In the shiny-penny case, fair pennies have a 1/2 chance to land Tails, but Tails pennies are twice as likely to be noticed. So among the pennies I actually notice, about 2/3 will be Tails. When I notice this penny, updating to (2/3) for Tails isn’t smuggling in a mysterious propensity; it’s just combining:

1) the base chance of Tails (1/2), and
2) the noticing rates (Tails noticed twice as often as Heads). — Pierre-Normand

You appear to be affirming the consequent. In this case, Tails is noticed twice as often because Tails is twice as likely to be noticed. It doesn't then follow that Tail awakenings happen twice as often because Tails awakenings are twice as likely to happen.

The Sleeping Beauty case in contrived in such a way that a Heads awakening is guaranteed to happen and two Tails awakenings are guaranteed to happen. This contrivance doesn't allow you to compare the likeliness of a Tails awakening compared to a Heads awakening.

1) Per run: most runs are 'non-six', so the per-run credence is P(6)=1/6 (the Halfer number).
2) Per awakening/observation: a 'six-run' spawns six observation-cases, a 'non-six' run spawns one. So among the observation-cases, 'six' shows up in a 6/5 ratio, giving P('six'|Awake)=6/11 (the Thirder number). — Pierre-Normand

This doesn't make sense.

She is in a Tails awakening if and only if she is in a Tails run.
Therefore, she believes that she is most likely in a Tails awakening if and only if she believes that she is most likely in a Tails run.
Therefore, her credence that she is in a Tails awakening equals her credence that she is in a Tails run.

You can't have it both ways.

Since she is only being rewarded with £100 for each sequence of six successful bets — Pierre-Normand

This isn't what's happening. There is only a single bet, placed before she is put to sleep. She is then given a 3 hour window in which she is able to change her bet, and can do so as many times as she likes. The same for Prince Charming, although he is never put to sleep.

In this situation, if either of their credences in the outcome genuinely changed to favour the die landing on a 6 then they would change their bet. Prince Charming does this when he learns that his die is loaded. So why doesn't Sleeping Beauty after having her memory wiped? Because despite Thirder word games, her credence in the outcome hasn't genuinely changed. She continues to know that if she changes her bet then she is most likely to lose.

If it helps, it's not a bet but a holiday destination. The die is a magical die that determines the weather. If it lands on a 6 then it will rain in Paris, otherwise it will rain in Tokyo. Both Prince Charming and Sleeping Beauty initially decide to go to Paris. If after being woken up Sleeping Beauty genuinely believes that the die most likely landed on a 6 then she genuinely believes that it is most likely to rain in Paris, and so will decide instead to go to Tokyo.

So, there are three "events" at issue: the coin toss, that occurs before the experiment, the awakenings, and the runs. — Pierre-Normand

But again, the paradox is only a paradox if the $H$ in $P(H|Awake) = {1\over3}$ denotes the same event as the $H$ in $P(H) = {1\over2}$.

The paradox is: this awakening gives me reason to believe that this coin toss most likely landed on a tails.

If this claim is false then Halfers are right and Thirders are either wrong or equivocating.

So i guess to increase her odds, she bets tails 100% of the time since she can't remember which phase of the experiment she's in, and the 2/3rds tailsers make a profit off the gambling? That's the thing: we need to talk about the drugs. How many times does she participate in the experiment, is it just one run through?

Do coins change their heads/tails bias based on the number of times tossed?

So i guess to increase her odds, she bets tails 100% of the time since she can't remember which phase of the experiment she's in, and the 2/3rds tailsers make a profit off the gambling? — ProtagoranSocratist

It doesn't increase her odds but it does increase her expected return in the long run.

On this point it's worth considering an extreme example I provided two years ago.

I am put to sleep and a coin is tossed 100 times. If it lands heads every time then I am woken up, interviewed, and put back to sleep 2¹⁰¹ times, otherwise I am woken up, interviewed, and put back to sleep once.

When being interviewed, I am asked a) my credence that the coin landed heads every time and b) to place a bet on the outcome.

All of these are true:

1. If I know that the experiment will be performed once
a. My credence is ${1\over{2^{100}}}$
b. I will bet that the coin did not land heads every time

2. If I know that the experiment will be performed 2¹⁰¹ times
a. My credence is ${1\over{2^{100}}}$
b. I will bet that the coin did land heads every time

I strongly believe that a perfectly rational agent like Sleeping Beauty will believe and do the same. Thirder reasoning seems to be that if (2b) results in twice as many successful bets then (1a) is false, and that simply doesn't follow, either for me or for Sleeping Beauty.

but doesn't the first coinflip is every phase of the experiment matter the most if it lands on heads, because then it ends? I understand the 1/3rd logic, but it simply doesn't apply here: the third flip, given the first two were heads (less likely than one tail and a head, but still very likely), is also unaffected by the other flips. You can't win with gambling logic here, as it never what phase the experiment is in, it's always 50/50. 1/3 argument is like arguing that heads is 2/3rds as likely because it ends the experiment...

I understand the 1/3rd logic, but it simply doesn't apply here: the third flip, given the first two were heads (less likely than one tail and a head, but still very likely), is also unaffected by the other flips. — ProtagoranSocratist

There is no third flip. The coin is only tossed once. When it lands Tails, Sleeping Beauty is awakened twice and when it lands Heads, she is awakened once. She also is being administered an amnesia inducing drug after each awakening so that she is unable to infer anything about the number of awakenings she may be experiencing from her memory, or lack thereof, of a previous awakening episode. It might be a good idea to either reread the OP carefully, or read the Wikipedia article on the problem: especially the description of the canonical form of the problem in the second section titled "The problem".

(For the record, my own "pragmatist" solution is an instance of what the Wikipedia article, in its current form, dubs the "Ambiguous-question position", although I think the exact formulation of this position in the article is faulty/misleading.)

Right. And this is they get the wrong answer, and have to come up with contradictory explanations for the probabilities of the days. See "double halfers." — JeffJo

Let me just note, for now, that I think the double halfer reasoning is faulty because it wrongly subsumes the Sleeping Beauty problem under (or assimilates it with) a different problem in which there would be two separate coin tosses. Under that scenario, a first coin would be tossed and if it lands Heads, then SB would be awakened Monday only. If it lands Tails, then a second coin would be tossed and SB would still be awakened Monday only if it lands Heads and be awakened Tuesday only if it lands Tails. Such a scenario would support of straightforward Halfer interpretation of SB's rational credence but it's different from the original one since it makes Monday-awakenings and Tuesday-awakenings mutually exclusive events whereas, in the original problem, SB could be experiencing both successively though not at the same time. The different awakening generation rules yield different credences. (I haven't read Mikaël Cozic's paper, where the double-halfer solution is being introduced, though.)

Why? How does something that is not happening, on not doing so on a different day, change her state of credence now? How does non-sleeping activity not happening, and not doing so on a different day, change her experience on this single day, from an observation of this single day, to an "experimental run?"

You are giving indefensible excuses to re-interpret the experiment in the only way it produces the answer you want. — JeffJo

Well, firstly, the Halfer solution isn't the answer that I want since my own pragmatist interpretation grants the validity of both the Halfer and the Thirder interpretations, but denies either one being the exclusively correct one. (I might as well say that Halfers and Thirders both are wrong to dismiss the other interpretation as being inconsistent with the "correct" one, rather than acknowledging its being incompatible but complementary.)

With this out of the way, let me agree with you that the arbitrary stringing up of discrete awakenings into composite experimental runs doesn't affect the Thirder credence in the current awakening being a T-awakening (which remains 2/3). However, likewise, the arbitrary breaking up of discrete experimental runs into several opportunities for being interviewed doesn't affect the Halfer credence in the current run being a T-run (which remains 1/2). The mistake that both Halfers and Thirders seem to make is to keep shouting at each other: "Your interpretative stance fails to refute my argument regarding the validity of my credence estimation." What they fail to see is that they are both right and that the "credences" that they are taking about are credences about different things.

There is no third flip. The coin is only tossed once. When it lands Tails, Sleeping Beauty is awakened twice and when it lands Heads, she is awakened once. — Pierre-Normand

okay, thanks for clearing that up, as i read the original description of the fake experiment more than once, and that part of it was unclear to me. To me it was saying that heads ends the experiment right away, tails produces two more flips. Maybe it's because i was reading it at work, but i doubt it, i remember the experiment confusing me, and i remember the article saying that it was confusing.

Still: the effects of one flip never effect the outcome of the other FLIPS, unless that is baked into the experiment, so it is a misleading hypothetical question (but interesting to me for whatever reason). The likelihood of the flips themselves are still 50/50, not accounting for other spooky phenomenon that we just don't know about. So, i'll think about it some more, as it has a "gamey" vibe to it...

Here's what would effect the outcome to skew the bias slightly in the tails direction: let's say the experimentor gives her the drug, keeps flipping the coin while she sleeps, and then wakes her up on the condition that there are 2/3 tails on the last 3 tosses...then asks "what was the last coin flip?". There would be exactly 1/3 odds of it being heads...but as it stands, even the "correct" way you describe it, i still can't side with 1/3 camp.
I guess at this point it's a game to see how long it will take before i get frustrated with talking about this, so go on...

Still: the effects of one flip never effect the outcome of the other FLIPS, unless that is baked into the experiment, so it is a misleading hypothetical question (but interesting to me for whatever reason). The likelihood of the flips themselves are still 50/50, not accounting for other spooky phenomenon that we just don't know about. So, i'll think about it some more, as it has a "gamey" vibe to it... — ProtagoranSocratist

There are no other flips. From beginning to end (and from anyone's perspective), we're only talking about the outcome of one single coin toss. Either it landed Heads or it landed Tails. We are inquiring about SB's credence (i.e. her probability estimation) in either one of those results on the occasion where she is being awakened. The only spooky phenomenon is her amnesia, but that isn't something we don't know about. It's part of the setup of the problem that SB is being informed about this essential part of the protocol. If there were no amnesia, then she would know upon being awakened what the day of the week is. If Monday (since she wouldn't remember having been awakened the day before) then her credence in Tails would be 1/2. If Tuesday (since she would remember having been awakened the day before) then her credence in Tails would be 1 (i.e. 100%). The problem, and competing arguments regarding what her credence should be, arise when she can't know whether or not her current awakening is the first one.

(Very roughly, Halfers argue that since she is guaranteed to be awakened once in any case, her being awakened conveys no new information to her and her estimation of the probability that the coin landed Tails should remain 1/2 regardless of how many times she is being awakened when the coin lands Tails. Thirders argue that she is experiencing one of three possible and equiprobable awakening episodes, two of which happen when the coin landed Tails, and hence that her credence in the coin having landed Tails becomes 2/3.)

I think the double halfer reasoning is faulty because it wrongly subsumes the Sleeping Beauty problem under (or assimilates it with) a different problem in which there would be two separate coin tosses. — Pierre-Normand

The point is that, like you, they construct the reasons in order to get the result they want. Not because the reasons are consistent in mathematics. But your explanation is wrong. They argue that there is a single, but somehow the Law of Total Probability does not apply. That Tails&Monday is both the same, and a different, outcome than Tails&Tuesday depending one which way you try to use that Law.

Well, firstly, the Halfer solution isn't the answer that I want since my own pragmatist interpretation grants the validity of both the Halfer and the Thirder interpretations, but denies either one being the exclusively correct one. — Pierre-Normand

Like I said, you want the halfer solution to have validity, so you manufacture reasons for it to be. There can't be two valid answers. Your logic fails to provide ANY solution to my last (repeated) variation.

Viewed from outside the experiment - i.e., not SB's viewpoint - there are two paths with two distinct days each (For this logic, I'm calling Heads&Mon and Tails&Mon different days). And each path has a 50% probability, and the days are not distinguished.

SB's viewpoint inside the experiment sees that only one day is happening. But recognizes that there are three other days, including one she would sleep through, that exist as possibilities but are not this day. I don't recall if I've used it here, but of course I have an equivalent version that clarifies this.

There are three Michelin three-star restaurants in San Francisco, where I'll assume the experiment takes place. They are Atelier Crenn, Benu, and Quince. Before the coin is tossed, a different restaurant is randomly assigned to each of Heads&Mon, Tails&Mon, and Tails&Tue. When she is awoken, SB is taken to the assigned restaurant for her interview. Since she has no idea which restaurant was assigned to which day, as she gets in the car to go there each has a 1/3 probability. (Note that this is Elga's solution.) Once she gets to, say, Benu, she can reason that it had a 1/3 chance to be assigned to Heads&Mon.

The point is that each restaurant represents one of three possible "waking" days, not the path that it is a part of. Outside the experiment there is a pair that represent the Tuesday path, but that is irrelevant to SB.

The point is that, like you, they construct the reasons in order to get the result they want. Not because the reasons are consistent in mathematics. But your explanation is wrong — JeffJo

It's not wrong, it's spooky just because of the gauranteed amnesia, which makes it weird and susceptible to forgetfulness. I think the problem was created more or less just to see what answers people would come up with, how they would project their logic onto what they read.

1/2 makes since, since theoretical coinflips always result in either or. 1/3 makes sense depending on how the experiment is manipulated, when she is asked to make a prediction and the information she is given etc. I gave a perfect example in my last post of how you could gaurantee one out of three heads, but whether sleeping beauty is given the true info is suspect. But the odd thing about coin flips, and this is what the question is exploiting, is that there will generally be a bias one way or the other, always with odd and normally with an even number of flips. It's less like a random number generator with thousands/millions/billions of possibilities like in real life gambling.

You appear to be affirming the consequent. In this case, Tails is noticed twice as often because Tails is twice as likely to be noticed. It doesn't then follow that Tail awakenings happen twice as often because Tails awakenings are twice as likely to happen. — Michael

Rather, the premiss I'm making use of is the awakening-episode generation rule. If the coin lands/landed Tails, two awakening episodes are being generated, else only one is. This premiss is available to SB since it's part of the protocol. From this premiss, she infers that, on average, when she participates in such an experiment (as she knows to be currently doing) the number of T-awakenings that she gets to experience is twice as large as the number of H-awakening. (Namely, those numbers are 1 and 1/2, respectively). So far, that is something that both Halfers and Thirders seem to agree on.

"1) Per run: most runs are 'non-six', so the per-run credence is P(6)=1/6 (the Halfer number).
2) Per awakening/observation: a 'six-run' spawns six observation-cases, a 'non-six' run spawns one. So among the observation-cases, 'six' shows up in a 6/5 ratio, giving P('six'|Awake)=6/11 (the Thirder number)."
— Pierre-Normand

This doesn't make sense.

She is in a Tails awakening if and only if she is in a Tails run.
Therefore, she believes that she is most likely in a Tails awakening if and only if she believes that she is most likely in a Tails run.
Therefore, her credence that she is in a Tails awakening equals her credence that she is in a Tails run.

You can't have it both ways.

This biconditional statement indeed ensures that her credences regarding her being experiencing a T-awakening, her experiencing a T-run, or her being in circumstances in which the coin landed (or will land) Tails, all match. All three of those statements of credence, though, are similarly ambiguous. All three of them denote three distinct events that can indeed only be actual (from SB's current epistemic situation on the occasion of an awakening) if and only if the other two are. The validity of those biconditionals doesn't resolve the relevant ambiguity, though, which is something that had been stressed by Laureano Luna in his 2020 Sleeping Beauty: An Unexpected Solution paper that we had discussed before on this thread (and that @fdrake had brought up, if I remember).

Under the Halfer interpretation of SB's credence, all three of those biconditionally related "experienced" events—by "experienced", I mean that SB is currently living those events, regardless of her knowing or not that she is living them—are actual on average 1/2 of the times that SB is experiencing a typical experimental run. Under the Thirder interpretation, all three of those biconditionally related "experienced" events are actual on average 2/3 of the times that SB is experiencing a typical awakening episode.

If it helps, it's not a bet but a holiday destination. The die is a magical die that determines the weather. If it lands on a 6 then it will rain in Paris, otherwise it will rain in Tokyo. Both Prince Charming and Sleeping Beauty initially decide to go to Paris. If after being woken up Sleeping Beauty genuinely believes that the die most likely landed on a 6 then she genuinely believes that it is most likely to rain in Paris, and so will decide instead to go to Tokyo.

This setup exactly mirrors some other variations I also had proposed (exiting the Left Wing or exiting the East Wing at the end of the experiment) that indeed warrant SB's reliance on her Halfer-credence to place her bet. But the original SB problem doesn't state what the "exit conditions" are. (If it did, there'd be no problem.) Rather than being offered to make a unique trip to Paris or Tokyo at the end of the current experimental run, SB could be offered to make a one day trip to either one of those destinations over the course of her current awakening episode, and then be put back to sleep. Her Thirder-credence would then be pragmatically relevant to selecting the destination most likely to afford her a sunny trip.

There are three Michelin three-star restaurants in San Francisco, where I'll assume the experiment takes place. They are Atelier Crenn, Benu, and Quince. Before the coin is tossed, a different restaurant is randomly assigned to each of Heads&Mon, Tails&Mon, and Tails&Tue. When she is awoken, SB is taken to the assigned restaurant for her interview. Since she has no idea which restaurant was assigned to which day, as she gets in the car to go there each has a 1/3 probability. (Note that this is Elga's solution.) Once she gets to, say, Benu, she can reason that it had a 1/3 chance to be assigned to Heads&Mon. — JeffJo

Yes, that is a very good illustration, and justification, of the 1/3 credence Thirders assign to SB given their interpretation of her "credence", which is, in this case, tied up with the experiment's "exit rules": one separate restaurant visit (or none) for each possible coin-toss-outcome + day-of-the-week combinatorial possibility. Another exit rule could be that SB gets to go the Atelier Crenn at the end of the experiment when the coin landed Heads and to Benu when it landed Tails. In that case, when awakened, she can reason that the coin landed Tails if and only if she will go to Benu (after the end of the experiment). She knew before the experiment began that, in the long run, after many such experiments, she would go to Atelier Crenn and to Benu equally frequently on average. When she awakens, from her new epistemic situation, this proportion doesn't change (unlike what was the case with your proposed exit rules). This supplies a sensible interpretation to the Halfer's 1/2 credence: SB's expectation that she will go to Atelier Crenn half the times (or be equally likely to go to Atelier Crenn) at the end of the current experimental run regardless of how many times she is pointlessly being asked to guess.