess are counted as a single one

This is the piece I don't understand about thirders (apologies if that sounds like aggressive wording): If you have a credence for the coin, why can't you use it?

>> (your second variant) SB is paid for both Monday and Tuesday when she bets on either Monday or Tuesday... if you repeat the experiment a large number of times, and then tag each wake-up event with the profit that SB made during the iteration that included that event, then the average of tags will be $18

Not exactly. I laid out a separate $6 on Mon and $6 on Tues to make this easier to think about. Combining into $12, you end up with the same amount of money being equally shared between two days. I don't think it's accurate to say that SB made $12 on Mon and also say that SB made $12 on Tues. So in other words, I think this scheme is doing double counting. Which also means to me that the $18 value doesn't really correspond to anything relevant and useful to the scenario?

[-]Vanilla_cabs20d20

This is the piece I don't understand about thirders (apologies if that sounds like aggressive wording): If you have a credence for the coin, why can't you use it?

No problem, I am often in that situation too of being puzzled by the other side of an argument :)

And, it's not like I'm confident that I fully understand the SB problem.

I'm guessing that the crux is this: for me, if (let's call it experiment 1) you interrogate SB once after the coin landed tails, she should say her credence (of heads) is 1/2. But if (experiment 2) you interrogate her twice (under amnesia), she should say her credence is 1/3. You seem to disagree with that, on the grounds that her answer means something else than what the "SB problem" is about? But if you run the experiments many times, in experiment 1 half of SB's answers will happen after a heads outcome, while in experiment 2 it's only a third, so it means something.

I don't think it's accurate to say that SB made $12 on Mon and also say that SB made $12 on Tues. So in other words, I think this scheme is doing double counting. Which also means to me that the $18 value doesn't really correspond to anything relevant and useful to the scenario?

The $18 value corresponds to how much money SB expects to make during this iteration on average.
I'm curious about your answer to the following experiment:

No coin. SB is put to sleep on Sunday, awoken on Monday, interrogated, and put to sleep+amnesia. Then the same happens on Tuesday: awoken, interrogated, and put to sleep+amnesia.
When interrogated (on Monday and Tuesday), they tell her: "Here is $10, how much do you think you will have gained at the end of the experiment on Wednesday?"
What should be her answer?
If you answer $20, do you see how the same reasoning is part of my $18 answer on the previous problem?

[-]Steff19d10

I take back one piece, when I said $18 doesn't correspond to anything. I believe that corresponds to how much Beauty should expect to be making per day on average: (1/2)(30) + (1/2)(6).

If we're talking about expectation per iteration, though, then it should be (1/2)(30) + (1/2)(12) = $21. If you're using the $18 value there, then you can get money pumped.

[-]Vanilla_cabs19d20

Let's be sure we agree on the parameters of the bet. If SB agrees to bet, she pays a fee, then:

she gains $30 if the coin came up heads.
she gains $12 if the coin came up tails.

And the question is: How much should she be willing to pay to bet? (assuming she's ok to break even)

Answer: depends on when you ask SB.

If you propose the bet to SB on Sunday or Wednesday, she should be willing to pay $21.

If you propose the bet during a wake-up event, she should be willing to pay only $18. That's the meaning of the $18. The difference comes from the fact that when she wakes up, the coin has only 1/3 chance of having landed heads. That's where the thirder position comes from. Test it if you don't believe it checks out.

[-]Steff18d10

Over many repetitions (to balance out bad luck streaks), it's rational to play the game for prices of $20 or $19. If you're saying that SB should reject those prices when presented it while waking, that's irrational, because an SB that accepts those prices will make money.

This is assuming that if asked twice, she'll always make the same decision because her decision-making is deterministic. (We could assume there's a chance she makes a different decision on Tuesday than on Monday but I think that's an unnecessary and irrelevant complication?)

[-]Vanilla_cabs18d10

Just to be sure: in your original formulation, the bet counted only once for the two wake-up events in the Tails case, and it was canceled if she didn't accept it on Monday and Tuesday.

That is not the scheme I described in previous comment. Each wake-up event is independent. She is awoken then asked to pay the fee now and gets the rewards now. If she pays twice she gets the rewards twice. It's a bet that can be done, it is no less valid than yours.
In this bet (when asked during a wake-up event), her gain expectation is $18. Accepting for $19 will lose money on average. Her decision-making process on the other day is irrelevant. Test it if you don't believe it.

[-]Steff2mo20

Thanks for taking a look! I'll be sure to go over what you've written - I ran out of time to spend any more time on this stuff for this weekend / this week so I won't be able to reply soon, just wanted to let you know that your response is appreciated

[+][comment deleted]2mo10

[-]JeffJo1mo10

I have liked this scenario ever since I found out that there are three Michelin 3-star restaurants in San Francisco. And that their names (almost) line up in a convenient order. They are Atelier Crenn (which I will call "A"), Benu ("B"), and Quince ("C" to fit in the order). (All of this is irrelevant, except it helps the narrative.)

I will use them to distinguish H&Mon, T&Mon, and T&Tue in Beauty's world, without affecting her credence in any way. Each will be randomly assigned one of these combinations at the same time the coin is flipped. When Beauty is awakened, she will be taken to the restaurant assigned to that day and coin result, and asked for her credence that the coin landed Heads. The point of all this is that there is no useful information in whether she dines at "A", "B", or "C"

There are also two important times in this process: T1 is after Beauty is woken but before the limo leaves to take her to a restaurant. And T2 is after dinner. So far, this is identical to the canonical problem. Since Beauty has no information about the assignments, it cannot matter if she is asked the question at T1 or T2. This IS the canonical Sleeping Beauty Problem regardless of the timing.

But I will now propose two variations. In Variation 1, Beauty is not woken on H&Tue, as in the canonical version. In Variation 2, she is told in advance that she will be woken both days. The same randomization applies to those three combinations, and Denny's ("D") is always assigned to H&Tue.

At V1T1 (that is, time T1 in Variation 1) or V1T2, there seems to be controversy about what credences Beauty should assign to each member of the sample space {H&Mon, T&Mon, T&Tue}. So I'll delay discussing it.

At V2T1, there can be no controversy about the credences for the sample space {H&Mon, T&Mon, H&Tue, T&Tue}. Each has a credence of 1/4. At V2T2, Beauty has received "new, centered evidence" about the sample space. If she is dining at "D", she knows the coin landed on Heads. If she is at "A", "B", or "C", she can use that "new, centered evidence" to eliminate H&Tue. She can update her credences for H&Mon, T&Mon, or T&Tue from 1/4 each to 1/3 each.

My point in this discussion is that the only difference between the two variations is when Beauty receives this information. The information itself is the same. In Variation 1, she receives it at time T1, when she is woken up. The equiprobable 1/4 credences apply to the time T0, when the experimenters are deciding if she should be woken. BEAUTY KNOWS THIS TIME WILL HAPPEN, AND WHAT INFORMATION APPLIES TO IT, EVEN IF SHE DOES NOT EXPERIENCE IT. In Variation 2, she receives it at time T2. Regardless, it is not her ability to perceive the state that determines her credence. It is the knowledge she gains - between T0 and T1 in Variation 1, or between T1 and T2 in Variation 2 - that constitutes "new, centered evidence" for the purpose of the update.

The answer to canonical Sleeping Beauty Problem is 1/3. Halfers confuse the perception of the information with the knowledge of when it is made evident.

[-]Steff2mo10

I don't like the verb "observes" since it can be confused between the notion of an observation that's happening right now (still 1/2) versus tallying over all total observations over repeated experiments (therefore 1/3), which is why I personally prefer the noun "guesses" for distinguishing them.

[-]Sarenard2mo31

Fun fact on the rolling dice part: even mathematicians as well known as Leibniz have made this error (see https://www.filosofie.info/wp-content/uploads/on-probable-grounds1.pdf, page 50: “for example, with two dice, it is as doable to throw a twelve as to throw an eleven for each can only be done in one way; but it is three times more doable to throw seven, for that can be done by throwing six and one, five and two, and four and three, and one of these combinations is as doable as the other”).

Probability is really hard to understand at an intuitive level !

[-]JeffJo26d10

The following is what no Halfer has ever responded to, except to say "Gee, that might be interesting to discuss. We'll get back to it." They never do.

The experiment:

Beauty will be put to sleep on Sunday Night, and a fair coin will be flipped. On each of the next two days (Monday and Tuesday), she will participate in one of two different procedures, named A and B. There are two requirements for them: #1: During either procedure, Beauty will be unaware of what may have, or will have, happened on the "other" day. #2: In procedure A she will be asked a probability (or credence, if you prefer) question about the coin flip. But no such question will be asked in procedure B.
On Monday, Beauty will participate in procedure A. On Tuesday, she will participate in procedure B if the coin landed on Heads, and in procedure A is it landed on Tails.
The question, that is only asked in procedure A, is "What do you believe is the probability that the coin landed on Heads?"

This is a simple conditional probability question. There are four situations during the experiment when Beauty could participate in a procedure. Before she is - or, if she won't be - asked the question, Requirement #1 makes her current situation an independent sampling into that set of four possibilities. If she is asked the question, Beauty knows that she is participating in procedure A, and so it is either Monday after Heads, Monday after Tails, or Tuesday after Tails. The probability that the coin landed on Heads is 1/3. If she is not (yet) asked, well, does it really matter?

The popular version of the Sleeping Beauty Problem fits this outline. Halfers don't want to acknowledge two facts about that implementation of my outline. First, that "new Information" does not mean that something happened that was not guaranteed to happen, or is all that Beauty can perceive. It means that something happened that eliminates a possibility from the sample space. Second, that the possibility of something happening on Tuesday makes everything that could happen on Tuesday a member of that sample space. EVEN IF SHE WOULD SLEEP THROUGH IT, Beauty knows that the events of Tuesday, after Heads, must be represented is the sample space.

There is no probability issue in the Sleeping Beauty Problem, just obfuscation about what belongs in a sample space.

[-]Steff21d10

I believe what you've described is just the original problem again with an extra possible waking of no question asked on Tuesday, but that doesn't impact the credences of her waking and observing the question. The answer is still 1/2.

[-]JeffJo5d10

<Sigh>. It is very difficult to get halfers to understand that "Monday, after Tails" and "Tuesday, after Tails," ARE DIFFERENT OUTCOMES. So are "Monday, after Heads" and "Tuesday, after Heads." All four are outcomes, even the one Beauty sleeps through. Yes, each pair occurs in the same timeline, but Beauty's amnesia makes the probability experiment be, to her, a sampling of the four possible situations. Monday, after Tails and Tuesday, after Tails, are independent outcomes. I can make that clearer, I just can't make halfers accept it.

This is a repeat of something I posted previously. But, as I predicted, that was ignored. Suppose there are three gourmet restaurants in town, randomly (and unknown to Beauty) designated "A", "B", and "C". Call Denny's "D". {A,B,C} are associated, in that order, with {T1,T2,H1}. D is associated with H2. Beauty is wakened both days (1 and 2, if that wasn't clear), and will be taken to the assigned restaurant.

Before being taken to a restaurant, Beauty can relate the sample space {A,B,C,D} with the probability distribution {1/4,1/4,1/4,1/4}. She can also use the equivalent sample space {T1, T2, H1, H2}, because she doesn't know which restaurant is which (except Denny's).

Steff's answer above uses two events, not four. The sample space is {A&B,C&D} or {T1&T2,H1&H2}. The probability distribution is {1/2,1/2}.

When Beauty arrives at a gourmet restaurant (remember, she doesn't know if it is A, B, or C), she can update the distribution for the four-event sample space to {1/3,1/3,1/3,0}. If I understand Steff's answer ("an extra possible waking ... doesn’t impact the credences of her waking and observing the question"), the two-event distribution is still {1/2,1/2}. But that can't be - Beauty knows she could be at C, but can't be at D. That it could be H1, but not H2.

The Tails path works differently. It could be A, or B, but not A&B. It could be T1, or T2, but not T1&T2. Most Halfers want us to believe that the events T1, T2, and T1&T2 are all the same event, with the same probability of 1/2.

I'm sorry, that is wrong. The probabilities for {A,B,C,D}={T1,T2,H1,H2} update to {1/3,1/3,1/3,0}. Steff's statement that the extra possible waking can't impact anything is wrong.

And my point from before is the nothing in this calculation changes if Beauty is left asleep on H2, but sees that she has arrived at a gourmet restaurant. Or is simply wakened and questioned. "New information" is an informal term in probability that, while not defined (because it is informal), does not mean "what happened was not guaranteed to happen." It means that we learned something which depends on certain elements of the probability experiment happening, and rules out others.

In the canonical version of the problem, Beauty knows that she will not be awake during certain portions of the experiment. That doesn't remove that outcome from the experiment, as Halfers imply, it removes its probability from the distribution requiring an update.

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