Does the halfer position in the Sleeping Beauty problem make for an irrational gambler?

Question

It's my understanding that the Sleeping Beauty problem doesn't have a consensus answer, with major camps along the lines of "halfers," "thirders," and "the-question-statement-is-ill-defined-ers." However, I'm curious what the halfer positions have to say about the following gambling-related interpretations of credence.

If P is a proposition, let G(P) be the gambling game consisting of paying $40 for the chance to win $100 iff the proposition P is actually true. The point being, if P is expected to be true with probability 1/2, a profit-motivated participant should play G(P), but not if P is expected to be true with probability 1/3.

(1) If one asks Sleeping Beauty to play G(P) where P = "The coin landed heads" upon waking up, I think even a halfer has to concede to act "like a thirder" and decline the game. (At least this is what my friend, a halfer, has conceded, and is the only reasonable choice in my eyes.) But this is also not a perfectly fair demonstration, since the nature of the game constrains Sleeping Beauty to play and lose twice if P fails to be true.

(2) If one instead considers the reformulation of Sleeping Beauty (known as the "Sailor's child problem" in the wiki) where one is a lifelong orphan but one knows one's parents made a 50% random choice to have either one or two children, one can consider a similar game, where P = "I am an only child." This game is such that one will not be constrained to lose twice if wrong. And yet, if an enterprising individual offered this wager to every halfer orphan in this circumstance, he would stand to make a $20 profit per three contestants.

The only way I can think to reconcile the contradiction is if halfers either reject the equivalence of these two thought experiments, or reject the connection of degree-of-belief to gambling strategy altogether.

Answer 1

TL;DR

Yes, the halfer position in all the Sleeping Beauty problems (such as the classical Sleeping Beauty or the Sailor's Child or the Orphaned Child, each of which is a variation on the identical underlying problem) makes for an irrational gambler who can be exploited by a bookie who is a thirder, giving the bookie an expected net gain (though not an assured net gain in each single experiment run).

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The Sleeping Beauty Problem (SBP) remains controversial among the "tutored". Phil Paper Survey shows that in 2020 about 28% of professional philosophers accepted (or leaned towards) the thirder position, while about 19% chose the halfer position. Unfortunately, the halfers will probably not want to put their money where their mouth is, otherwise a thirder bookie can be expected to get rich. (Expected in 'the long run' in repeated experiments, though not guaranteed in any single round.)

I will assume the SBP description as given on Wikipedia (the coin toss of heads leads to one awakening, tails leads to two). At each awakening Beauty, who is fully aware of the experiment setup, is asked what her credence is that the coin came up heads. Credence is defined as degree of belief, degree of certainty about an event.

The question then is really what event? What are we talking about when describing the event as "the coin came up heads". In particular, what does Beauty know about this and what is she (un)certain about?

A probability measure requires a partitioning (by some agent) of the world into basic possible events, where each basic possible event gets some probability assigned. This partitioning (with probabilities) is an epistemic universe. The epistemic universe for Beauty during the run of the experiment is the set of (possible) awakenings which we can describe as {MoH, MoT, TuT}. This is the set of three possible basic events. Those events can only be distinguished from each other by the Experimenter, but not by Beauty (during the experiment). Beauty knows the universe of possible events, she knows the names of the events, but does not know how to identify or recognize them. Her uncertainty arises from this epistemic inability.

By the principle of indifference Beauty can only assign an equal probability of 1/3 to each of the basic events. Note that (as @Feryll pointed out in a comment) this only applies if the coin is indeed a fair coin and if Beauty knows this (also knows it in her epistemic universe). If Beauty has no information at all about whether or not the coin is fair, she can still only assign 1/3 equal probability to each of the basic events (to start with, as a priori priors), but knowing that she has no knowledge about the fairness of the coin implies that she will have no certainty about her own credence answers. In a way, this would make it "impossible" for her to answer the question. Any answer would be rationally justified. ^C

Now, by design, the event "the coin came up heads" is in this epistemic universe equal to the subset {MoH}, while "the coin came up tails" is equal to {MoT, TuT}. This implies, directly, without further necessary calculations, that for Beauty, during the awakenings, the probability she will assign to "the coin came up heads" is 1/3. (She will assign this if she is rational, based on her probability assignment to the basic events.)

It really does not matter that the whole experiment was started off by a coin flip. The initial coin flip is (mostly) an irrelevant nuisance aspect of the problem. (It's a little similar to a "nuisance parameter" on which we sometimes condition in statistical modeling.) It's also a nuisance in the sense that, in my opinion, it's the main reason why even the tutored can get confused by the problem statement.

The initial coin flip has only one function, which is to set up the three possible events in Beauty's epistemic world (that is: to ensure there are only three possible events, and to give them a name, similar to what I did above). Any other way to set up the branching (or initialize this world) would also be fine, as long as Beauty has no knowledge (and can not discover any new knowledge) that could make her more (or less) certain that one of those three possibilities is the one she is dealing with. (However, the a priori assignment of probabilties to the basic events is part of the experiment setup -- as known to Beauty during the run. In the classical version it's crucial that the coin is fair. The information that the coin is fair gives Beauty the knowledge that each basic event has the same probability of 1/3. In alternative versions, where the branching might be done very differently, Beauty still needs to be informed somehow about those basic probabilities. See note C. In other words, the fact that branching is used or that a coin is flipped is irrelevant to the problem, but if the setup is determined by coin flip, then it is very relevant that a fair coin is used -- relevant for getting to the 1/3 credence answer. Perhaps it's really this distinction that makes the problem confusing to some...)

If the thirder position is correct, then the halfer position is incorrect. It's also possible to show that the halfer position is incorrect by using a (kind of) Dutch Book argument, as the OP suggested in point (1).

A Dutch Book is defined as a betting strategy, usually over a series of bets, in which a gambler has a guaranteed loss. Note that all the offered bets need to apply to the same epistemic universe.

Playing as bookie against a halfer, it's not possible to force a guaranteed gain for the bookie in each single experiment run. But it is possible to get an expected net gain.^A It's not possible to set up a similar (expected gain) Dutch Book against the thirder position. (This follows from the fact that there is a Dutch Book of a thirder bookie against the halfer position and against other non-thirder positions.)

A Dutch Strategy is far more complicated stratagem in which a bettor who changes their credence about the odds is offered a series of bets that seem fair to the bettor -- given their credence at a particular time -- but that guarantees an overall loss to the bettor. (It's a contentious issue whether or not acceptance of a Dutch Strategy implies irrationality on the side of the gambler. Personally, I don't think it's a persuasive kind of argument. The whole approach seems very dubious to me. See below.)

Titelbaum's Ten Reasons to Care about the Sleeping Beauty Problem quotes Hitchcock who gives (not a Dutch Book but) a Dutch Strategy argument against the thirder position:

Hitchcock (2004) proposes that on Sunday night, the bookie sells Beauty a bet that costs $15 and pays $30 if the coin comes up heads. Since Beauty is 1/2 confident in heads on Sunday night, she will accept this bet as fair. The bookie also tells Beauty on Sunday night that when she awakens Monday morning, he will sell her a bet for $20 that pays $30 if the coin comes up tails. If Beauty plans on being a thirder, she is certain that she will accept this bet as fair on Monday morning. (Notice that the bookie places this bet only once – on Monday morning – however the coin flip comes out.) Yet now Beauty is guaranteed to shell out a total of $35 for two bets which together will pay her $30, no matter the flip outcome. Planning to be a thirder exposes Beauty to a sure loss of $5.

The first bet expresses odds of 1:1, has an expected gain of $0 so is acceptable for someone who believes the odds are 1:1. The second bet expresses odds of 2:1 for tails up. This bet is not acceptable, of course, if it is only offered once on Mondays. The bet is in principle acceptable to a thirder (since the thirder believes in those odds), but only if the bet is offered either at each awakening or at a randomly selected awakening. Only in the last two cases will this bet have an expected non-negative gain (of $0) and thus be acceptable for the thirder.

If a bet is only offered on Mondays, however, then the thirder will only accept odds of 1:1! (See also the answer on the math stackexchange by Qiaochu Yuan.)

To come back to the OP's question - The Orphaned Child problem as sketched by the OP is a little bit different from the Sailor's Child Problem as described on Wikipedia, but essentially the same. Both problems are, as far as the abstract problem goes, also identical to Sleeping Beauty Problem.

In the Orphaned Child Problem there are again, not two, but three distinct basic 'events':

• I (the child) am the first born (only) child (MoH)
• I am the first born child, but have a younger sibling (MoT)
• I am the second born child, and have an older sibling (TuT)

This is the assumed epistemic universe. The problem in this case is that I don't know which of these three events describes me now. So, I can only assign 1/3 probability to each (this uses the assumption that my parents flipped a fair coin to decide whether to have only one or two children). The event then of "being the only child" is exactly analogous to "the coin having come up heads", with a probability of 1/3. So, contrary to what the OP wrote

This game is such that one will not be constrained to lose twice if wrong

you will lose twice when wrong, since you don't know if you are the younger or older sibling (if you have a sibling).^B

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(A) Dutch Book against halfer (not as guaranteed, but as expected)

A halfer may accept a bet on "heads up" with odds 1:1. A rational halfer will therefore consider a bet with stake $1 and prize (at least) $2 a fair bet (the expected gain is in this case greater than or equal to $0; note that the prize includes the stake which was payed when accepting the bet). In the sample bet of the OP, the stake is $40 and the prize is $100 ($40 + $60). If the odds would truly be 1:1, the expected gain would then be $10 (per run): about half of the time you gain $60, half of the time you lose $40. However, it's easy to see, by taking the average over repeated runs, that the actual expected gain is -$10.

Note that there is a rather fundamental difference between this Dutch Book-like argument and actual Dutch Books. In an actual Dutch Book, the bookie is guaranteed a net gain no matter what. There is no risk involved. The promised expected gain here is "merely" what can be expected "in the long run" (and as we know, in the long run we'll all be dead). For someone who is risk averse, it is perfectly rational to refuse these bets. The risk might be infinitesimal small 'in the long run', but there is still some risk.

"Perfectly rational" is of course an overstatement. In "real life" we would probably not consider someone to be "perfectly rational", but rather "phobic" or "not in their right mind", if they would not be willing to accept any risk whatsoever ever.

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(B) After writing the first version of my post, I noticed that Radford M. Neal (professor emeritus at the Dep. of Statistics and Computer Science in Tortonto) comes to the same conclusion as me in his article
Puzzles of Anthropic Reasoning Resolved Using Full Non-indexical Conditioning (2006):

Do the Sailor’s Child and Sleeping Beauty problems differ in any important way? In the Sleeping Beauty problem, the instances of Beauty awakening on Monday and Tuesday can be visualized as “children” of the Beauty who existed on Sunday. That these “children” are much more closely related than are real children of the same father seems inessential, particularly since the only information transferred from Beauty-on-Sunday to Beauty-awakened-later is “common knowledge” about the setup, such as that the coin is fair. In this light, we can see that contrary to many treatments in the literature, the Sleeping Beauty problem is not really about updating of beliefs as new information is received — a procedure that in any case seems dubious when actual or suspected memory loss is an issue.

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(C) If Beauty doesn't know whether or not the coin is fair, then to start with she can only assign equal probabibilites of 1/3 to each of the basic events {MoH, MoT, TuT}. However, her most honest answer to the question would then be "If I'm forced to make a bet, then I'd say that the probability that this is a MoH awakening is 1/3, but really I cannot tell, it could be anything from 0 to 1". While if the coin is fair and she knows the coin if fair, she can tell with certainty "I know that this being a MoH awakening has a probability of 1/3, so my credence for heads having come up is 1/3. I'm in principle willing to take a bet on that". ("In principle willing..." means that if Beauty is willing to take fair bets at odds 1:1 for $1 (stake $1, prize $2), then, being rational, she is or should be willing to also take the bet on MoH with a stake of $1 and prize of at least $3. If Beauty is totally risk averse, she would not take any bets, of course, but "in principle" only expresses a conditional willingness.)

If Beauty's amnesia only applies within each experiment run, and if she can remember results of repeated runs, then she can through repeated experiment, with the same coin, gain (or lose) certainty about the probability of heads having come up. She can in that case apply Bayes' rule, given new evidence. She would only be able to update the priors between experiment runs. Each new experiment would then simply start with the updated priors.

If the coin is unfair and Beauty knows in advance (and during the experiment) what the probability is of it coming up heads (on Sunday night), then that will determine the probabilities she can/should assign to the basic events.

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UPDATE

It turns out that above answer is essentially the same as the answer given by Berry Groisman in The End of Sleeping Beauty's Nightmare (2008). Groisman gives a very intuitive, clean description of the solution and explains the confusion of the halfers. Halfers conflate "The coin landed Heads under the setup of coin tossing" (which has probability 1/2) with "This awakening is a Head-awakening under the setup of wakening" (which has probability 1/3). Thirders (even Elga) are also not completely clear about that. In fact, anyone who thinks that some kind of "update" of credences is happening missed the crucial point that we're talking about different probabilistic events. Quote from Groisman's summary:

The concept of an event is central and crucial in Probability Theory. The Sleeping Beauty Problem arises due to improper use of the notion of an event. The setup under which the event takes place must always be taken into account. If we do so, then we realize that the original question posed to SB can be interpreted in two different ways. The first interpretation is ‘What is your credence that the coin landed Heads under the setup of coin tossing?’, and the answer should be 1⁄2. The second interpretation is ‘What is your credence that this awakening is a Head-awakening under the setup of wakening?’, and the answer should be 1/3. Thus there is no paradox!

The only criticism I would have is that according to me, the most natural, default interpretation of the question, it the second one. But that's a minor quibble.

Even though some people still contest this claim, I believe that Groisman's reduction of the SBP to a balls-in-urns problem (also decribed on the Wikipedia page) is completely correct. In fact, we can also represent the filling of the urn in such a way that the probability of adding a green ball in any randomly selected step of the procedure is 1/3. If we define the protocol of filling the urn as

Toss a fair coin and observe the outcome. On H, select a green ball, s = {g}. On T, select two red balls, s = {r1, r2}. Then, while you still have balls in s, remove a ball from s and drop it into the urn.

Then define the "steps" as those events that remove one ball from the current selection, and drop them into the urn. It's clear then that, given that the coin toss was fair, the probability that at any randomly selected "step" a green ball was dropped into the urn, is 1/3.

Now imagine that the balls are conscious little creatures who don't know their own color, but are fully aware of the whole setup and this protocol. At each step in our protocol we ask them, "What is your certainty that you are green?" The only rational answer is 1/3. -- This is the "anthropic" version of the question. And it's clear that this anthropic version is just a colorful way of presenting the question, without further meaning (apart from making the question more confusing to some people).

If we describe the filling differently -- in other words, define the sample space of events differently --, the probability is different. If we define a step as "drop a green ball into the urn or drop two red balls into it", then there are only two possible steps, {g} and {r1, r2}, rather than three possibilities g, r1, r2. So, in that case "the probability that at any random step a green ball is added" is 1/2. The difference between the two views is simply that in one case we look at the sets and in the other we look at the elements. The confusing part of SBP is that people fail to take into account that a set of one element is not the same "thing" as that element (a conundrum with which, surprisingly, logicians and mathematicians also still struggeled around 1900).

Another way of explaining the confusing is: People tend to forget that when they are counting or selecting items, they are implicitly performing a mapping. Selecting a ball (also selecting just one ball) can be thought of as creating a set, making a mapping, applying a function. Then in one context, we can only consider the elements (to determine Beauty's credence during the experiment), while in another context we look at the sets that were initially created by the coin toss. When we equate g (prob 1/3) with {g} (prob 1/2) we get confused.

Answer 2

The Sleeping Beauty Problem, as usually stated, breaks with conventional probability theory by allowing the number of observations of an outcome to depend on that outcome. This is why gambling arguments will never resolve it. There is no established way that one can interpret the meaning of this difference. While I agree with your interpretation, it is not provable because there is no basis for such a proof. I believe that the original intent of the thought problem was to help establish such precedent, but that can only be done by proving an answer without relying on interpreting the difference.

That can be done. The Sleeping Beauty Problem, as usually stated, is not the actual Sleeping Beauty Problem. It was originally posed in private discussions by Arnold Zuboff, and made public by Adam Elga in his 2000 paper "Self-locating Belief and the Sleeping Beauty Problem." Both are referenced in the Wikipedia article you linked. Elga stated it this way (Zuboff's is different only in the number of potential wakings):

Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?

Notice that Monday, Tuesday, and most importantly the difference in what can happen on each named day, are not mentioned. Elga introduced those elements, in how he implemented the experiment, to create his thirder solution. Halfers question the validity of his treatment. And it is all unnecessary. There is a better implementation.

I'll keep the day names, since they don't matter. The random elements are that on Sunday night, after SB is put to sleep, we flip two coins. Call them C1 and C2. On Monday Night, regardless of what has happened so far, we turn coin C2 over to show the opposite side. So regardless of the day, there are four equally-likely combinations for what is shown: {HH, HT, Th, TT}.

The observation elements are that on both Monday and Tuesday morning, the researchers examine the two coins. If both are showing Heads, they let SB sleep thru the rest of the day. But if either is showing Tails, then (1) they wake SB, (2) they ask SB "What is your credence now for the proposition that coin C1 is currently showing Heads?", (3) after SB answers, they put her to sleep with amnesia.

This way, an awake SB knows that her observation is dependent on the combination HH being eliminated. She would not be awake if the coins were showing HH, and the remaining three all remain equally likely. Since coin C1 is showing Heads in only one of those three, her "credence" in that circumstance must be 1/3.

This proof is inarguable. Which is probably why the only argument which is made against it has been that I have changed the problem. The fact is, that I changed it less than Elga did, and neither changes any of pertinent details. (Mine essentially uses coin C2 to determine whether Elga's Monday, or Tuesday, could be the day SB sleeps thru.)

The answer is proven to be 1/3. Halfer's need to come to grip with why their arguments are invalid.

Answer 3

The only way I can think to reconcile the contradiction is if halfers either reject the equivalence of these two thought experiments, or reject the connection of degree-of-belief to gambling strategy altogether.

My previous answer assumed that the proposition could be freely chosen, but even if it is always "head wins" you kinda changed the scenario by making it a bet or taking the perspective of the bookie.

Like in the case of (1) you could compute the different outcomes and it would be:

• +$60 if you bet on heads and it comes up heads
• -$80 if you bet on heads and it comes up tails

So if the odds are indeed evenly distributed you'd lose $10 per game on average. So you would not play the game, but unlike the thirders you'd not do it because you think that heads is less likely, but because the payout scheme is stacked against heads.

In terms of (2) it's slightly different because now it's not the same person being asked in case of tails. So from the perspective of the individual being asked it's just:

• +$60 if the coin flip landed heads
• -$40 if the coin flip landed tails

That the bookie gets +$80 for tails as he gets to prompt more people is insignificant to the perspective of the individual being asked.

So there is no contradiction, there are no halfers acting non-halfer-like and their action isn't less rational than that of the thirders.

Answer 4

NO, the halfer position does not make for an irrational gambler.

TL/DR:

(1) The halfer correctly distinguishes between the bet offered and the payout function of the resulting game and acts rationally and in accordance with the halfer position by rejecting the bet.

(2) The two problems ask different questions.

As of right now, I would put myself somewhere between a "halfer" and a "the-question-statement-is-ill-defined-er". Focusing on the former part (or half?) of this position, I would answer with a resounding no to the question in the title.

We can see the original bet in the question, but let us look at the game that is actually offered here and how it differs from that bet. For the payout function that the potential gambler faces, f(b,r), where b is the bet on the toss and r is the result of it, we get the following four possibilities of heads ("H") and tails ("T") when using your example of a $40 bet for the chance of a $100 payout:

• f(H,H) = +$60 (pay $40, then win $100)
• f(H,T) = -$80 (pay $40 and lose) twice
• f(T,H) = -$40 (pay $40 and lose)
• f(T,T) = +$120 (pay $40, then win $100) twice

Notice, as you have already pointed out yourself, that the bet is run and evaluated twice (for the same result!) if and only if the coin comes up T, and only once for a result of H. Also notice, that your game G(P) is represented by the first two lines of the payout function above, f(H,H) and f(H,T), where the gambler choses to bet on your proposition P ("the coin landed heads"). It is of special importance to note that the possible payouts are not +60 and -40, as the question seems to assume here!

(1) A rational halfer would look at those payouts, apply his credence of 1/2 to both of the two outcomes of the coin toss, and get an expected payout for the whole game of:

• 1/2 * f(H,H) + 1/2 * f(H,T) = (1/2 * $60 + 1/2 * -$80) = -$10

So, unless the halfer is risk-seeking and the joy of playing is worth at least $10 per game to them, the rational choice for a halfer is not to play the game. This also means that the halfer does indeed act like a halfer by rejecting the gamble, even if a thirder would also reject the game.

So the halfer rejects the gamble not due to his flawed assesment of the probability, but because he uses all the information he has been given about the whole of the game, as opposed to the simple bet. Or, looking back at the game in the question, it demonstrates that the G(P) in the question does in fact not model the Sleeping Beauty problem in its entirety, as it is lacking the component of the forced repeat in case of T and therefore uses an incorrect payout function (with respect to the Sleeping Beauty problem).

(2) The two problems ask two different questions. Let me demonstrate by expanding the setup slightly:

a) For each of 100 Beauties flip a coin and, according to the result, put them either into hospital ward H or hospital ward T for their regular experiment. Both wards are identical, they do not know in which of the two wards they are, each has their own room. They are then asked their credence for being in ward H whenever woken up. On average, there will be 50 Beauties in ward H and 50 Beauties in ward T.

b) 100 Sailors flip a coin and then go their baby making ways. Much later in life, they all find themselves in the very same hospital, either in ward H or ward T according to their earlier coin toss. They regret never having met their children, invite them, and are visited by all their children, one at a time. The children are then asked their credence for being in ward H when meeting their father. On average, there will be 50 children in ward H and 100 children in ward T.

To conclude, I can not really see many ways to reconcile both approaches either, but rather find one side a lot more convincing than the other. For the gambling approach in particular, I think it could work. However, if one wants to gain insight into the credence of the gambler, one should be carefull to correct for the possible distortion from the payout function on the action taken or avoid it alltogether by keeping it simple. Possibly by just offering a single bet with equal amounts of money won or lost, using the credence stated during questioning, after the experiment. ;)

Addendum: Does this mean the thirder position makes for an irrational gambler?

Possibly somewhat surprisingly, I would again say no. Or at the very least not neccessarily.

Let me try to explain why I believe that. The way the problem is discussed above shows that the thirder misunderstands, or at least misrespresents, the game theoretical problem as it is posed in this question. This is of course not good for the potential gambler, but i would not call this irrational in and of itself. I would argue, instead, that the thirder explicitly picks their credence of 1/3 in order to correct for the error made in modelling the game, with the explicit aim of not ending up acting irrationally with respect to the game's outcomes.

I also like @Hudjefa's answer's attempt at conflict resolution. :)

Since this is my first contribution on this site, feedback of any kind is greatly appreciated!

Answer 5

The fact that there are halfers (who believe the probability of heads is 1/2) and thirders (who believe that the probablity of heads is 1/3) indicates that there is confusion over whether the two events - heads on the coin toss and Sleeping Beauty being awake - are probabilistically dependent/independent. The two events are independent if the probability of heads given that Sleeping Beauty finds herself awake is 1/2 (no change in the probablity of heads on the coin toss, its initial probability is 1/2 and it stays 1/2 even after Sleeping Beauty is woken up) and dependent if the probability of heads is anything other than 1/2 (there's a change in the probability of heads from 1/2 to 1/3 in the case of thirders).

Bayes' Theorem
P(H|A) = Probability of heads given Sleeping Beauty is awake
P(A|H) = Probability that Sleeping Beauty is awake given heads = 1 (she's woken up if the coin shows a heads)
P(H) = Initial probability of heads = 1/2
P(A) = Probability that Sleeping Beauty is woken up = 1 (she is woken up)
P(A & H) = P(H & A) = The probability that Sleeping Beauty is woken up and the coin toss was a heads.
P(A & H) = P(H & A) = P(H) * P(A|H) = P(A) * P(H|A)

That out of the way, let's calculate ...
P(H|A) = [P(H & A)]/P(A) = [P(H) * P(A|H)]/P(A) = (1/2 * 1)/1 = 1/2
P(H) = P(H|A) = 1/2. That is to say the two events (heads on the coin toss and Sleeping Beauty awake) are independent events.

So far so good ...

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P(H & A) = P(A & H) = P(T)

1. P(T) = P(H) * P(A|H) = 1/2 * 1 (already explained)

2. P(T) = P(A) * P(H|A) = 1 * 1/3 [There are 3 outcomes and they are (Heads, Monday, Awake), (Tails, Monday, Awake) and (Tails, Tuesday, Awake) and Sleeping Beauty can't tell which of these is the case just by being awake]

As you can see, P(H) * P(A|H) != P(A) * P(H|A). This is the issue!!!

P(A & H) = P(A) * P(H|A) = 1 * 1/3 = 1/3
P(H & A) = P(H) * P(A|H) = 1/2 * 1 = 1/2

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Attempt at Conflict Resolution
P(H|A) != 1/3. In other words, that there are 3 events/outcomes (mentioned above) is an illusion. There are only 2 viz. (Heads, Monday, Awake) and (Tails, Monday, Tuesday, Awake). Mathematically, we would say that we've overcounted, in this case the possibilities.
We have to establish the identity relation (Tails, Monday, Awake) = (Tails, Tuesday, Awake). I suppose that in terms of coin toss outcome, both are identical (both are tails) i.e. it doesn't seem to matter whether it's Monday/Tuesday for coin toss outcome. In a sense then, (Tails, Monday, Awake) = (Tails, Tuesday, Awake) with respect to coin toss outcome (both tails). The conclusion seems inevitable ... (Tails, Monday, Awake) and (Tails, Tuesday, Awake) are not 2 events but actually the same 1 event, and once more ... there'll be peace in the galaxy.

😊

Answer 6

From a decision theory perspective, what incentive does Sleeping Beauty have, to say any particular number? None. She could say it's an 80% chance the coin is heads, and nothing bad would happen to her.

So what we need in order to make sense of the situation and give her a reason to answer one probability over another, is to give her some incentive. For this we can just apply a proper scoring rule. For instance, we may say that we give her some amount to participate, but take away the logarithmic score ln(r_i) every time she answers, where r_i is the probability she assigned to the outcome that actually occurred. If x is the probability she reports for heads, her expected payoff is 0.5 ln(x) + 0.5 * 2 * ln(1 - x). This is maximized for x = 1/3, a thirder position.

Or we may use the quadratic scoring rule. We give 2r_i - sum(r_j^2) = 2r_i - r_i^2 - (1 - r_i)^2 where r_i is the correct answer. Her expected payoff where x is her probability of heads is then 0.5 (2x - x^2 - (1-x)^2) + 0.5 * 2 * (2(1-x) - x^2 - (1-x)^2). This is also maximized for x = 1/3.

Or we may play a different game: we may give the reward only after her guess on Monday, not on Tuesday. In that case, her expected payoff with the logarithmic score is 0.5 ln(x) + 0.5 * ln(1 - x), maximized when x = 0.5, a halver position. With the quadratic score, again it is maximized when x = 0.5.

Probability is fundamentally a tool that an agent uses to make decisions. The reason the agent makes decisions is to maximize payoff. So, the agent should choose a probability assignment that maximizes payoff; probability is subordinate to payoff. If there is no payoff involved, then it does not matter what probability assignment is chosen.

Answer 7

If one is waking up, then one does not know if it is Monday or Tuesday.

The probability of waking Monday is 100% and the coin options are both H and T. This gives us two events, M-H, and M-T where one awakes. The probability of waking on Tuesday is only 50%, as T-H will not involve waking, but T-T will.

When SB wakes up, it will therefore be one of three options, each with the same probability: M-H, M-T, T-T. The probability of the coin being heads is 1/3. This is the thirder solution.

Note: The reason there is confusion over Sleeping Beauty is because of confusion over both the question, and the nature of the statistics. THIS question, which is what one should SAY about the coin flip, is different from this other question, which is what one should BELEIVE about the flip: Is there a reasonably up-to-date (as of early 2020s) review article on the Sleeping Beauty problem?

The different question matters for the statistics involved. The flip is 50-50. But the statistics of observations have a sampling bias driven by the flip value, such that the sampling leads to a skewed set of observations relative to the flip. Unlike the assumption in both questions, I consider the statistics straightforward. The flip is halfer, and the observations are thirder.

Answer 8

However, I'm curious what the halfer positions have to say about the following gambling-related interpretations of credence.

The problem with the halfer position is there is no way to bet as a halfer given the way the problem is stated.

The thirder position is that on average there are 3 interviews for every 2 coin tosses with twice the number of tails interviews. If SB can only place a bet during interviews then SB has no choice and bet as a thirder: one bet per interview.

A halfer is only betting on the state of the coin. In order to bet like a halfer, SB must be allowed to place a bet before she is put to sleep the first time and place no bets during the interview process. So there is one bet per coin toss.

The difference between the halfer and the thirder then becomes the size of the bet. Since a thirder bets 1.5 times more often than a halfer. The halfer bet must be 1.5 times larger to be equivalent.

Answer 9

The Sleeping Beauty problem has a very natural Ramsey-style solution, which is often overlooked in Dutch Book discussions:

Don’t take the bet, ever, no matter what odds are offered, on the grounds that the bookmaker is taking advantage of the knowledge gap for the purposes of their own profit at your expense.

Let’s look at it like this. You wake up in a standard, coin-toss sleeping beauty problem. You know that, in 50% of the worlds you wake up in, your interlocutors will offer you this deal once, and if you guess it is heads you will win, and in 50% they will offer you it twice, and if you guess it it heads you will lose. So, rationally, you will know that there is a kind of a “dead zone” of 50%+ which you should never take.

Now, you might think it makes rational sense to take the 1:10 bet, right? After all, if this same game repeats enough times, 50% of the time you will guess heads and win 10x your stake, and the other 50% you will lose your stake twice, but in the long run you would anticipate a 5x profit.

But this raises a simple problem. Your interlocutor is administering a drug that causes you to forget your memories. Moreover, they’re doing so in such a way that you can’t tell the difference between the two times you’re woken up.

Any position where you accept this challenge is a deeply irrational one in the first place. Basic self-preservation instincts should be screaming at you not to give this bookmaker this much authority. Because, to be blunt, when you wake up, do you know how many times around the cycle you have really been?

The assumption that the bookmaker is looking to profit from your ignorance ought, in itself, to dramatically overpower the offered odds in this situation.