You write: My position is very similar to Ata’s. I don’t believe that the term “probability” is completely unambiguous once we start including weird scenarios that fall outside the scope which standard probability was intended to address.
The usual Sleeping Beauty problem is only mildly fantastic, and does NOT fall outside the scope which standard probability theory addresses. Ata argues that probabilities are not meaningful if they cannot be used for a decision problem, which I agree with. But Ata’s argument that Sleeping Beauty is a situation where this is an issue seems to be based on a simple mistake.
Ata gives two decision scenarios, the first being
Each interview consists of Sleeping Beauty guessing whether the coin came up heads or tails, and being given a dollar if she was correct. After the experiment, she will keep all of her aggregate winnings.
In this situation, Beauty makes the correct decision if she gives probability 1⁄3 to Heads, and hence guesses Tails. (By making the payoff different for correct guesses of Heads vs. Tails, it would be possible to set up scenarios that make clear exactly what probability of Heads Beauty should be using.)
The second scenario is
Each interview consists of Sleeping Beauty guessing whether the coin came up heads or tails. After the experiment, she will be given a dollar if she was correct on Monday.
In this scenario, it makes no difference what her guess is, which Ata says corresponds to a probability of 1⁄2 for Heads. But this is simply a mistake. To be indifferent regarding what to guess in this scenario, Beauty needs to assign a probability of 1⁄3 to Heads. She will then see three equally likely possibilities:
Heads and it’s Monday
Tails and it’s Monday
Tails and it’s Tuesday
The differences in payoff for guessing Heads vs. Tails for these possibilities are +1, −1, and 0. Taking the expectation with respect to the equal probabilities for each gives 0, so Beauty is indifferent. In contrast, if Beauty assigns probability 1⁄2 to Heads, and hence probability 1⁄4 to each of the other possibilities, the expected difference in payoff is +1/4, so she will prefer to guess Heads. By using different payoffs for correct guesses of Heads vs. Tails, it is easy to construct an “only Monday counts” scenario in which Beauty makes a sub-optimal decision if she assigns any probability other than 1⁄3 to Heads. See also my comment on an essentially similar variation at https://www.lesswrong.com/posts/u7kSTyiWFHxDXrmQT/sleeping-beauty-resolved
I think that 1⁄3 probablity for Heads in fact leads to the correct decision with any betting scheme in the usual Sleeping Beauty problem. There is no difficulty in applying standard probability and decision theory. 1⁄3 is simply the correct answer. Other answers are the result of mistakes in reasoning. Perhaps something more strange happens in more fantastic versions of Sleeping Beauty, but when the only fantastic aspect is memory erasure, the answer is quite definitely 1⁄3.
That the answer is 1⁄3 is even more clear for the Sailor’s Child problem. But you say: I don’t agree that the probability for Sailor’s child is necessarily 1⁄3. It depends on whether you think this effect should be handled in the probability or the decision theory. I would like to emphasize again that the Sailor’s Child problem is completely non-fantastical. It could really happen. It involves NOTHING that should cause any problems in reasoning out the answer using standard methods. If standard probability and decision theory can’t be unambiguously applied to this problem, then they are flawed tools, and the many practical applications of probability and decision theory in fields ranging from statistics to error-correcting codes would be suspect. Saying that the answer to the Sailor’s Child problem depends on some sort of subjective choice of whether “this effect should be handled in the probability or the decision theory” is not a reasonable position to take.
“I think that 1⁄3 probability for Heads in fact leads to the correct decision with any betting scheme in the usual Sleeping Beauty problem”—As does 1⁄2 probability with decisions weighted by number of repeats.
“In contrast, if Beauty assigns probability 1⁄2 to Heads, and hence probability 1⁄4 to each of the other possibilities, the expected difference in payoff is +1/4, so she will prefer to guess Heads”—Actually the chance of occurring(<Tails and it’s Monday>) in the half solution is 1⁄2, as is the occurring(<Tails and it’s Tuesday>) ⇒ expected value of 0. The probabilities will only be 1⁄4 if when two possibilities overlap, we randomly choose one to be “selected”. So it is selected(<Tails and it’s Monday>) that is 1⁄4. We get to the equivalent occurring value by doubling.
“I would like to emphasize again that the Sailor’s Child problem is completely non-fantastical. It could really happen”—Whether or not it is fantastical is irrelevant, standard probability doesn’t support indexicals. That doesn’t make it a flawed tool as we can usually substitute in exact values to avoid such issues. Even when that fails, there is still the possibility to extent it.
Each time Beauty guesses, it is either Monday or Tuesday. She doesn’t know for sure which, but that’s only because she has decided to follow the rules of the game.. If instead she takes out her ax, smashes a hole in the wall of her room, goes outside, and asks a passerby what day of the week it is, she will find out whether it is Monday or Tuesday. Ordinary views about the reality of the physical world say she should regard it as being either Monday or Tuesday regardless of whether she actually knows which it is. For each decision, there are no “repeats”. She either wins a dollar as a result of that decision or she does not.
This should all be entirely obvious. That it is not obvious to you indicates that you are insisting on addressing only some fantastic version of the problem, in which it can somehow be both Monday and Tuesday at the same time, or something like that. Why are you so reluctant to figure out the answer to the usual, only mildly-fantastic version? Don’t you think that might be of some interest?
Similarly, you seem to be inexplicably reluctant to admit that the answer for the Sailors Child problem is 1⁄3. Really, unless I’ve just made some silly mistake in calculation (which I highly doubt), the answer is 1⁄3. Your views regarding indexicals are not relevant. The Sailor’s Child problem is of the same sort as are solved every day in numerous practical applications of probability, The standard tools apply. They give the answer 1⁄3.
I was using “repeats” to simply mean that she is interviewed twice.
“If instead she takes out her ax, smashes a hole in the wall of her room, goes outside, and asks a passerby what day of the week it is, she will find out whether it is Monday or Tuesday”—yes, but these possibilities aren’t exclusive. She may smash a hole in the wall and discover in is Monday, then do the same thing and discover that it is Tuesday when she is memory wiped. I’ll assume that we want our probabilities to sum to 1 (though I can imagine someone responding to the possibility of multiple interviews by allowing them to be more than 1). In that case we have to make the cases exclusive; one way to do that is to selected(<Tails and it’s Monday>) instead of occurring(<Tails and it’s Monday>). Otherwise, we can follow the thirders. But neither way is necessarily wrong.
“That it is not obvious to you indicates that you are insisting on addressing only some fantastic version of the problem, in which it can somehow be both Monday and Tuesday at the same time, or something like that”—no, I’m just saying that occurs(<Tails and it’s Monday>) overlaps with occurs(<Tails and it’s Tuesday>). Admittedly, I previously wrote “occurring” instead of “occurs” which may have confused the matter. Here “occurs” means happens at some point—whether past, present or future.
“The Sailor’s Child problem is of the same sort as are solved every day in numerous practical applications of probability”—sure, but that doesn’t mean that it falls into the scope of standard probability theory without any translation. How are you removing the indexicals?
But the fact that she is interviewed twice is of no relevance to her calculations regarding what to guess in one of the interviews, since her payoffs from guessing in the two interviews are simply added together. The decision problems for the two interviews can be solved separately; there is no interaction. One should not rescale anything according to the number of repetitions.
When she is making a decision, it is either Monday or Tuesday, even though she doesn’t know which, and even though she doesn’t know whether she will be interviewed once or twice. There is nothing subtle going on here. It is no different from anybody else making a decision when they don’t know what day of the week it is, and when they aren’t sure whether they will face another similar decision sometime in the future, and when they may have forgotten whether or not they made a similar decision sometime in the past. Not knowing the day of the week, not remembering exactly what you did in the past, and not knowing what you will do in the future are totally normal human experiences, which are handled perfectly well by standard reasoning processes.
The probabilities I assign to various possibilities on one day when added to the probabilities I assign to various possibilities on another day certainly do not have to add up to one. Indeed, they have to add up to two.
The event of Beauty being woken on Monday after the coin lands Tails and the event of Beauty being woken on Tuesday after the coin lands Tails can certainly both occur. It’s totally typical in probability problems that more than one event occurs. This is of course handled with no problem in the formalism of probability.
If the occurrence of an indexical in a probability problem makes standard probability theory inapplicable, then it is inapplicable to virtually all real problems. Consider a doctor advising a cancer patient. The doctor tells the patient that if they receive no treatment, their probability of survival is 10%, but if they undergo chemotherapy, their probability of survival is 90%, although there will be some moderately unpleasant side effects. The patient reasons as follows: Those may be valid probability statements from the doctor’s point of view, but from MY point of view they are invalid, since I’m interested in the probability that I will survive, and that’s a statement with an indexical, for which probability theory is inapplicable. So I might as well decline the chemotherapy and avoid its unpleasant side effects.
In the Sailor’s Child problem, there is no doubt that if the child consulted a probabilist regarding the chances that they have a sibling, the probabilist would advice them that the probability of them having a sibling is 2⁄3. Are you saying that they should ignore this advice, since once they interpret it as being about THEM it is a statement with an indexical?
“But the fact that she is interviewed twice is of no relevance to her calculations regarding what to guess in one of the interviews, since her payoffs from guessing in the two interviews are simply added together”—Let’s suppose you can buy a coupon that pays $1 if the coin comes up heads and $0 otherwise. Generally, the fair price is p, where p is the probability of heads. However, suppose there’s a bug in the system and you will get charged twice if the coin comes up tails. Then the fair price (c) can be calculated as follows:
Find c such that: Expected expenditure = expected value of coupon
c(p + 2(1-p)) = p
c(2-p)= p
c = p/(2-p)
If p=1/2, then c=1/3
What’s wrong with using this system to translate between p and c so that we can figure out how to bet? The exact same system works for sleeping beauty.
“The probabilities I assign to various possibilities on one day when added to the probabilities I assign to various possibilities on another day certainly do not have to add up to one”—I was claiming that there are two options for how to extend probability to cover these situations assuming we want to make the events non-overlapping. One is to allow probabilities more than one (ie. your chance of being interviewed is 1.5 and your chance of experiencing each of the three states is 0.5).
Alternatively, you can maintain sum of probabilities being one by asking about the probability of events that are exclusive. The easiest way to do this in the standard sleeping beauty is to randomly choose only one interview to “count” in the case where there are multiple interviews. This gives the following probabilities: heads 0.5, tails + Monday selected = 0.25, tails + Tuesday selected = 0.25. The question isn’t so much whether you can construct this formalism, but whether this it is something that we care about.
“Those may be valid probability statements from the doctor’s point of view, but from MY point of view they are invalid, since I’m interested in the probability that I will survive”—If you are the patient, we can remove the indexical by asking about whether Radford Neal will survive.
Let’s suppose you can buy a coupon that pays $1 if the coin comes up heads and $0 otherwise. Generally, the fair price is p, where p is the probability of heads. However, suppose there’s a bug in the system and you will get charged twice if the coin comes up tails.
In the scenarios that Ata describes, Beauty simply gets paid for guessing correctly. She does not have to pay anything. More generally, in the usual Sleeping Beauty problem where the only fantastic feature is memory erasure, every decision that Beauty takes is a decision for one point in time only. If she is woken twice, she make two separate decisions. She likely makes the same decision each time, since she has no rational basis for deciding differently, but they are nevertheless separate decisions, each of which may or may not result in a payoff. There is no need to combine these decisions into one decision, in which Beauty is deciding for more than one point in time. That just introduces totally unnecessary confusion, as well as being contrary to what actually happens.
The easiest way to do this in the standard sleeping beauty is to randomly choose only one interview to “count” in the case where there are multiple interviews.
That’s close to Ata’s second scenario, in which Ata incorrectly concludes that Beauty should assign probability 1⁄2 to Heads. It is of course a different decision problem than when payoffs are added for the two days, but the correct result is again obtained when Beauty considers the probability of Heads to be 1⁄3, if she applies decision theory correctly. The choice of decision problem has no effect on how the probabilities should be calculated.
If you are the patient, we can remove the indexical by asking about whether Radford Neal will survive.
Yes indeed. And this can also be done for the Sailor’s Child problem, giving the result that the probability of Heads (no sibling) is 1⁄3.
a) Regardless of how sleeping beauty makes their decision, we can model it as an algorithm decided ahead of time. If part of the decision is random, we can program that in too. So we can assume they make the same meta-level decision, so have the same expected pay-off for both interviews.
b) I don’t follow the argument here? You seem to just be assuming that I am wrong?
c) We can’t just say, “Radford Neal” for sailors child without defining who will have that name.Is it one particular mother who will call their child Radford Neal if they have one? Or is a random child assigned that name?
a) But Beauty is actually a human being. If your argument depends on replacing Beauty by a computer program, then it does not apply to the usual Sleeping Beauty problem. Why are you so reluctant to actually address the usual, only-mildly-fantastic Sleeping Beauty problem?
In any case, why is it relevant that she has the same expected payoff for both interviews (which will indeed likely be the case, since she is likely to make the same decision)? Lots of people make various decisions at various times that happen to have the same expected payoff. That doesn’t magically make these several decisions be actually one decision.
b) If I understand your setup, if the coin lands Heads, Beauty gets one dollar if she correctly guesses on Monday, which is the only day she is woken. If the coin lands Tails, a ball is drawn from a bag with equal numbers of balls labeled “M” and “T”, and she gets a dollar if she makes a correct guesses on the day corresponding to the ball drawn, with her guess the other day being ignored. For simplicity, suppose that the ball is drawn (and then ignored) even if the coin lands Heads. There are then six possible situations of coin/ball/day when Beauty is considering her decision:
1) H M Monday
2) H T Monday
3) T M Monday
4) T T Monday
5) T M Tuesday
6) T T Tuesday
If Beauty is a Thirder, she considers all of these to be equally likely (probability 1⁄6 for each). In situations 4 and 5, her action has no effect, so we can ignore these in deciding on the best action. In situations 1 and 2, guessing Heads results in a dollar reward. In situations 3 and 6, guessing Tails results in a dollar reward. So she is indifferent to guessing Heads or Tails.
c) Really, can you actually not suppose that in the Sailor’s Child problem, which is explicitly designed to be a problem that could actually occur in real life, the child has not been given a name? And if so, do you also think that if the child gets cancer, as in the previous discussion, that they should refuse chemotherapy on the grounds that since their mother did not give them a name, they are unable to escape the inapplicability of probability theory to statements with indexicals? I’m starting to find it hard to believe that you are actually trying to understand this problem.
a) Even if sleeping beauty is a human, they are still a deterministic (or probabilistically deterministic) machine, so their responses in any scenario can be represented by an algorithm.
b) The halfer gets the same solution (indifference) too as 1), 2), 5) and 6) are all assigned a probability of 1⁄4; whilst 3) and 4) are ignored.
c) My point isn’t that the child might not have a name. My point is that in order to evaluate the statement: “Radford Neal has a half-sibling” we have to define the scheme in which someone comes to be called Radford Neal.
So, suppose the two potential mothers are Amy and Barbara. The first possibility is that Amy calls their child, if they have one, “Radford Neal”. However, if this is the case, it may come to pass that Amy doesn’t have a child so no-one is called Radford Neal and the reference fails. Alternatively, we might want to ensure that there is someone always called Radford Neal. If they only have one child, this is trivial, if there’s two, we could pick randomly. My point is that there isn’t a unique way of assigning the name, so I don’t know what scheme you want to use to replace the indexical.
a) You know, it has not actually been demonstrated that human consciousness can be mimicked by Turing-equivalent computer. In any case, the only role of mentioning this in your argument seems to be to push your thinking away from Beauty as a human towards a more abstract notion of what the problem is in which you can more easily engage in reasoning that would be obviously fallacious if your thoughts were anchored in reality.
b) Halfer reasoning is invalid, so it’s difficult to say how this invalid reasoning would be applied in the context of this decision problem. But if one takes the view that probabilities do not depend on what decision problem they will be used for, it isn’t possible for possibilities 5) and 6) to have probability 1⁄4 while possibilities 3) and 4) have probability zero. One can imagine, for example, that Beauty is told about the balls from the beginning, but is told about the reward for guessing correctly, and how the balls play a role in determining that reward, only later. Should she change her probabilities for the six possibilities simply because she has been told about this reward scheme? I suspect your answer will be yes, but that is simply absurd. It is totally contrary to normal reasoning, and if applied to practical problems would be disastrous. Remember! Beauty is human, not a computer program.
c) You are still refusing to approach the Sallor’s Child problem as one about real people, despite the fact that the problem has been deliberately designed so that it has no fantastic aspects and could indeed be about real people, as I have emphasized again and again. Suppose the child is considering searching for their possible sibling, but wants to know the probability that the sibling exist before deciding to spend lots of money on this search. The child consults you regarding what the probability of their having a sibling is. Do you really start by asking, “what process did your mother use in deciding what name to give you”? The question is obviously of no relevance whatsoever. It is also obvious that any philosophical debates about indexicals in probability statements are irrelevant—one way or another, people solve probability problems every day without being hamstrung by this issue. There is a real person standing in front of you asking “what is the probability that I have a sibling”. The answer to this question is 2⁄3. There is no doubt about this answer. It is correct. Really. That is the answer.
Thanks for taking the time to write all of these responses, but I suspect that we’ve become stuck. At some point I’ll write up some posts aimed at trying to argue for my position, rather than primarily aimed at addressing rebuttal and perhaps it will clear up some of these issues.
You write: My position is very similar to Ata’s. I don’t believe that the term “probability” is completely unambiguous once we start including weird scenarios that fall outside the scope which standard probability was intended to address.
The usual Sleeping Beauty problem is only mildly fantastic, and does NOT fall outside the scope which standard probability theory addresses. Ata argues that probabilities are not meaningful if they cannot be used for a decision problem, which I agree with. But Ata’s argument that Sleeping Beauty is a situation where this is an issue seems to be based on a simple mistake.
Ata gives two decision scenarios, the first being
Each interview consists of Sleeping Beauty guessing whether the coin came up heads or tails, and being given a dollar if she was correct. After the experiment, she will keep all of her aggregate winnings.
In this situation, Beauty makes the correct decision if she gives probability 1⁄3 to Heads, and hence guesses Tails. (By making the payoff different for correct guesses of Heads vs. Tails, it would be possible to set up scenarios that make clear exactly what probability of Heads Beauty should be using.)
The second scenario is
Each interview consists of Sleeping Beauty guessing whether the coin came up heads or tails. After the experiment, she will be given a dollar if she was correct on Monday.
In this scenario, it makes no difference what her guess is, which Ata says corresponds to a probability of 1⁄2 for Heads. But this is simply a mistake. To be indifferent regarding what to guess in this scenario, Beauty needs to assign a probability of 1⁄3 to Heads. She will then see three equally likely possibilities:
Heads and it’s Monday
Tails and it’s Monday
Tails and it’s Tuesday
The differences in payoff for guessing Heads vs. Tails for these possibilities are +1, −1, and 0. Taking the expectation with respect to the equal probabilities for each gives 0, so Beauty is indifferent. In contrast, if Beauty assigns probability 1⁄2 to Heads, and hence probability 1⁄4 to each of the other possibilities, the expected difference in payoff is +1/4, so she will prefer to guess Heads. By using different payoffs for correct guesses of Heads vs. Tails, it is easy to construct an “only Monday counts” scenario in which Beauty makes a sub-optimal decision if she assigns any probability other than 1⁄3 to Heads. See also my comment on an essentially similar variation at https://www.lesswrong.com/posts/u7kSTyiWFHxDXrmQT/sleeping-beauty-resolved
I think that 1⁄3 probablity for Heads in fact leads to the correct decision with any betting scheme in the usual Sleeping Beauty problem. There is no difficulty in applying standard probability and decision theory. 1⁄3 is simply the correct answer. Other answers are the result of mistakes in reasoning. Perhaps something more strange happens in more fantastic versions of Sleeping Beauty, but when the only fantastic aspect is memory erasure, the answer is quite definitely 1⁄3.
That the answer is 1⁄3 is even more clear for the Sailor’s Child problem. But you say: I don’t agree that the probability for Sailor’s child is necessarily 1⁄3. It depends on whether you think this effect should be handled in the probability or the decision theory. I would like to emphasize again that the Sailor’s Child problem is completely non-fantastical. It could really happen. It involves NOTHING that should cause any problems in reasoning out the answer using standard methods. If standard probability and decision theory can’t be unambiguously applied to this problem, then they are flawed tools, and the many practical applications of probability and decision theory in fields ranging from statistics to error-correcting codes would be suspect. Saying that the answer to the Sailor’s Child problem depends on some sort of subjective choice of whether “this effect should be handled in the probability or the decision theory” is not a reasonable position to take.
“I think that 1⁄3 probability for Heads in fact leads to the correct decision with any betting scheme in the usual Sleeping Beauty problem”—As does 1⁄2 probability with decisions weighted by number of repeats.
“In contrast, if Beauty assigns probability 1⁄2 to Heads, and hence probability 1⁄4 to each of the other possibilities, the expected difference in payoff is +1/4, so she will prefer to guess Heads”—Actually the chance of occurring(<Tails and it’s Monday>) in the half solution is 1⁄2, as is the occurring(<Tails and it’s Tuesday>) ⇒ expected value of 0. The probabilities will only be 1⁄4 if when two possibilities overlap, we randomly choose one to be “selected”. So it is selected(<Tails and it’s Monday>) that is 1⁄4. We get to the equivalent occurring value by doubling.
“I would like to emphasize again that the Sailor’s Child problem is completely non-fantastical. It could really happen”—Whether or not it is fantastical is irrelevant, standard probability doesn’t support indexicals. That doesn’t make it a flawed tool as we can usually substitute in exact values to avoid such issues. Even when that fails, there is still the possibility to extent it.
Each time Beauty guesses, it is either Monday or Tuesday. She doesn’t know for sure which, but that’s only because she has decided to follow the rules of the game.. If instead she takes out her ax, smashes a hole in the wall of her room, goes outside, and asks a passerby what day of the week it is, she will find out whether it is Monday or Tuesday. Ordinary views about the reality of the physical world say she should regard it as being either Monday or Tuesday regardless of whether she actually knows which it is. For each decision, there are no “repeats”. She either wins a dollar as a result of that decision or she does not.
This should all be entirely obvious. That it is not obvious to you indicates that you are insisting on addressing only some fantastic version of the problem, in which it can somehow be both Monday and Tuesday at the same time, or something like that. Why are you so reluctant to figure out the answer to the usual, only mildly-fantastic version? Don’t you think that might be of some interest?
Similarly, you seem to be inexplicably reluctant to admit that the answer for the Sailors Child problem is 1⁄3. Really, unless I’ve just made some silly mistake in calculation (which I highly doubt), the answer is 1⁄3. Your views regarding indexicals are not relevant. The Sailor’s Child problem is of the same sort as are solved every day in numerous practical applications of probability, The standard tools apply. They give the answer 1⁄3.
I was using “repeats” to simply mean that she is interviewed twice.
“If instead she takes out her ax, smashes a hole in the wall of her room, goes outside, and asks a passerby what day of the week it is, she will find out whether it is Monday or Tuesday”—yes, but these possibilities aren’t exclusive. She may smash a hole in the wall and discover in is Monday, then do the same thing and discover that it is Tuesday when she is memory wiped. I’ll assume that we want our probabilities to sum to 1 (though I can imagine someone responding to the possibility of multiple interviews by allowing them to be more than 1). In that case we have to make the cases exclusive; one way to do that is to selected(<Tails and it’s Monday>) instead of occurring(<Tails and it’s Monday>). Otherwise, we can follow the thirders. But neither way is necessarily wrong.
“That it is not obvious to you indicates that you are insisting on addressing only some fantastic version of the problem, in which it can somehow be both Monday and Tuesday at the same time, or something like that”—no, I’m just saying that occurs(<Tails and it’s Monday>) overlaps with occurs(<Tails and it’s Tuesday>). Admittedly, I previously wrote “occurring” instead of “occurs” which may have confused the matter. Here “occurs” means happens at some point—whether past, present or future.
“The Sailor’s Child problem is of the same sort as are solved every day in numerous practical applications of probability”—sure, but that doesn’t mean that it falls into the scope of standard probability theory without any translation. How are you removing the indexicals?
But the fact that she is interviewed twice is of no relevance to her calculations regarding what to guess in one of the interviews, since her payoffs from guessing in the two interviews are simply added together. The decision problems for the two interviews can be solved separately; there is no interaction. One should not rescale anything according to the number of repetitions.
When she is making a decision, it is either Monday or Tuesday, even though she doesn’t know which, and even though she doesn’t know whether she will be interviewed once or twice. There is nothing subtle going on here. It is no different from anybody else making a decision when they don’t know what day of the week it is, and when they aren’t sure whether they will face another similar decision sometime in the future, and when they may have forgotten whether or not they made a similar decision sometime in the past. Not knowing the day of the week, not remembering exactly what you did in the past, and not knowing what you will do in the future are totally normal human experiences, which are handled perfectly well by standard reasoning processes.
The probabilities I assign to various possibilities on one day when added to the probabilities I assign to various possibilities on another day certainly do not have to add up to one. Indeed, they have to add up to two.
The event of Beauty being woken on Monday after the coin lands Tails and the event of Beauty being woken on Tuesday after the coin lands Tails can certainly both occur. It’s totally typical in probability problems that more than one event occurs. This is of course handled with no problem in the formalism of probability.
If the occurrence of an indexical in a probability problem makes standard probability theory inapplicable, then it is inapplicable to virtually all real problems. Consider a doctor advising a cancer patient. The doctor tells the patient that if they receive no treatment, their probability of survival is 10%, but if they undergo chemotherapy, their probability of survival is 90%, although there will be some moderately unpleasant side effects. The patient reasons as follows: Those may be valid probability statements from the doctor’s point of view, but from MY point of view they are invalid, since I’m interested in the probability that I will survive, and that’s a statement with an indexical, for which probability theory is inapplicable. So I might as well decline the chemotherapy and avoid its unpleasant side effects.
In the Sailor’s Child problem, there is no doubt that if the child consulted a probabilist regarding the chances that they have a sibling, the probabilist would advice them that the probability of them having a sibling is 2⁄3. Are you saying that they should ignore this advice, since once they interpret it as being about THEM it is a statement with an indexical?
“But the fact that she is interviewed twice is of no relevance to her calculations regarding what to guess in one of the interviews, since her payoffs from guessing in the two interviews are simply added together”—Let’s suppose you can buy a coupon that pays $1 if the coin comes up heads and $0 otherwise. Generally, the fair price is p, where p is the probability of heads. However, suppose there’s a bug in the system and you will get charged twice if the coin comes up tails. Then the fair price (c) can be calculated as follows:
Find c such that: Expected expenditure = expected value of coupon
c(p + 2(1-p)) = p
c(2-p)= p
c = p/(2-p)
If p=1/2, then c=1/3
What’s wrong with using this system to translate between p and c so that we can figure out how to bet? The exact same system works for sleeping beauty.
“The probabilities I assign to various possibilities on one day when added to the probabilities I assign to various possibilities on another day certainly do not have to add up to one”—I was claiming that there are two options for how to extend probability to cover these situations assuming we want to make the events non-overlapping. One is to allow probabilities more than one (ie. your chance of being interviewed is 1.5 and your chance of experiencing each of the three states is 0.5).
Alternatively, you can maintain sum of probabilities being one by asking about the probability of events that are exclusive. The easiest way to do this in the standard sleeping beauty is to randomly choose only one interview to “count” in the case where there are multiple interviews. This gives the following probabilities: heads 0.5, tails + Monday selected = 0.25, tails + Tuesday selected = 0.25. The question isn’t so much whether you can construct this formalism, but whether this it is something that we care about.
“Those may be valid probability statements from the doctor’s point of view, but from MY point of view they are invalid, since I’m interested in the probability that I will survive”—If you are the patient, we can remove the indexical by asking about whether Radford Neal will survive.
Let’s suppose you can buy a coupon that pays $1 if the coin comes up heads and $0 otherwise. Generally, the fair price is p, where p is the probability of heads. However, suppose there’s a bug in the system and you will get charged twice if the coin comes up tails.
In the scenarios that Ata describes, Beauty simply gets paid for guessing correctly. She does not have to pay anything. More generally, in the usual Sleeping Beauty problem where the only fantastic feature is memory erasure, every decision that Beauty takes is a decision for one point in time only. If she is woken twice, she make two separate decisions. She likely makes the same decision each time, since she has no rational basis for deciding differently, but they are nevertheless separate decisions, each of which may or may not result in a payoff. There is no need to combine these decisions into one decision, in which Beauty is deciding for more than one point in time. That just introduces totally unnecessary confusion, as well as being contrary to what actually happens.
The easiest way to do this in the standard sleeping beauty is to randomly choose only one interview to “count” in the case where there are multiple interviews.
That’s close to Ata’s second scenario, in which Ata incorrectly concludes that Beauty should assign probability 1⁄2 to Heads. It is of course a different decision problem than when payoffs are added for the two days, but the correct result is again obtained when Beauty considers the probability of Heads to be 1⁄3, if she applies decision theory correctly. The choice of decision problem has no effect on how the probabilities should be calculated.
If you are the patient, we can remove the indexical by asking about whether Radford Neal will survive.
Yes indeed. And this can also be done for the Sailor’s Child problem, giving the result that the probability of Heads (no sibling) is 1⁄3.
a) Regardless of how sleeping beauty makes their decision, we can model it as an algorithm decided ahead of time. If part of the decision is random, we can program that in too. So we can assume they make the same meta-level decision, so have the same expected pay-off for both interviews.
b) I don’t follow the argument here? You seem to just be assuming that I am wrong?
c) We can’t just say, “Radford Neal” for sailors child without defining who will have that name.Is it one particular mother who will call their child Radford Neal if they have one? Or is a random child assigned that name?
a) But Beauty is actually a human being. If your argument depends on replacing Beauty by a computer program, then it does not apply to the usual Sleeping Beauty problem. Why are you so reluctant to actually address the usual, only-mildly-fantastic Sleeping Beauty problem?
In any case, why is it relevant that she has the same expected payoff for both interviews (which will indeed likely be the case, since she is likely to make the same decision)? Lots of people make various decisions at various times that happen to have the same expected payoff. That doesn’t magically make these several decisions be actually one decision.
b) If I understand your setup, if the coin lands Heads, Beauty gets one dollar if she correctly guesses on Monday, which is the only day she is woken. If the coin lands Tails, a ball is drawn from a bag with equal numbers of balls labeled “M” and “T”, and she gets a dollar if she makes a correct guesses on the day corresponding to the ball drawn, with her guess the other day being ignored. For simplicity, suppose that the ball is drawn (and then ignored) even if the coin lands Heads. There are then six possible situations of coin/ball/day when Beauty is considering her decision:
1) H M Monday
2) H T Monday
3) T M Monday
4) T T Monday
5) T M Tuesday
6) T T Tuesday
If Beauty is a Thirder, she considers all of these to be equally likely (probability 1⁄6 for each). In situations 4 and 5, her action has no effect, so we can ignore these in deciding on the best action. In situations 1 and 2, guessing Heads results in a dollar reward. In situations 3 and 6, guessing Tails results in a dollar reward. So she is indifferent to guessing Heads or Tails.
c) Really, can you actually not suppose that in the Sailor’s Child problem, which is explicitly designed to be a problem that could actually occur in real life, the child has not been given a name? And if so, do you also think that if the child gets cancer, as in the previous discussion, that they should refuse chemotherapy on the grounds that since their mother did not give them a name, they are unable to escape the inapplicability of probability theory to statements with indexicals? I’m starting to find it hard to believe that you are actually trying to understand this problem.
a) Even if sleeping beauty is a human, they are still a deterministic (or probabilistically deterministic) machine, so their responses in any scenario can be represented by an algorithm.
b) The halfer gets the same solution (indifference) too as 1), 2), 5) and 6) are all assigned a probability of 1⁄4; whilst 3) and 4) are ignored.
c) My point isn’t that the child might not have a name. My point is that in order to evaluate the statement: “Radford Neal has a half-sibling” we have to define the scheme in which someone comes to be called Radford Neal.
So, suppose the two potential mothers are Amy and Barbara. The first possibility is that Amy calls their child, if they have one, “Radford Neal”. However, if this is the case, it may come to pass that Amy doesn’t have a child so no-one is called Radford Neal and the reference fails. Alternatively, we might want to ensure that there is someone always called Radford Neal. If they only have one child, this is trivial, if there’s two, we could pick randomly. My point is that there isn’t a unique way of assigning the name, so I don’t know what scheme you want to use to replace the indexical.
a) You know, it has not actually been demonstrated that human consciousness can be mimicked by Turing-equivalent computer. In any case, the only role of mentioning this in your argument seems to be to push your thinking away from Beauty as a human towards a more abstract notion of what the problem is in which you can more easily engage in reasoning that would be obviously fallacious if your thoughts were anchored in reality.
b) Halfer reasoning is invalid, so it’s difficult to say how this invalid reasoning would be applied in the context of this decision problem. But if one takes the view that probabilities do not depend on what decision problem they will be used for, it isn’t possible for possibilities 5) and 6) to have probability 1⁄4 while possibilities 3) and 4) have probability zero. One can imagine, for example, that Beauty is told about the balls from the beginning, but is told about the reward for guessing correctly, and how the balls play a role in determining that reward, only later. Should she change her probabilities for the six possibilities simply because she has been told about this reward scheme? I suspect your answer will be yes, but that is simply absurd. It is totally contrary to normal reasoning, and if applied to practical problems would be disastrous. Remember! Beauty is human, not a computer program.
c) You are still refusing to approach the Sallor’s Child problem as one about real people, despite the fact that the problem has been deliberately designed so that it has no fantastic aspects and could indeed be about real people, as I have emphasized again and again. Suppose the child is considering searching for their possible sibling, but wants to know the probability that the sibling exist before deciding to spend lots of money on this search. The child consults you regarding what the probability of their having a sibling is. Do you really start by asking, “what process did your mother use in deciding what name to give you”? The question is obviously of no relevance whatsoever. It is also obvious that any philosophical debates about indexicals in probability statements are irrelevant—one way or another, people solve probability problems every day without being hamstrung by this issue. There is a real person standing in front of you asking “what is the probability that I have a sibling”. The answer to this question is 2⁄3. There is no doubt about this answer. It is correct. Really. That is the answer.
Thanks for taking the time to write all of these responses, but I suspect that we’ve become stuck. At some point I’ll write up some posts aimed at trying to argue for my position, rather than primarily aimed at addressing rebuttal and perhaps it will clear up some of these issues.