• Sleeping Beauty Problem
    They ruled that out before the experiment begun. You might as well say that they can rule out it being the case that the coin landed heads and that this is day 3. — Michael

    Before the experiment begins, neither John Doe nor the sitter can rule out a possible future in which we are on Day2 of the experiment and John Doe remains asleep. As soon as John Doe wakes up, they can both rule out the proposition "Today is Day2 of the experiment and the coin landed tails". This is one crucial change to their epistemic situation. They now are able to refer the the current day at issue with the indexical "today" and rationally update their credences by means (self-)reference to it.
  • Sleeping Beauty Problem
    Yes. — Michael

    So, as they await the interviewer, John Doe and the sitter contemplate the probability that the coin landed tails. The coin might be right there on a nightstand with piece of cardboard covering it. Both John Doe and his sitter are equally informed of the full details of the protocol. After John Doe has woken up, they both have access to exactly the same relevant information. There credences target the proposition: "This coin landed tails". Since they are evaluating the exact same proposition from the exact same epistemic perspective, why don't they have the same credences on your view?
  • Sleeping Beauty Problem
    The question is whether or not it is rational for the participant to reason this way. Given that the experiment doesn't work by randomly assigning an interview to them from the set of all interviews, I don't think it is. The experiment works by randomly assigning an interview set from the set of all interview sets (which is either the head set or the tail set), and so I believe it is more rational to reason in this way. — Michael

    Do you agree that from the sitter point of view, the probability that the coin landed tails is 2/3?
  • Sleeping Beauty Problem
    Why? — Michael

    Sorry, I meant to say that he can rule out it being the case the the coin landed heads and that this is Day2.
  • Sleeping Beauty Problem
    A more comparable example would be if there are four doors, two containing a goat and two containing a car. You pick a door (say 1) and then Monty opens one of the two doors that contain a goat (say 2). What is the probability that your chosen door contains a car? What is the probability that the car is behind door 3, or door 4? — Michael

    In that case, the probability that my chosen door contains a car remains 1/2. The probabilities that a car is behind door 3 or behind door 4 get updated from 1/2 to 3/4 each.
  • Sleeping Beauty Problem
    To apply this to the traditional problem: there are two participants; one will be woken on Monday, one on both Monday and Tuesday, determined by a coin toss.

    I am one of the participants. What is the probability that I am woken twice?

    Do I reason as if I am randomly selected from the set of all participants, and so that I am equally likely to be woken twice, or do I reason as if my interview is randomly selected from the set of all interviews, and so that I am more likely to be woken twice?

    Halfers do the former, thirders the latter.

    Which is the most rational?

    Given the way the experiment is conducted I think the former (halfer) reasoning is the most rational.     — Michael

    The halfer and thirder responses, as you frame them here, correspond to different questions answered from different epistemic perspectives.

    Consider this: let's say you are being hired as an observer in these experiments. Each observer is tasked to attend one session with one participant. The participants are woken up and interviewed once on Day 1, or twice on Day 1 and Day 2, which is determined by a coin toss. As an observer, you are assigned to a randomly chosen participant, and you don't know whether this is their only awakening or one of two.

    In the experiment facility, there are, on average, twice as many rooms occupied by participants waking twice (due to tails on the coin toss) as there are rooms with participants waking once (heads on the coin toss). Now suppose you had access to the participant registry where all active participants are listed. You spot the name 'John Doe.' What are the chances he will be woken up twice? You credence is 1/2, and this would also be the case for John Doe's credence before he undergoes the first sleep session.

    Now, let's say that by a stroke of luck, you are assigned to John Doe on that particular day. Your job is to measure his vitals as he awakens and get him breakfast as he waits for the interview. You arrive in John's room and wait for him to wake up. What are the chances that the coin toss resulted in tails, indicating this could be one of two awakenings rather than the only one?

    Once you have been assigned to John Doe, your credence (P(T)) in this proposition should be updated from 1/2 to 2/3. This is because you were randomly assigned a room, and there are twice as many rooms with participants who wake up twice as there are rooms with participants waking up once. Once John Doe awakens, he can rule out the possibility of it being Day 2 of his participation, and so can you. His credence (P(T)) then aligns with yours because both your credences are targeting the exact same proposition, and both of you have the same epistemic access to it.
  • Sleeping Beauty Problem
    I think this is basically a Monty Hall problem. I would say that the probability that I will be put to sleep is 1/2, that the probability that the person to my left will be put to sleep is 1/2, that the probability that the person to my right will be put to sleep is 1/2, and that the probability that one of the other two will be put to sleep is 1/2. — Michael

    Your variation of the problem indeed appears to me to contain elements reminiscent of the Monty Hall problem, but with a key difference in the selection process.

    In the original Monty Hall scenario, Monty's action of revealing a goat behind a door you didn't choose is non-random. This non-random revelation influences your strategy, prompting you to switch doors. (The probabilities of your initial choice and the remaining door hiding the car that were 1/3 and 1/3, respectively, get updated to 1/3 and 2/3.

    Your scenario can be compared to a modified Monty Hall scenario. In this modified scenario, imagine Monty picks one of the unchosen doors at random and reveals a goat by chance. Because this revelation is a random event, it doesn't provide the same targeted information as in the traditional Monty Hall scenario. Consequently, the initial probability estimates (1/3) for your chosen door and the remaining unchosen door hiding a car are updated to 1/2 each. Therefore, you would be indifferent about whether to stick with your original door or switch to the remaining door.

    Similarly, in your variation, if a participant is revealed in random way to have been selected to be put to sleep, much like Monty randomly revealing a goat, the probabilities for the remaining three participants (including yourself) would need to be updated similarly. If it's known that exactly one of the remaining three participants had been selected, then the probabilities for each of those three would increase to 1/3.
  • Sleeping Beauty Problem
    The double halfer approach does entail:

    P(Heads & Monday) = P(Tails & Monday) = P(Tails and Tuesday) = 1/2     — Michael

    From the external point of view of the experimenter, it makes sense that the tree probabilities add up to more than one since the three outcomes are non exclusive. The events (Tails & Monday) and (Tails and Tuesday) can (and indeed must) be realized jointly. The intersection zone in the Venn diagram highlights the fact that those two events aren't mutually exclusive. The perspective (and epistemic position) of Sleeping Beauty is different. When she is awoken, those three possible states are mutually exclusive since it is not possible for her to believe that, in her current state of awakening, the current day is both Monday and Tuesday. Her credences in the three mutually exclusive states that she can be in therefore must add up to 1.
  • Sleeping Beauty Problem
    Finding out that today is Monday just removes the blue circle. — Michael

    I agree with the idea that Sleeping Beauty's credence in H is updated to 1/2 after she learns that her current awakening is occurring on a Monday. The question is what was this credence updated from, before she learned that the current day was Monday. At that point she could not rule out that the current day was Tuesday. Does that not imply that her credence in H was lower than 1/2 before she learned that the day was Monday? Before she learned that, all three mutually exclulsive areas of the Venn diagram represent possible states for her to be in. She therefore should have non-zero credences for each one of them.
  • Sleeping Beauty Problem
    I now realize that in the OP's stipulation of the problem, and in line with most discussions of it, it is when the fair coin lands tails that Sleeping Beauty is awoken twice, and else she is awoken once. I all my examples I had assumed the opposite. This is just an arbitrary convention but I hope my unfortunate reversal of it didn't generated any confusion.

    In any case, in the future, I'll revert back to the conventional practice.
  • Sleeping Beauty Problem
    This isn't comparable to the Sleeping Beaty problem because being a participant isn't guaranteed. That makes all the difference. — Michael

    I appreciate your viewpoint, but I can modify my analogies to meet the condition in your first scenario.

    Scenario 3 (Lottery study)

    Imagine that tickets numbered from one to 2^100 are randomly mixed. We have a waitlist of 2^101 participants. The experimenter selects tickets one by one from the mixed pile. When they draw the 'winning' ticket numbered 2^100, they select (2^100)+1 participants from the waitlist and assign them this ticket. Otherwise, they select just one participant and allocate the losing ticket drawn. This protocol guarantees that the waitlist is exhausted, and everyone gets a chance to participate.

    As one of the 2^101 participants, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, as half the participants (plus one) have been assigned this ticket.

    Scenario 4 (Sleeping Beauty)

    Let's again take tickets numbered from one to 2^100, randomly mixed. We have a waitlist of 2^100 participants. Each participant is assigned a ticket, put to sleep, and the participant with the 'winning' ticket is awoken and interviewed (2^100)+1 times, while everyone else is awoken and interviewed only once.

    As one of the 2^100 participants who has just been awoken, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, given that half the participant interviews (plus one) are with participants assigned the winning ticket. Or so I still would argue; here I am just addressing your most recent objection regarding guaranteed participation.

    Scenario 5 (Bonus for @fdrake)

    Returning to the original Sleeping Beauty problem: upon awakening and awaiting her interview, Sleeping Beauty reflects on her credence that the coin landed on heads. As she checks her smuggled-in cellphone and learns that today is Monday, her halfer position would require her to maintain her credence at 1/2, despite ruling out the possibility of it being Tuesday. The thirder position, on the other hand, allows her to adjust her credence from 2/3 to 1/2 after acquiring this new piece of information. This seems more plausible, and I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all.
  • Sleeping Beauty Problem
    Regarding betting, expected values, and probability:

    Rather than one person repeat the experiment 2^100 times, the experiment is done on 2^100 people, with each person betting that the coin will land heads 100 times in a row. 2^100 - 1 people lose, and 1 person wins, with the winner's winnings exceeding the sum of the losers' losses. The expected value of betting that the coin will land heads 100 is greater than the cost, but the probability of winning is still 1/2 ^100

    Even though I could win big, it is more rational to believe that I will lose.     — Michael

    For sure, but your new variation doesn't mirror the Sleeping Beauty problem anymore. You earlier version was better. We must rather imagine that in the unlikely event that the coin lands heads 100 times in a row, 2^101 persons get pulled from the waitlist (or else, only one person does). In that case, if you are one random person who has been pulled from the waitlist, it is twice as likely that your have been so pulled as a result of the coin having landed heads 100 times in a row than not. It is hence rational for you to make this bet and in the long run 2/3 of the people pulled from the waitlist who make that bet with be right.
  • Sleeping Beauty Problem
    She's certainly able to update it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. I'm saying that it's irrational of her to.

    The only rational approach, upon waking, is to recognize that it landing heads 100 times in a row is so unlikely that it almost certainly didn't, and that this is her first and only interview.     — Michael

    She does recognize that for the coin to land 100 times in a row is unconditionally unlikely. But why would it not be rational for her to condition her credence in the truth of this proposition on her knowledge that her awakening isn't an event that she can distinguish in respect of its cause, and that a very large number of such indistinguishable awakening events can stem from such an unlikely cause?
  • Sleeping Beauty Problem
    The difference is that the unconditional probability of being called up is very low, and so just being called up at all affects one's credence. In the Sleeping Beauty case (both the normal and my extreme version), she's guaranteed to be awoken either way. — Michael

    I can easily adjust my lottery study example such that I am guaranteed to be selected but, once selected, the very (unconditionally) low event that led to my selection (alongside a very large number of people) still is more likely than not to have been the cause of my selection. All that is needed is to shorten the waitlist by about 98%.
  • Sleeping Beauty Problem
    There's actually two spaces. See here. — Michael

    Yes, I agree with your representation.
  • Sleeping Beauty Problem
    Then you have to say the same about my extreme example. Even when she knows that the experiment is only being run once, Sleeping Beauty's credence that the coin landed heads 100 times in a row is greater than here credence that it didn't.

    And I think that's an absurd conclusion, showing that your reasoning is false.     — Michael

    I'm not sure why you think this is absurd. Compare again my lottery study example. Suppose there are one billion people on the waiting list. If a coin lands heads 20 times in a row, then, 100 million people get pulled from the list. Else, one single person gets pulled from the list. I am then informed that I got pulled from the list (but not whether I am alone or one from 100 million). Is it absurd for me to believe that the coin landed heads 20 times in a row? My credence in this proposition should be roughly 99%

    In Sleeping Beauty's case, your intuition that her credence in the high probability of the sequence of heads ought to be absurd apparently stems from your unwillingness to contemplate the possibility of her being able to updating it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. The need for her to update her credence upon being awakened stems from her lacks of power of distinguishing those events in respect of their causes (i.e. the coin flip results). Another reason why you are neglecting the need she has for updating her credence might be a result of your characterizing the experiment that is being run once as starting with a sequence of coin flips. But from Sleeping Beauty's perspective, it really begins with a random awakening.
  • Sleeping Beauty Problem
    There is a space of possible awakening/interview events A that are being characterised by the day in which they occur ((M)onday or (T)uesday) and by the state of a coin that has been flipped prior to them occurring ((H)eads or (T)ails). P(H) = P(T) = 0.5. The rest of the conditional dependencies are part of the stipulation of the problem or can be inferred from them.
  • Sleeping Beauty Problem
    I never buy betting arguments unless the random variables are set up! — fdrake

    They are!
  • Sleeping Beauty Problem
    They describe completely different approaches to modelling the problem. That doesn't immediately tell us which SB ought to model the situation as, or whether they're internally coherent. — fdrake

    One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating.
  • Sleeping Beauty Problem
    1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?

    2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads?

    Thirders answer the second question, which I believe is the wrong answer to the first question. The experiment doesn't work by randomly selecting an interview from a set of interviews after repeating the experiment several times and then dropping Sleeping Beauty into it.     — Michael

    I would rather say that the experience works by ensuring that Sleeping Beauty finds herself being awoken in circumstances that she knows to be twice as likely to occur (because twice a frequent) as a result of a coin having landed heads than as a result of this coin having landed tails. This formulation entails no metaphysical queerness.
  • Sleeping Beauty Problem
    My reasoning is that P(Awake) = 0.5 given that there are 6 possible outcomes and I will be awake if one of these is true:

    1. Heads and I am 1
    2. Tails and I am 2
    3. Tails and I am 3     — Michael

    If you mean P(Awake) to refer to the probability of your being awakened at all (on at least one day) then P(Awake) is indeed 0.5. But in that case P(Awake|Heads) should, consistently with this interpretation, refer to your being awakened at all conditioned on the case where the coin landed heads. This is (1/3+2/3)/2 = 0.5
  • Sleeping Beauty Problem
    I don't think it makes sense to say P(Awake) = 3/4. P(Awake) is just the probability that she will be woken up, which is 1. — Michael

    But I'm not saying that. What I'm saying is that she is being awoken every Mondays and she is awoken half the time on Tuesdays. So, on average, on a random day, she is being awoken 3/4 times. I am then using Bayes' theorem to deduce the probability of a random awakening having occurred on a Tuesday. But the result was rather trivial.
  • Sleeping Beauty Problem
    (I woke up early)

    The question which has been eating me is "What is the probability of the day being Tuesday?". I think it's necessary to be able to answer that question for the thirder position. But I've not found a way of doing it yet that makes much sense. Though I'm sure there is a way! — fdrake

    P(Tuesday|Awoken) = (P(Awoken|Tuesday) / P(Awoken)) * P(Tuesday)

    Sleeping Beauty is awoken with probability 3/4 on an average day (Monday or Tuesday). On Tuesdays, she is awoken with P = 1/2. Therefore, P(Awoken|Tuesday) / P(Awoken) = (1/2)/(3/4) = 2/3.

    This (2/3) is the Bayesian updating factor. The unconditioned probability of the day being Tuesday is 1/2. The updated probability therefore is P(Tuesday|Awoken) = (2/3)*(1/2) = 1/3, as expected.
  • Sleeping Beauty Problem
    I think you numbers there are wrong. See this. — Michael

    In the quoted post you say: "P(Awake|Heads) is just the prior probability that she will be woken up if the coin lands heads"

    I think my lottery study analogy suggests a better interpretation of the P(Awoken|Heads)/P(Awoken) Bayesian updating factor. But I must go to sleep now. Thanks for engaging! I'll be back with equal probability on one of my next two awakenings.
  • Sleeping Beauty Problem
    Also this makes no sense. You can't have a probability of 2. — Michael

    This is not a probability. It's a ratio of probabilities that I have expressed as a ratio of corresponding frequencies. The updated probability P(Heads|Awoken) is 2/3. The quoted ratio being larger than one just reflects the fact that Bayesian updating results in a probability increase in this case.
  • Sleeping Beauty Problem
    Being able to bet twice if it lands tails, and so make more money, doesn’t make it more likely that it landed tails; it just means you get to bet twice.

    You might as well just say: you can place a £1 bet on a coin toss. If you correctly guess heads you win £1; if you correctly guess tails you win £2.

    Obviously it’s better to bet on tails, but not because tails is more probable.     — Michael

    It makes it twice as likely that individual bets are winning bets. Right? Likewise in Sleeping Beauty's problem, the fact that she is being awoken twice when the coin lands heads makes it more likely that a randomly selected awakening is the result of a coin having landed heads. When she if afforded the opportunity to make one singe bet on any given awakening, her expected value when making this bet is conditioned on the probability that this particular awakening is the result of the coin having landed heads. Do you agree that her expected value for this single bet (in my scenario) is $120? If she would rather judge that the probability for the coin having landed heads is 1/2, she should expect the expected value of her bet to be $90 and would be rationally justified to decline waging $100.
  • Sleeping Beauty Problem
    This is a follow-up to my previous post.

    How do you condition on such a thing? What values do you place into Bayes' theorem?

    P(Heads|Questioned)=P(Questioned|Heads)∗P(Heads) / P(Questioned)     — Michael

    In the case of Sue's selection to participate in the lottery study, we have

    P(Heads|Selected)=P(Selected|Heads)∗P(Heads) / P(Selected)

    Since on each fair coin toss, 1.5 participants are being selected on average and when the coin lands on heads 2 participants are selected, P(Selected|Heads) / P(Selected) is 2/1.5 = 4/3.

    P(Heads|Selected) therefore is 4/3 * P(Heads) = (4/3)*(1/2) = 2/3

    Likewise, in the case of Sleeping Beauty

    P(Heads|Awoken) =(P(Awoken|Heads) / P(Awoken)) * P(Heads) = 2/1.5 * 1/2 = 2/3
  • Sleeping Beauty Problem
    The simplest "experiment" is just to imagine yourself in Sleeping Beauty's shoes. — Michael

    Wasn't that rather the Cinderella problem?

    You're inviting us to imagine ourselves in Sleeping Beauty's shoes to support the halfer position. However, there are other ways to imagine her situation that can be argued to support the thirder position. Consider the following thought experiment:

    Suppose we introduce a character, Sue, who signs up for a lottery study. She joins a waitlist where participants are selected one or two at a time based on a coin toss: two participants when it lands heads and one when it lands tails. Upon being selected, they're given the option to buy a ticket for $100 that could potentially be worth $180 if the coin had landed heads (or else is worth nothing).

    The expected value of the ticket, and whether Sue should purchase it, depends on her credence about the coin toss. If Sue holds the halfer position, her credence is 1/2, and the expected value of the ticket is $90. Hence, she shouldn't buy it. However, if Sue holds the thirder position, her credence in the proposition that the coin landed heads is 2/3, making the ticket's expected value $120. Hence, she should buy it.

    Sue could argue for the thirder position as follows: if she has been selected from the waiting list, it is twice as likely that she has been chosen (together with another participant) as a result of the coin landing heads. As a frequent participant in the study, Sue would find, over time, that she profits if she always buys the ticket (an average net gain of $20 per participation), which corroborates the thirder position.

    To make this scenario more directly analogous to the original problem, let's imagine that Sleeping Beauty, upon each awakening, can not only express her belief about the coin toss but also place a bet on it. In the long run, she would profit from taking the bet as a thirder, further reinforcing this position.

    The pitfall of the 'imagine-yourself-in-her-shoes' argument lies in conflating Sue's perspective with the experimenter's by focusing only on Sue's situation before the coin toss. Eventually, everyone on the waitlist will be selected, just as Sleeping Beauty is guaranteed to be awoken at least once. Her credence that the coin will land heads is 1/2 before being put to sleep, and the same is true for the study participants before they're selected. However, once the coin has been tossed and they've been chosen, their credence about the value of their tickets being $180 (and that the coin landed on heads) should be updated to 2/3. The same applies to Sue's credence that her current awakening was due to a coin landing heads up.
  • Two envelopes problem
    There's the question of whether the "Bivariate Distribution Specification" reflects the envelope problem. It doesn't reflect the one on Wiki. The reason being the one on the wiki generates the deviate (A,A/2) OR (A,2A) exclusively when allocating the envelope, which isn't reflected in the agent's state of uncertainty surrounding the "other envelope" being (A/2, 2A).

    It only resembles the one on the Wiki if you introduce the following extra deviate, another "flip" coinciding to the subject's state of uncertainty when pondering "the other envelope":     — fdrake

    In the Wikipedia article, the problem is set up thus: "Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?"

    Your setup for the bivariate distribution specification is a special case of the problem statement and is perfectly in line with it. Let's call our participant Sue. Sue could be informed of this specific distribution, and it would represent her prior credence regarding the contents of the initially chosen envelope. If she were then to condition the Expected Value (EV) of switching on the hypothetical situation where her initially chosen envelope contains $10, the EV for switching, in that particular case, would be positive. This doesn't require an additional coin flip. She either is in the (5, 10) case or the (10, 20) case, with equal prior (and equal posterior) probabilities in this scenario. However, this is just one hypothetical situation.

    There are other scenarios to consider. For instance, if Sue initially picked an envelope containing $5, she stands to gain $5 with certainty by switching. Conversely, if she initially picked an envelope with $20, she stands to lose $10 with certainty by switching.

    Taking into account all three possibilities regarding the contents of her initially chosen envelope, her EV for switching is the weighted sum of the updated (i.e. conditioned) EVs for each case, where the weights are the prior probabilities for the three potential contents of the initial envelope. Regardless of the initial bivariate distribution, this calculation invariably results in an overall EV of zero for switching.

    This approach also underlines the flaw the popular argument that, if sound, would generate the paradox. If we consider an initial bivariate distribution where the potential contents of the larger envelope range from $2 to $(2^m) (with m being very large) and are evenly distributed, it appears that the Expected Value (EV) of switching, conditioned on the content of the envelope being n, is positive in all cases except for the special case where n=2^m. This would suggest switching is the optimal strategy. However, this strategy still yields an overall EV of zero because in the infrequent situations where a loss is guaranteed, the amount lost nullifies all the gains from the other scenarios. Generalizing the problem in the way I suggested illustrates that this holds true even with non-uniform and unbounded (though normalizable) bivariate distributions.

    The normalizability of any suitably chosen prior distribution specification (which represents Sue's credence) is essentially a reflection of her belief that there isn't an infinite amount of money in the universe. The fallacy in the 'always switch' strategy is somewhat akin to the flaw in Martingale roulette strategies.
  • Two envelopes problem
    You can conclude either strategy is optimal if you can vary the odds (Bayes or nonconstant probability) or the loss function (not expected value). Like if you don't care about amounts under 20 pounds, the optimal strategy is switching. Thus, I'm only really interested in the version where "all results are equally likely", since that seems essential to the ambiguity to me. — fdrake

    If we assume that all results are equally likely, the EV of switching given that the chosen envelope was seen to contain n is (2n + n/2)/2 - n = 1.5n. Hence whatever value n might be seen in the initially chosen envelope, it is irrational not to switch (assuming only our goal is to maximize EV). This gives rise to the paradox since if, after the initial dealing, the other envelope had been chosen and its content seen, switching would still be +EV.

    As I wrote, the prior probabilities wouldn't be assigned to the numbers (5,10,20), they'd be assigned to the pairs (5,10) and (10,20). If your prior probability that the gameshow host would award someone a tiny amount like 5 is much lower than the gigantic amount 20, you'd switch if you observed 10. But if there's no difference in prior probabilities between (5,10) and (10,20), you gain nothing from seeing the event ("my envelope is 10"), because that's equivalent to the disjunctive event ( the pair is (5,10) or (10,20) ) and each constituent event is equally likely — fdrake

    I did indeed first assigned priors to the two cases—(5, 10) and (10, 20)—and only derivatively calculated priors regarding the possible contents of the first chosen envelope (or of the other one).

    Edit: then you've got to calculate the expectation of switching within the case (5,10) or (10,20). If you specify your envelope is 10 within case... that makes the other envelope nonrandom. If you specify it as 10 here and think that specification impacts which case you're in - (informing whether you're in (5,10) or (10,20), that's close to a category error. Specifically, that error tells you the other envelope could have been assigned 5 or 20, even though you're conditioning upon 10 within an already fixed sub-case; (5,10) or (10,20).

    The conflation in the edit, I believe, is where the paradox arises from. Natural language phrasing doesn't distinguish between conditioning "at the start" (your conditioning influencing the assignment of the pair (5,10) or (10,20) - no influence) or "at the end" (your conditioning influencing which of (5,10) you have, or which of (10,20) you have, which is totally deterministic given you've determined the case you're in).

    I am not conditioning within an already fixed sub-case. Seeing that there is $10 in one envelope still leaves it open that there might be $5 or $20 dollars in the other one. To be clear, you seems to be emphasizing the role of conditioning and timing, arguing that my observation of the $10 should not affect my decision about whether I'm in the (5,10) or (10,20) case. You appear to argue that, by the time I've observed the $10, the case is already determined, and I can't switch between cases based on my observation.

    However, it seems that you've misunderstood my use of Bayesian updating. I am not arguing that observing the $10 allows me to switch between cases. Rather, I'm saying that, given an observation of $10, I can update my beliefs about the probability of being in the (5,10) case or the (10,20) case.

    (Some further discussion with GPT-4)
    Reveal
    GPT-4

    The only suggestion I would make is to add a sentence or two at the end summarizing your main point, to ensure it is clear to fdrake and others following the discussion. Something like:

    "Essentially, I'm arguing that upon observing a value in the chosen envelope, we can revise our beliefs about the probabilities of the different cases, which subsequently alters the expected value of switching envelopes. This can resolve the two-envelope paradox without assuming a non-uniform prior or altering the objective of maximizing expected value."

    Other than that, your reply seems to cover all necessary points.

    PN:

    I think what ultimately resolves the paradox is that when we don't actually observe the content of the initially chosen envelope, then the EV if switching is the weighted average of the EV of switching for all the suitably conditioned (updated) cases regarding the possible contents of the other envelope. And this EV is zero. Is that clear?

    GPT-4

    Yes, your point is clear.

    When we don't observe the content of the initially chosen envelope, the Expected Value (EV) of switching indeed comes down to the weighted average of the EV for each possible content of the other envelope. This, however, requires an important assumption: the weights used in this average should represent our prior beliefs about the probabilities of each case (i.e., each possible content of the other envelope). The probabilities have to be updated (conditioned) based on the information available, which in this case is none as we don't observe the content of the initially chosen envelope.

    If these conditions are met, and given that one envelope has exactly double (or half) the money of the other, the overall EV of switching is indeed zero, which aligns with our intuition that without any extra information, both envelopes are equally likely to contain the larger amount.

    PN:

    About your remark that "the probabilities have to be updated (conditioned) based on the information available, which in this case is none as we don't observe the content of the initially chosen envelope," what I am rather assuming is that, for whatever content n of the initially chosen envelope, the EV of switching in that particular case, conditionally on our being in that case, is equal to the EV of switching calculated by means of Bayesian updating if we were to gain knowledge that we are in that case. Hence, it is not being assumed that we actually have the information.

    GPT-4

    Ah, I see what you're getting at now. You're right. I apologize for misunderstanding your point.

    In the case of the Two Envelopes Problem, you are considering the expected value of switching under each possible value of the initially chosen envelope (i.e., you're taking a "case by case" approach) and then averaging these values according to their prior probabilities. In other words, you are considering hypothetical situations in which you would know the value of the initially chosen envelope, even though you do not actually observe its content.

    This approach effectively evaluates the average EV of switching without actually knowing the content of the initially chosen envelope, but considering all possible outcomes.

    To rephrase and clarify: The expected value (EV) of switching needs to be calculated conditionally for each possible outcome. Then, these conditional EVs are averaged over, using the prior probabilities associated with each outcome (not necessarily uniform). As long as the expected values of switching are calculated properly, taking into account the specific conditions of each case, the overall expected value of switching will turn out to be zero, regardless of the specific shape of the prior distribution. This reflects the intuitive idea that without additional information, switching or not switching should yield the same overall expected outcome, hence resolving the paradox.
  • Gender is a social construct, transgender is a social construct, biology is not
    [...]This battle you define is therefore one over authority, meaning it is a political battle between the progressives and the orthodox (lower case), but it is not, as you claim, just a foolish error by the transexuals in not appreciating the old rule that sex and gender correlate. They wish to overthrow that old rule — Hanover

    This is an very enlightening analogy.
  • Exploring the artificially intelligent mind of GPT4
    Seems to me that one of the big players who’s completely failed to catch this train, is Amazon. I’ve been using Alexa devices for about eighteen months, and they’re pretty lame - glorified alarm clocks, as someone said. — Wayfarer

    They are in hot pursuit: Amazon plans to rework Alexa in the age of ChatGPT
  • Two envelopes problem
    Nevertheless, if they observe n=10 in the first envelope, I still think there's a problem with assigning a probability distribution on the values (5, 20) in the other envelope. This is because that stipulates there being three possible values in the envelopes combined; (5, 10, 20); whereas the agent knows only two are possible. [...] — fdrake

    Your assertion that 'only two values are possible' for the contents of the envelopes in the two-envelope paradox deserves further exploration. If we consider that the potential amounts are $(5, 10, 20), we might postulate some prior probabilities as follows:

    P_1 = P(a) = P(($5, $10)) = 3/4,
    P_2 = P(b) = P(($10, $20)) = 1/4,

    which translates into priors for the unopened envelope:

    P_3 = P(A) = P(($5)) = 3/8,
    P_4 = P(B) = P(($10)) = 1/2,
    P_5 = P(C) = P(($20)) = 1/8.

    This distribution could reflect an informed guess about Joe, the envelope-filler, who is more likely to risk a smaller rather than a larger amount.

    Suppose Ann chooses an envelope. If it contains either $5 or $20, she can unambiguously update her priors to 1 and 0, or 0 and 1, respectively. The decision to switch or not becomes trivial. If, however, her envelope contains $10, she must update her beliefs about the contents of the other envelope using Bayes' theorem:

    P_updated(A) = P_updated((unseen=$5)) = P((unseen=$5) | (seen=$10)) = (1 * 3/8) / (1/2) = 3/4.

    Given this posterior, if Ann sees $10 in her envelope, the expected value (EV) for switching is negative:

    (3/4)$5 + (1/4)$20 - $10 = -$1.25.

    Therefore, she should retain her $10, as her prior for Joe having included $20 is sufficiently low. Regardless, before she inspects the second envelope, both outcomes ($5 or $20) remain possible.

    If we return to the original problem scenario (addressing @Michael's concern), where the first envelope remains sealed, the initial value probabilities become (3/8, 1/2, 1/8) for $5, $10, and $20 respectively. This gives an initial expected value of:

    3/8 * $5 + 1/2 * $10 + 1/8 * $20 = $9.375.

    The expected value if Ann switches relies on the weighted sum of the expected values for the unopened envelope, conditional on the potential contents of the chosen envelope. As choices of $5 and $20 guarantee $10 in the other envelope, while a choice of $10 leads to an expected value of $8.75 for the other envelope, this calculates to:

    3/8 * $10 + 1/2 * $8.75 + 1/8 * $10 = $9.375. (Lo and behold!)
  • Two envelopes problem
    And given that the larger number is twice the value of the smaller number, the probability that the other side is half the value is 1/2 and the probability that the other side is twice the value is 1/2.

    Which step in this line of reasoning do you disagree with?     — Michael

    There is nothing there that I disagree with. But I don't think the paradox arises if the values of the two envelopes are stipulated in advance ($10 and $20, say). The paradox arises when we stipulate that the probability of the second envelope containing the largest amount is 1/2 and is not conditionally dependent on the value of the initially chosen envelope. In the example that you give, the probability of the second envelope containing $20 conditionally on the first envelope containing $10 is 1 (and vice versa).
  • Two envelopes problem
    ↪Michael


    Indeed, I concur that the paradox can also manifest when the option to switch envelopes is offered prior to opening the initially chosen one. The resolution I (and @sime also, I think) proposed also applies in this scenario. The apparent rationality of switching, in this case, is predicated on the concept that the Expected Value (EV) of the decision is expressed as (10n + n/10)/2 - n, a value that remains positive irrespective of n. This line of thought, however, is based on the assumption that the probabilities for the second envelope containing either 10n or n/10 are independent of the value of n.

    If we adjust this assumption to reflect that these probabilities are indeed conditional upon n (in the manner that Bayesian updating with an informative prior would suggest), then it becomes plausible to hypothesize — and likely not too challenging to demonstrate — that the EV of switching remains zero.
  • Exploring the artificially intelligent mind of GPT4
    Thanks! Actually as far as I know, it’s still ChatGPT - I’m signing in via OpenAI although whether the engine is the same as GPT-4, I know not. Also appreciate the ref to Haugeland. — Wayfarer

    Unless you are subscribing to ChatGPT Plus (for $20 per month), it's GPT-3.5 you have access to. When you subscribe to ChatGPT Plus, you can then select the GPT-4 model when you start a new conversation. You can also interact with another version of GPT-4 for free by using Microsoft's new Bing through the Edge browser.
  • Exploring the artificially intelligent mind of GPT4
    It might by chance find a correct reference. But Equally it might make up a new reference. — Banno

    In my experience, GPT-3.5 is much more liable to make up references whenever there is any sort of gap in its knowledge. GPT-4 very seldom does so when the topic under discussion isn't very arcane and there is a sufficient amount of material in its training data for it to have been able to memorise it and/or extract the gist of it. GPT-4 is much more likely to spontaneously acknowledge that it doesn't know something. The big liability of LLMs is that, in those cases where (1) their knowledge and understanding of a topic is tenuous or nebulous, and (2) they ends up making stuff up about it, they are quite unable to become aware on their own that the opinion they expressed isn't derived from external sources. They don't know what it is that they know and what it is that they don't know. Their training data isn't directly accessible to them and their don't have meta-cognitive strategies that might enable them to distinguish recall from confabulation.
  • Two envelopes problem
    A Bayesian analysis reveals that the culprit of the paradox is the assignment of a non-informative prior to the distribution that generates the envelopes contents. — sime

    My understanding and resolution of the paradox is somewhat aligned with this perspective. The paradox was first introduced to me about 30 years ago by a friend who was a professor in statistics at UQAM (Université du Québec à Montréal). After further thought (and only after I was introduced to Bayes' theorem) I realized that the situation where it appears beneficial to switch the initially chosen envelope arises when we make an unrealistic assumption: that our belief about the probability distribution over possible envelope contents is that it is both uniform and infinite.

    However, given any reasonably well-defined (and bounded) prior, opening one envelope may indeed inform our decision to either switch or stick to the original choice. This decision would be guided by the Expected Value (EV) of switching, which in turn is dictated by the revised probabilities concerning the potential contents of both envelopes. Notably, there's only one unique amount in the initially chosen envelope that would result in a zero EV for switching, rendering the choice indifferent.

    The paradox seems to emerge from the assumption that opening the initial envelope provides equal probabilities for the second envelope containing either 10n or n/10 the amount in the first one, irrespective of the value of n. This is where I believe the core misunderstanding lies.
  • Exploring the artificially intelligent mind of GPT4
    Maybe Heidegger got it from there. — Jamal

    In his paper, 'Truth and Rule Following', John Haugeland artfully fuses Kant's concept of 'synthetic a priori' and the phenomenological/existential idea of 'always already there'. Although he does not extensively use these specific terminologies – I would need to revisit the paper for precise references – it's evident that both concepts heavily influence his ideas about 'constitutive rules' and 'existential commitments to constituted domains'. Haugeland acknowledges his indebtedness to both Kant (via Sellars) and Heidegger. I enjoyed how 'Wayfarer' engaged 'ChatGPT' (presumably GPT-4) to elaborate on this intricate connection. In future conversations, I might be inclined to explore with GPT-4 to what extent those notions highlight features its own artificial mind.
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    Imagine feeling obliged to defend this degenerate. — Mikie

    Some people find his performances merely laughable, others find them merely repugnant. Another false dichotomy.
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