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.
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.