It's truly remarkable that a question which is of no philosophical significance or interest could generate so many responses on a philosophy forum! — Janus

The OP asks, "What should you do?"

Think of the problem as being in the same family as Pascal's Wager, involving decision theory and epistemology.

This still looks like you're considering what would happen if we always stick or always switch over a number of repeated games. I'm just talking about playing one game. — Michael

You seem to be saying that with an unknown distribution, there is an expected gain from switching for one game even though over repeated games (with an unknown distribution) there isn't.

For one game, would you be willing to pay up to 1.25 times the amount in the chosen envelope to switch (say, 1.2 times the amount)?

If it's in the middle then there's an expected gain of £2.50 for switching — Michael

There is with that distribution but not the {{5,10},{5,10},{5,10},{5,10},{10,20}} distribution. In this case, there is an expected loss of $2 for switching from $10.

Without knowing the distribution, the player only knows that the unconditional expected gain is zero (i.e., the sum of the expected gains for each possible amount weighted against their probability of being observed). With the above distribution, that is (4 * $5 + 5 * -$2 + 1 * -$10) / 10 = ($20 - $10 - $10) / 10 = $0.

If we then play repeated games then I can use the information from each game to switch conditionally, as per this strategy (or in R), to realize the .25 gain over the never-switcher. — Michael

The first game does not have any information from previous games so the player should not expect a .25 gain without that information. The player should only expect a gain (or loss) from switching when they know the distribution which is what the information built up over repeated games would provide.

You just telescoped the step of multiplying by the chance of picking that number.

Could put & where you have |. — Srap Tasmaner

Yes, I messed up the math. Have edited...

If I'm told that one envelope contains twice as much as the other, and if I pick one at random, am I right in saying before I open it that there's a 50% chance that my envelope contains the smaller amount?1 If so, I must also be right in saying after I open it and see the amount that there's a 50% chance that my envelope contains the smaller amount (assuming I don't know how the values are selected). — Michael

You are referring to two different probabilities. The first is P(lower) which is 50%, the second is P(lower|amount) which depends on the specifics of the distribution.

To see this, suppose the distribution is {{5,10},{5,10},{5,10},{5,10},{10,20}} with an equal probability of the host selecting any one of those envelope pairs.

There is a 50% chance of the player choosing the lower envelope from the envelope pair that the host selected. When the amount is unknown, there is also a 50% chance that the selected envelope is the lower envelope.

However if the player observes the amount to be 10, P(higher|10) = 4/5 and P(lower|10) = 1/5. The expected value from switching is 10/2 * P({5,10}) + 10*2 + P({10,20}) = 5 * 4/5 + 20 * 1/5 = 8. So there is a negative expected gain from switching.

If the player does not know the initial distribution and simply assumes P(lower|10) = P(higher|10) = 1/2 then the calculated expected value from switching will be 12.50 which is wrong. However it will still be the case that P(lower) = 1/2 since that does not depend on their mistaken assumption.

So the difference depends on whether one is conditioning on the specific amount or not. Without knowing the distribution, P(lower|amount) cannot be calculated, only P(lower) can. Opening the envelope and learning the amount constitutes a kind of context switch where the player switches from considering P(lower) to considering P(lower|amount).

Edit: fixed math

Do we need to assume that X is not continuous? If it is, all these probabilities are just 0, aren't they?

*** Urk. Forgetting that at least one of them has to be non-zero. — Srap Tasmaner

Yes, I think it's reasonable to assume discreteness (with a finite but unspecified precision and range).

Wouldn't that information be useful? — Pierre-Normand

Yes. But I think the OP is asking for a general solution for one run with no special assumptions about the context (such as whole dollar amounts or million dollar limits).

Considering multiple runs as blind runs with no knowledge acquired from previous runs should be OK. Perhaps using amnesia per the Sleeping Beauty problem...

To remove that option, I recast the problem with the envelopes containing IOUs rather than cash, for an amount that is a real number of cents, with an arbitrary but large number of decimal places shown. The amount is only rounded to the nearest cent (or dollar) when the IOU is cashed in. — andrewk

:up:

Do we know that both P(X = a) and P(X = a/2) are non-zero? We know that at least one of P(X = a | Y = a) and P(X = a/2 | Y = a) is non-zero, but we do not know that both are. Without Y = a, we wouldn't even know that at least one of P(X = a) and P(X = a/2) are non-zero. Without knowing that both are non-zero, we can't even safely talk about the odds P(X = a):P(X = a/2). — Srap Tasmaner

Yes. You learn something about the distribution when you open an envelope (namely, that it had an envelope with that seen amount). But not enough to calculate anything useful. It's like getting a bicycle with one wheel. You might wonder whether you could get somewhere with it, but you probably can't.

I’m going to assume that A means your envelope, B means the “other” envelope, and X means the lower value. But A, B, and X mean the abstract concept of the random variable; it is a, b, and x that mean values. So “2X” is meaningless. What I think you mean here is:

E(B∣A=a) = Pr(X=a∣A=a)*(2a) + Pr(X=a/2∣A=a)*(a/2) — JeffJo

Isn't 2X just a transformation of X that doubles the possible values in X? So Pr(2X=a) would be equivalent to Pr(X=a/2). Here's an academic example using P(X^2 <= y).

Personally, I think in terms of modeling the actual OP the inclusion of distributions or switching strategies is misguided. If our goal is to assess the situation before us then they move outside that scope. However, these models do provide a useful platform to investigate properties that may be applied to other probabilistic aspects. — Jeremiah

I agree. Regarding the OP, the math provides no reason to be anything other than indifferent to sticking or switching.

(Of course, people may wish to stick or switch for individual reasons related to the specific context, such as utility, but that is a separate issue.)

And that's because of what I explained here. — Michael

Yes your account of objective probabilities explains it. But that is just probability simpliciter where the player's expected gain calculations are based on her knowledge of the distribution.

The problem is when knowledge of the distribution is absent. There isn't a symmetry between {a/2,a} and {a,2a} (as there is with, say, the two sides of a fair coin) because one of those envelope pairs may not be part of the distribution. And even if they were both present, there might be other relevant differences that make one envelope pair more likely to be selected than the other.

It's like treating a potentially biased coin as fair. Actual results won't reflect calculated expectations if assumptions about the coin weighting are wrong.

So what am I supposed to do if I don't know how the values are selected? As I said here, if I don't have any reason to believe that X = 5 is more likely than X = 10 then why wouldn't I switch? I think the principle of indifference is entirely appropriate in this circumstance. There's more to gain than there is to lose, and a loss of £5 is an acceptable risk. — Michael

You can make a similar argument for keeping. Suppose you choose an envelope but, in this game, instead of opening your envelope the host opens the other envelope and shows you the amount which is £10. Now your choice is to switch to the £10 envelope or keep your chosen envelope. It seems that there is more to gain by keeping, so you keep.

The arguments are symmetrical. That could be an argument to always select the unknown quantity. Or it could be an argument to be indifferent to keeping or switching. The latter would be an application of the indifference principle based on a symmetry.

That conclusion is also supported by running the game many times which shows that the actual gain from always switching (or always keeping) is more or less zero.

And what you seem to be avoiding with that attitude, is that the expectation formula (v/2)/2 + (2v)/2 is already assuming: — JeffJo

I am not an advocate for that expectation formula, so I don't see why you'd think I am avoiding those objections to it. - andrewk
Maybe I was mixing Andrews up. I apologize. — JeffJo

I don't advocate it either. I advocate E = ((v/2) * Pr(v/2,v) + (v*2) * Pr(v,2v)) / (Pr(v/2,v) + Pr(v,2v)).

But suppose you don't do this. Suppose you just select some X at random from {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, put it in one envelope, and then put 2X in the other envelope. I select one at random and open it to see £10. Am I right in saying that {£5, £10} and {£10, £20} are equally likely? — Michael

Yes. That is because you know what the distribution is.

If we don't have this information then we should apply the principle of indifference, which will have us say that {£5, £10} and {£10, £20} are equally likely. — Michael

Without knowing the distribution you don't know whether both the envelope pairs were possible and, if they were, whether they were equally weighted.

Given that knowledge you can apply the principle of indifference (based on the envelope pair weightings being identical). And you can also calculate the expected gain. Otherwise all bets are off, so to speak.

*** You might have (3) and (4) a little wrong but I can't judge. The McDonnell & Abbott paper makes noises about the player using Cover's strategy having no knowledge of the PDF of X. — Srap Tasmaner

There is an expected gain of X from strategic switching if the algorithm generates a random number between the X and 2X envelope amounts. In this case, the player will switch only when the chosen amount is less than the random number (thus ending up with the other envelope with 2X) and stick otherwise (thus keeping her chosen envelope with 2X).

However there needs to be a distribution to generate the random number from. If that distribution is too wide or too dissimilar to the envelope distribution, the probability of the above situation occurring could be vanishingly small such that it is never realizable in practice. So for the strategy to be useful, at least a ballpark estimate of the maximum possible amount would be needed.

It seems to me that there are some claims about the Two Envelopes problem that everyone might agree on.

1. If the player does not know the amount in the chosen envelope then the expected gain from switching is zero.

2. If the player knows the amount in the chosen envelope and also knows everything about the distribution then the expected gain can be calculated and used to strategically switch on (i.e., switch if the expected gain is positive else stick).

For example, if the player knows that there are only two equally likely envelope pairs of { $5, $10) and { $10, $20 } respectively and sees $10 then they would know that there is a $2.50 expected gain from switching. Whereas if they see $20, they would know that there is a $10 expected (and actual) loss from switching.

3. If the player knows the amount in the chosen envelope but knows nothing else about the distribution then there is no general strategy that would increase their expected gain above zero.

4. If the player knows the amount in the chosen envelope and knows or can estimate information about the distribution then there are strategic switching strategies (e.g., Cover's strategy) that can increase their expected gain above zero.

Yes, assuming that the player does accurately estimate the maximum possible payout (and the procedure for generating the envelope amounts). If she doesn't, then the positive expected gain calculation is invalid.

So we cannot talk meaningfully about 'the real value of c' — andrewk

Can you give a concrete example where such a value would be used?

2. Treat Y as known and model X using a Bayesian prior. This leads to a rule under which the player can calculate a value c such that her expected switch gain is positive if Y<c and negative if Y>c. — andrewk

I think the issue is that even if you know Y from opening the initial envelope, the expected gain from switching is still zero if you don't also know c.

You could potentially calculate c if you observed many runs of the experiment and c remained constant. In which case you could condition on amounts less than or equal to c and in those cases calculate a positive expected gain from switching.

However for a single run, c just is the lower amount in the two envelopes and is, per the problem definition, unknown.

I'm just doing this:

1. We pick an envelope at random
2. There's a 50% chance that my envelope is the X envelope and a 50% chance that my envelope is the 2X envelope.
3. I open my envelope and see £10
4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope.
5. From 4, there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5.

We seem to agree on 1-3, but you disagree with 4 and/or 5? — Michael

I disagree that the 50% chance can be assumed. If the sample space is finite, then there is a non-zero chance that the $10 amount is the maximum amount. If it is, the actual gain from swapping will always be negative. Accounting for both the minimum and maximum amounts exactly cancels out the 1.25X expected gain for other amounts.

For an infinite sample space, assuming 50% chance and conditioning on any specific amount will always conclude a 1.25X expected gain (the grass always appears greener on the other side). Yet unconditionally, there is no reason to prefer either envelope. That again implies that the 50% chance assumption is mistaken. A higher probability weighting (say 2/3 or a graded weighting) for lower amounts (or perhaps for amounts closer to 1) may bring agreement, but I haven't checked the math.

I have numbered the steps from 0 to 4. Which step(s) do you not believe? — andrewk

See above per the 50% chance assumption.

Why not? I've opened the envelope and seen that I have $10. That's in the rules as specified in the OP. And knowing the rules of the game I know there's a 50% chance that the other envelope contains $5 and a 50% chance that the other envelope contains $20. — Michael

The problem is that you use the $10 starting amount to generate the random half ($5) or double ($20) amount for the second envelope. That is equivalent to emptying the second envelope and refilling it with half or double the amount of the first chosen envelope. Which doesn't reflect the rules in the OP.

Is there some rule of statistics that says that one or the other is the proper way to assess the best strategy for a single instance of the game (where you know that there's $10 in your envelope)? — Michael

It's that the switching strategy gains nothing if the starting envelope was randomly chosen and the amount of the second envelope remains unchanged. Whereas switching would be correct if the second envelope amount was randomly decided on the basis of the first chosen envelope amount.

I would have thought that if we want to know the best strategy given the information we have then the repeated games we consider require us to have that same information, and the variations are in the possible unknowns – which is what my example does. — Michael

The problem is making the randomizing element dependent on the first chosen envelope amount instead of simply as the choice of the envelope.

Interesting. The problem is different. Unlike in the OP, in the Wiki case, the envelope has not been opened before the option to swap. So the player has no new information in the wiki case. I think their analysis in the Simple Case is wrong, because it assumes the existence of an expected value that may not exist, but I think the conclusion that there is no reason to switch may in spite of that error be correct in that case, but not in this one. — andrewk

Before choosing, there is no reason to prefer one envelope to the other. On choosing an envelope and learning the amount in it, there is still no reason to prefer one envelope to the other. So there is no reason to switch.

However if on opening the first envelope, the second envelope were then emptied and randomly filled with either half or twice the amount of the first envelope then the expected value of the second envelope would be (X/2 + X*2)/2 = 1.25X. So the player should switch.

Operationally, they are two different scenarios (requiring different strategies).

In the absence of knowing the distribution of X, any calculations based on expected values prior to opening the envelope are meaningless and wrong. — andrewk

The Wikipedia entry uses the expected value being the same for both envelopes as their simple resolution. Do they have that wrong, in your view?

Like Baden you're conflating different values of X. — Michael

No, but I can make the same argument using concrete amounts (see below).

Given a starting envelope of $10 — Michael

You can't assume you have a starting envelope of, for example, $10, and that the other envelope has either $5 or $20. That doesn't reflect the problem as specified in the OP.

The OP instead assumes that you have two envelopes of, for example, $10 and $20, and that you randomly choose one of them. So half the time, the starting envelope would have $10 and half the time the starting envelope would have $20.

Over 100 runs with a switching strategy, you would switch from the $20 to the $10 envelope 50 times (earning $500) and switch from the $10 to the $20 envelope 50 times (earning $1000) for a total of $1500.

Over 100 runs with a keeping strategy, you would keep the $20 envelope 50 times (earning $1000) and keep the $10 envelope 50 times (earning $500) for a total of $1500.

So you earn $1500 (on average) on either strategy.

That's exactly what I did here and here. If in each game we see that we have £10 we win more by switching than by not switching. — Michael

As Jeremiah points out, your code doesn't reflect the problem in the OP. Before an envelope is picked, the expected value of each envelope is the same. That doesn't change when you choose an envelope and find it has $10 in it. The expected value of the other envelope is also $10 and so the expected gain from switching is $0.

To see this, suppose you do the experiment 100 times where you always switch. 50 times (on average) you choose the $2X envelope and switch to $1X. So you earn $50X. 50 times you choose the $X envelope and switch to $2X. So you earn $100X. In total, you earn $150X.

Now you do the same experiment 100 times where you never switch. 50 times you choose the $2X envelope and don't switch. So you earn $100X. 50 times you choose the $1X envelope and don't switch. So you earn $50X. In total, you earn $150X.

So the expected value in both sets of experiments is the same.

The experiment you're modeling with your code is where the second envelope is emptied and randomly filled with either half or twice the amount of the first chosen envelope amount. In which case you should switch.

It looks like you're assuming the player's utility function is the identity function, which would be unusual and unrealistic. Even if we assume that, the quoted calculation doesn't take account of all the available information. There is new information, which is the known dollar value in the opened envelope. That changes the utility calculation. — andrewk

Yes it does. But considering myself, my utility function would be roughly linear for the amounts being talked about here (around $20). Which is to say, I would accept an offer of $1 to either switch or keep since my expected gain from doing so would be $1.

What should you do? — Jeremiah

The expected value of each envelope is $(X + 2X)/2 = $3X/2 = $1.5X. You should be indifferent between switching or not.

However consider this variation. If you choose an envelope, and then the other envelope is emptied and randomly filled with either half or double the amount of the envelope you chose, then the expected value of the other envelope is $(X/2 + X*2)/2 = $5X/4 = $1.25X. Since $1.25X > $X, you should switch.

//ps// from the Fuchs article:

These views have lately been termed “participatory realism” to emphasize that rather than relinquishing the idea of reality (as they are often accused of), they are saying that reality is more than any third-person perspective can capture. — Wayfarer

So the argument is that is necessary to index the quantum state to a participant (broadly conceived). As an analogy, there is nothing mystical about ordinary statements like, "The apple is red" or "I am in pain". But they are statements that are only meaningful when indexed to individual sentient creatures with particular sensory capabilities. There is no intrinsic "redness" or "pain" in the world.

The way QBism conceives of indexing compared to RQM is different (Bayesian probabilities versus relative reference frames). But the general idea is that there needs to be a natural integration of first and third person perspectives - a view from somewhere - to make proper sense of quantum mechanics.

Which, as it happens, is something that the natural and holistic approaches of Aristotelian hylomorphism and Peircean pragmatism, to name but two, both do.

So I think that is what so disturbed Einstein about the so-called ‘quantum leap’, uncertainty, non-locality and the other aspects of quantum physics - they all tend to undermine the concept of an ultimately existent object. And that has deep philosophical ramifications: it is why the essay in the OP is about ‘the war over reality”. — Wayfarer

More recent philosophical discussion has moved on from the reality versus mysticism debates between the founders of quantum mechanics. "The war over reality" essay is about whether the quantum state describes the underlying world (intrinsic realism) or information about observers (participatory realism). Fuch's QBism and Rovelli's RQM are examples of the latter (characterized as Type-II in the paper linked below).

Type-II interpretations do not deny the existence of an objective world but, according to them, quantum theory does not deal directly with intrinsic properties of the observed system, but with the experiences an observer or agent has of the observed system. — Interpretations of quantum theory: A map of madness

What I want to emphasize at the moment is that I cannot see any way in which the program of QBism has ever contradicted what Einstein calls the program of “the real.” — On Participatory Realism - Chris Fuchs

Again - the role of the observer is inextricable; you can't assume 'a view from nowhere'. — Wayfarer

Agreed.

What I think all of this is showing is the role of the observing mind in the establishment of duration. After all, time exists on a scale - if you were a being who lived for a billion years, your sense of duration would be completely different from that of the human. But which is the most accurate? Well, it's a meaningless question; 'accuracy' can only be judged, given a scale. — Wayfarer

Yes.

objectivity cannot be absolute. — Wayfarer

Do you mean there can be different standards for measurement, depending on the context?

The climax of all of that was the EPR paradox, which of course was never able to be made subject to experimental analysis in Einstein’s lifetime, but was to become the subject of the famous Alain Aspect experiments which proved once and for all ‘spooky action at a distance’. — Wayfarer

It hasn't been proven - it's an interpretational issue. Per the quantum interpretations table on Wikipedia, roughly half are local interpretations, including QBism.

But this undermines that view, because it illustrates the sense in which the observer is inextricably part of the picture. We don't, actually, stand outside of, or apart from, the Universe which we are analysing; so what we're analysing cannot be absolutely objective.

Andrei Linde makes this exact point at 3:16 in this Closer To Truth interview. — Wayfarer

What Linde is saying is that, taken as a whole, the universe is predicted to be static and unchanging (per the Wheeler-DeWitt equation). In order to predict a dynamic and changing universe, as we all observe, you have to split the universe into subsystems, i.e., you, the observer + the rest of the universe.

For some experimental results on this, see Quantum Experiment Shows How Time ‘Emerges’ from Entanglement

That doesn't challenge the idea of objective reality. It just means that what is measured depends on one's frame of reference.

I think the potential philosophical problem is actually at 6:04 in the video where Linde is asked how it is that the universe seems to have been around a lot longer than sentient creatures. He says, "This brings me to the interpretation of quantum mechanics ... Everything becomes real at the moment it is observed ... Before you make an observation there is no such thing as real existence of anything there. But once you make an observation everything looks as if it existed all the time before it happens."

Still some things to puzzle through, but I'm convinced. My sojourn in the land of halferism is at its end. — Srap Tasmaner

Welcome back!

I do still disagree about how to interpret this thing though. The failure rate of my tails-guessing Beauties is still 1/2, no matter how much they pat themselves on the back. The argument you give here totally justifies conditioning on being interviewed, so the epistemic issue isn't there; it's in this conflict between the two ways of measuring success. — Srap Tasmaner

I regard them as two distinct epistemic perspectives. One is Beauty's on Sunday (or Wednesday) who doesn't condition on being interviewed and the other is Beauty's on Monday or Tuesday who does.

If you take a step back, SB looks a bit like a fucked up way of doing two trials of a single experiment. (No worries about the single coin flip -- the trial is asking different subjects for their credence.) But whichever way you split, by toss outcome or by day, it's not two trials: it's one trial each for two different experiments and which experiment is being run is determined by the coin toss, and is thus the source of Beauty's uncertainty. — Srap Tasmaner

Yes. You could say the setup is that there is one of two experiments which is randomly selected. In one experiment, only one random person gets interviewed. In the other, everyone does. You find yourself getting interviewed. So what is the probability that you're in the single interview experiment? 1/(1+N) where N is the number of people.

As a result, it is structurally unlike a biased coin or regular betting scenario. There is no one-to-one mapping when the coin toss outcome itself has a different number of interviews (or agents) associated with it. Instead the sample space must properly account for those degrees of freedom with probability being the measure of which state an agent is currently in.

I toss fair coin twice. I ask for your credence that the first toss landed heads only on {HH, TH, TT}.

My question is this: do you think this is equivalent to SB? And why or why not? — Srap Tasmaner

Yes, in the sense that one should condition on being interviewed.

Sorry, I'm not getting your experiment, or its equivalence to SB. — Srap Tasmaner

He's simply saying that if you interview different people for each permutation instead of just one person, then the thirder-style (1/(1+N)) result follows. Which is equivalent to the SB scenario.

In terms of your M&M example, if two people guess tails and they are correct, they both get an M&M. To reflect Sleeping Beauty, the experiment is set up such that only one person gets to guess when the outcome is heads. If the person conditions on the fact that they are getting to guess at all, then they will know that they are more likely to be in the tails track.

Here's the spontaneous version of guessing-SB:
Suppose I'm going to teach Andy & Michael a little about probability. I'm going to flip a coin a bunch of times, but before each flip, they each guess. When they're right they get an M&M, and when we're done we'll count the M&M's and stuff. Now suppose before one toss, Michael guesses "Heads! Heads heads heads heads heads!!!!" If the coin lands heads, do I give him 1 M&M or 6? — Srap Tasmaner

Normally you would just give 1 M&M. But it is ultimately a question about what sample space and rules are appropriate in the circumstances.

100 tosses, presumed result of 50 heads and 50 tails, 150 interviews.
If the Beauties all guess tails all the time, they will get 100 right out of their 150 answers.

That looks like a 2/3 success rate, right? But is it?

Out of the 100 tosses, they got 50 of them wrong. Looked at this way, that's a 50% success rate. — Srap Tasmaner

Yes that's the nature of the experiment. There are two ways of looking at it.

In my view, probability is a measure of the state that the agent is in. Unconditionally, there is a 1/2 chance that Beauty will be in a state associated with heads.

However when conditioning on being awake and interviewed, there is a 1/3 chance that Beauty will be in a state associated with heads.

So the issue between double-halfers and thirders is whether conditioning is valid here.

Here's a variation of the experiment. Suppose that for Tuesday and Heads Beauty is also awakened and interviewed. At every interview she is informed whether or not it is a Tuesday and Heads interview. She knows these rules prior to the experiment. Naturally if she is informed that it is Tuesday and Heads at the interview, she can conclude with certainty that she is in a state associated with heads.

However if Beauty is told that it is not Tuesday and Heads at her interview, should she condition on that information or not?

Yeah that one was too easy. — Srap Tasmaner

It happens! Here's how I see it. Lewis' halfer view fails because it gives an absurd result of 2/3 when conditioning on Monday. But the double-halfer view also fails because on informing Beauty that it is Monday the result of 1/2 violates conditionalization.

Whereas accepting conditionalization and reasoning from the Monday result of 1/2 leads naturally to the thirder view. So that's one argument in favor of it.

Note also that all the real action takes place after the Monday interview since the coin toss can occur that evening. So suppose the Monday interview were removed from the experiment entirely. If Beauty were interviewed at all, then she would know that the outcome is tails (i.e., she conditions on being interviewed on Tuesday).

If she can condition on being interviewed in that scenario and reallocate probability from Tuesday/Heads to the remaining state, then she can also condition on being interviewed in the standard Sleeping Beauty scenario and reallocate probability from Tuesday/Heads to the remaining three states. The condition has obtained just the same whether or not she has learned something new.

I have tossed the coin and given you a box.
What are the chances there's a red marble in the box? — Srap Tasmaner

1/2.

Which in Beauty's case is zilch, isn't it? — Srap Tasmaner

Beauty knows beforehand that she will be awakened and interviewed. But P(Heads|Awake) does not become relevant until she is actually awakened in the experiment, in which case P(Heads) consequently changes from 1/2 to 1/3 for her. Before and after the experiment, P(Heads) = 1/2.

I agree about your double-header example, but don't see the similarity to SB at all. Interviewing here clearly gives you information. — Srap Tasmaner

Yes, but what is actually relevant is that the interview condition has obtained. It is on that basis that P(First coin heads|Interviewed) = 1/3 becomes relevant and P(First coin heads) consequently changes from 1/2 to 1/3 for Alice.

Both Beauty and Alice can calculate all the applicable conditionals beforehand. The only difference is that Beauty knows with certainty that the interview condition will obtain in her future whereas Alice does not. But what makes the conditional relevant is whether the condition currently obtains for the agent, not whether it constitutes new or old information.

No. I'm not sure how to formalize this — Srap Tasmaner

I don't think the halfer view ultimately flies. I think it conflates the question of the nature of fair coins (which we all agree come up heads half the time) with the question of what an agent knows about the outcome of a particular fair coin toss.

For a straightforward example where those two answers differ, suppose that the experimenter will toss two fair coins in sequence. Alice (in Wonderland) will not know the outcome of the coin tosses until after the experiment has completed. However the rule of the experiment, which Alice knows beforehand, is that she will only be interviewed if the outcome is not a double-header. If she is interviewed, she will be asked the probability that the first coin came up heads.

When being interviewed, Alice should condition on being interviewed which results in a sample space of { HT, TH, TT } and an answer of 1/3.

I think the Sleeping Beauty experiment is analogous to this. The theater is just that she is interviewed twice if the first coin toss comes up tails and amnesia is added to make the interviews indistinguishable.

Now I think that's wrong. There is no discounting. None of the 1/2's should be reduced to 1/4's. Monday is not 1/2:1/4 either. — Srap Tasmaner

When Beauty is being interviewed, what probability should she assign to Monday and Tails? If 1/2 then Tuesday and Tails would be 0 which doesn't seem right.

I'm just following the principal principle. If I can figure out what the objective chances are, so can Beauty, and she can set her credences accordingly. — Srap Tasmaner

That is of no use to her. When awakened, she doesn't know whether she is in an awake state that she should assign a probability of 1/2 to or 1/4 to. She can only condition on being awake and thus assign 1/3.