RE: R�p : Thought Experiment #269-G (Duplicates)
Lee Corbin writes: [quoting Bruno Marchal] Why not choose D, that is I will see 0 on the wall OR I will see 1 on the wall. Okay, now you have switched back to the prior (prediction) level. Here is the reason not to say that. As the person who is about to be duplicated knows all the facts, he is aware (from a 3rd person point of view) that scientifically there will be *two* processes both of which are very, very similar. It will be false that one of them will be more him than the other. Therefore he must identify equally with them. Therefore, it is wrong to imply that he I will be one of them but not the other of them. But if you answer I will see 0 on the wall OR I will see 1 on the wall then it makes it sound as though one of those cases will obtain but not the other. (This is usually how we talk when Bruno admits, for example, that tonight he either will watch TV *or* he will not watch TV. But the case of duplicates is not like that. In the case of duplicates, it is a scientific fact that Bruno will watch TV (in one room) and will not watch TV (in the other room). In short, it will be true that Bruno will watch TV and will not watch TV---simply because there will be two instances of Bruno.) Is there any way of asking the question such that the answer is there is an even chance that I will see either a 1 or a 0? For example, every time I flip a coin it *seems* that I get either heads or tails, and not both. The objective truth may well be that coin-tossing causes duplication and I do, in fact, experience both, but don't realise it. I am interested in asking and/or answering the question assuming this sort of ignorance. Can it be done, or is it linguistically as well as physically and logically impossible? --Stathis Papaioannou _ FREE pop-up blocking with the new MSN Toolbar get it now! http://toolbar.msn.click-url.com/go/onm00200415ave/direct/01/
Rép : Thought Experiment #269-G (Duplicates )
Le 08-juil.-05, à 00:59, Lee Corbin a écrit : Bruno writes Each Lee-i is offered 5$ each time his bet is confirmed, but loses 5$ if he makes a wrong bet. And yes, it would be possible to emphasize to each instance that he is to attempt to maximize his own instance's earnings. Quite correct. What will be your strategy in each version? Will your strategy differ? Now if the Lees know all these facts, then they'll anticipate being in both rooms upon each iteration. Therefore, they'll anticipate losing $5 in one room and gaining $5 in the other. They'll also realize that all bit sequences are being carried out. Therefore, it doesn't make any difference whatsoever. The expectation of each sequence is exactly the same number of dollars: zero. I don't get the significance of this. I don't understand your answer, and actually you did not answer. It looks like you are forgetting I give you the choice between A, B, C, D. Sorry. You are asked to bet on your immediate and less immediate future feeling. Precisely: we ask you to choose among the following bets: Immediate: A. I will see 0 on the wall. B. I will see 1 on the wall. C. I will see 0 on the wall and I will see 1 on the wall. D. I will see 0 on the wall or I will see 1 on the wall. I choose C: insofar as I consider myself as a program, then the program will see 0 on the wall and the program will also see 1 on the wall. The program will experienced both. I will experience both. I guess you did choose C, without saying. Right. In that case you are correct the expectation will be zero. Are you sure there is not a better strategy among A, B, C, D? Why do you think that there is a better strategy? C. will comport with all the facts. And afterwards, when a poll is conducted among all those who can prove that they are Lee Corbin it will be found that half of them saw a 1 and half saw a zero. It is preposterous to finger *any* of them and accuse them of not being me. They will each believe that they are me (i.e., the me here in the past). That is, for each Lee', they will assert Lee' = Lee. So also will Lee'' assert that Lee'' = Lee. So IT'S FREAKING OBVIOUS THAT Lee'' = Lee'. Yet substitute someone else's name for mine in those equations, and they'll demur. As I said I have no problem with accepting Lee = Lee' = Lee'' (although I think this will entail Lee = Bruno at some point, but I have no problem with that and we can come back to this notion later). But I was not argumentating on personal identity, only on the problem you face when predicting your immediate future (or less future) experience. It is a different matter. I duplicate you iteratively, by annihilating (painlessly!) you and reconstituting you in the 0-room and the 1-room which differs from having a 0 (resp. 1) painted on a wall. And I let you choose between the bets A, B, C, D described above. You choose C, that is: I will see 0 on the wall and I will see 1 on the wall. Now, as I said this is ambiguous. So if I am in a bad mood, asking the first 0-Lee' about its immediate apprehension if he answers me I am seeing 0 and I am seeing 1 I consider it as false (0-Lee' sees only 0!), and the same for the other Lee, so all the 2^n Lee must give 5$. If I am in a good mood, I accept your reasoning and you loose nothing, but also you win nothing. Why not choosing D, that is I will see 0 on the wall OR I will see 1 on the wall. I recall you that p or q is true if p is true or q is true. So with D all the Lee will win. D consists into admitting that you are ignorant about your immediate apprehension after the duplication. It has nothing to do with the fact that you are the two Lee. OK? Bruno http://iridia.ulb.ac.be/~marchal/
RE: Rép : Thought Experiment #269-G (Duplicates)
Bruno writes You are asked to bet on your immediate and less immediate future feeling. Precisely: we ask you to choose among the following bets: Immediate: A. I will see 0 on the wall. B. I will see 1 on the wall. C. I will see 0 on the wall and I will see 1 on the wall. D. I will see 0 on the wall or I will see 1 on the wall. As I said I have no problem with accepting Lee = Lee' = Lee'' (although I think this will entail Lee = Bruno at some point, but I have no problem with that and we can come back to this notion later). But I was not argumentating on personal identity, only on the problem you face when predicting your immediate future (or less future) experience. It is a different matter. You asked me to *predict*. I did. I duplicate you iteratively, by annihilating (painlessly!) you and reconstituting you in the 0-room and the 1-room which differs from having a 0 (resp. 1) painted on a wall. And I let you choose between the bets A, B, C, D described above. You choose C, that is: I will see 0 on the wall and I will see 1 on the wall. Now, as I said this is ambiguous. So if I am in a bad mood, asking the first 0-Lee' about its immediate apprehension if he answers me I am seeing 0 and I am seeing 1 I consider it as false (0-Lee' sees only 0!), and the same for the other Lee, so all the 2^n Lee must give 5$. Now you are asking an instance a question (since there are two of me), and it seems that you are playing on the ambiguity of the term you. When you ask an instance---now, *after* the copying has been done---whether he is seeing a 1 or seeing a 0 or seeing both, he has to stop you (I mean I have to stop you) and ask exactly what kind of information you are after. Clearly, if you are talking to one instance (so far as that instance knows) he'll say that he is seeing a 1 or he will say that he is seeing a 0. This is because he'll take the usual meaning of terms. When you then inform him that he has actually been copied and that there is another instance of him in the other room, then naturally he should say Okay, here I am seeing a 0 and in the other room the opposite. We know all the facts. What we want are two things: (1) we want to speak clearly. (2) we want to know whether or not to regard our duplicates as selves. I think that you've heard all my arguments. Why not choose D, that is I will see 0 on the wall OR I will see 1 on the wall. Okay, now you have switched back to the prior (prediction) level. Here is the reason not to say that. As the person who is about to be duplicated knows all the facts, he is aware (from a 3rd person point of view) that scientifically there will be *two* processes both of which are very, very similar. It will be false that one of them will be more him than the other. Therefore he must identify equally with them. Therefore, it is wrong to imply that he I will be one of them but not the other of them. But if you answer I will see 0 on the wall OR I will see 1 on the wall then it makes it sound as though one of those cases will obtain but not the other. (This is usually how we talk when Bruno admits, for example, that tonight he either will watch TV *or* he will not watch TV. But the case of duplicates is not like that. In the case of duplicates, it is a scientific fact that Bruno will watch TV (in one room) and will not watch TV (in the other room). In short, it will be true that Bruno will watch TV and will not watch TV---simply because there will be two instances of Bruno.) I recall you that p or q is true if p is true or q is true. So with D all the Lee will win. D consists into admitting that you are ignorant about your immediate apprehension after the duplication. It has nothing to do with the fact that you are the two Lee. OK? Nope. :-) With D, I am pretending that it is like you watching television---either 0 or 1. But with duplicates it is not like that: instead it is like 0 AND 1. Lee
Re: Thought Experiment #269-G (Duplicates)
Le 06-juil.-05, à 02:44, Lee Corbin a écrit : Bruno wrote about whether or not we are all the same person. Sent: Tuesday, July 05, 2005 1:59 AM Subject: Re: What does ought mean? (was RE: Duplicates Are Selves) I have changed the subject line once again, because this is no longer about what ought ought to mean. Le 04-juil.-05, à 22:18, Lee Corbin a écrit : Yes, but I contend that while there are two organisms present, there is only one person. It's much as though some space aliens kidnapped you and tried to say that Pete at spacetime coordinates (X1,T1) could not possibly be the same person as Pete at coordinates (X1,T2) because the times weren't the same. You'd have to get them to wrap their heads around the idea that one person could be at two different times in the same place. They might find this bizarre. I'm trying to tell you a possibility that you think equally bizarre: namely that Pete(X1,T1) is the same person as Pete(X2,T1), namely that the same person may be at two different locations at the same time. That's all. I like that idea, but if they are the *same* person then we are all the same person. Or, perhaps you were just meaning that they are very close/similar; That is so: but moreover, being very very close/similar is what should be meant by the same person. in which case you can say Pete(X2,T1) is much closer to Pete(X1,T1) than Bruno(x, now) is close to Lee(y, now). But then, strictly speaking Pete(X1,T1) is not the same person as Pete(X2,T1). Well, that's up for discussion! That's what we are trying to decide. I say that it gets pretty silly to formulate our ideas so that we turn out not to be the same person from second to second. Now, yes, in order to evade the notion that one is the same person as one's duplicate across the room, people will try anything, even denying that they have any identity whatsoever. They are, apparently, more comfortable with the notion that they are not the same person from second to second that the shocking idea that they and their duplicates are the same person. In any case I am not sure that those distinctions have any bearing on the existence of first person indeterminacy and the problem to quantify that indeterminacy. (Yes, maybe it is detached from the question you are trying to answer.) Imagine you are duplicated iteratively. At the start you are in room R. You are scanned and destroyed, painlessly, given some of the discussions we have, :-) this is rather pleasant to entertain and we tell you that you will be reconstituted in room 0 and in room 1. Then Lee0 and Lee1 are invited in room R again and the experience is repeated. Rooms 0 and 1 are identical and quite separate. The only difference is that in room 0 there is a big 0 drawn on the wall and in room 1 there is a big 1 drawn on the wall. So as this is repeated, there are 2, then 4, then 8, etc., copies, and each of them remembers a different sequence of 0's and 1's. Yes. You are asked to bet on your immediate and less immediate future feeling. Precisely: we ask you to choose among the following bets: Immediate: A. I will see 0 on the wall. B. I will see 1 on the wall. C. I will see 0 on the wall and I will see 1 on the wall. D. I will see 0 on the wall or I will see 1 on the wall. Less immediate: A'. I will always see 0 on the wall. B'. I will always see 1 on the wall C'. I will see as many 0 and 1 on the wall D'. I will see an incompressible sequence of 0 and 1 on the wall And there are three versions of the experiences. In a first version you are always reconstituted in the two rooms. Okay, let's handle just that for now. Good idea. We suppose obviously that you want maximize your benefit(s). Well, since you are asking *me*, then naturally I'll want a global maximum for me, and so a maximal sum for each instance. Good idea. And quite coherent with your idea that we are our duplicates. Each Lee-i is offered 5$ each time his bet is confirmed, but loses 5$ if he makes a wrong bet. And yes, it would be possible to emphasize to each instance that he is to attempt to maximize his own instance's earnings. Quite correct. What will be your strategy in each version? Will your strategy differ? Now if the Lees know all these facts, then they'll anticipate being in both rooms upon each iteration. Therefore, they'll anticipate losing $5 in one room and gaining $5 in the other. They'll also realize that all bit sequences are being carried out. Therefore, it doesn't make any difference whatsoever. The expectation of each sequence is exactly the same number of dollars: zero. I don't get the significance of this. I don't understand your answer, and actually you did not answer. It looks like you are forgetting I give you the choice between A, B, C, D. I guess you did choose C, without saying. In that case you are correct the expectation will be zero. Are you sure there is not a better strategy among A, B, C, D? And
RE: Thought Experiment #269-G (Duplicates)
Bruno writes Each Lee-i is offered 5$ each time his bet is confirmed, but loses 5$ if he makes a wrong bet. And yes, it would be possible to emphasize to each instance that he is to attempt to maximize his own instance's earnings. Quite correct. What will be your strategy in each version? Will your strategy differ? Now if the Lees know all these facts, then they'll anticipate being in both rooms upon each iteration. Therefore, they'll anticipate losing $5 in one room and gaining $5 in the other. They'll also realize that all bit sequences are being carried out. Therefore, it doesn't make any difference whatsoever. The expectation of each sequence is exactly the same number of dollars: zero. I don't get the significance of this. I don't understand your answer, and actually you did not answer. It looks like you are forgetting I give you the choice between A, B, C, D. Sorry. You are asked to bet on your immediate and less immediate future feeling. Precisely: we ask you to choose among the following bets: Immediate: A. I will see 0 on the wall. B. I will see 1 on the wall. C. I will see 0 on the wall and I will see 1 on the wall. D. I will see 0 on the wall or I will see 1 on the wall. I choose C: insofar as I consider myself as a program, then the program will see 0 on the wall and the program will also see 1 on the wall. The program will experienced both. I will experience both. I guess you did choose C, without saying. Right. In that case you are correct the expectation will be zero. Are you sure there is not a better strategy among A, B, C, D? Why do you think that there is a better strategy? C. will comport with all the facts. And afterwards, when a poll is conducted among all those who can prove that they are Lee Corbin it will be found that half of them saw a 1 and half saw a zero. It is preposterous to finger *any* of them and accuse them of not being me. They will each believe that they are me (i.e., the me here in the past). That is, for each Lee', they will assert Lee' = Lee. So also will Lee'' assert that Lee'' = Lee. So IT'S FREAKING OBVIOUS THAT Lee'' = Lee'. Yet substitute someone else's name for mine in those equations, and they'll demur. Lee
Thought Experiment #269-G (Duplicates)
Bruno wrote about whether or not we are all the same person. Sent: Tuesday, July 05, 2005 1:59 AM Subject: Re: What does ought mean? (was RE: Duplicates Are Selves) I have changed the subject line once again, because this is no longer about what ought ought to mean. Le 04-juil.-05, à 22:18, Lee Corbin a écrit : Yes, but I contend that while there are two organisms present, there is only one person. It's much as though some space aliens kidnapped you and tried to say that Pete at spacetime coordinates (X1,T1) could not possibly be the same person as Pete at coordinates (X1,T2) because the times weren't the same. You'd have to get them to wrap their heads around the idea that one person could be at two different times in the same place. They might find this bizarre. I'm trying to tell you a possibility that you think equally bizarre: namely that Pete(X1,T1) is the same person as Pete(X2,T1), namely that the same person may be at two different locations at the same time. That's all. I like that idea, but if they are the *same* person then we are all the same person. Or, perhaps you were just meaning that they are very close/similar; That is so: but moreover, being very very close/similar is what should be meant by the same person. in which case you can say Pete(X2,T1) is much closer to Pete(X1,T1) than Bruno(x, now) is close to Lee(y, now). But then, strictly speaking Pete(X1,T1) is not the same person as Pete(X2,T1). Well, that's up for discussion! That's what we are trying to decide. I say that it gets pretty silly to formulate our ideas so that we turn out not to be the same person from second to second. Now, yes, in order to evade the notion that one is the same person as one's duplicate across the room, people will try anything, even denying that they have any identity whatsoever. They are, apparently, more comfortable with the notion that they are not the same person from second to second that the shocking idea that they and their duplicates are the same person. In any case I am not sure that those distinctions have any bearing on the existence of first person indeterminacy and the problem to quantify that indeterminacy. (Yes, maybe it is detached from the question you are trying to answer.) Imagine you are duplicated iteratively. At the start you are in room R. You are scanned and destroyed, painlessly, given some of the discussions we have, :-) this is rather pleasant to entertain and we tell you that you will be reconstituted in room 0 and in room 1. Then Lee0 and Lee1 are invited in room R again and the experience is repeated. Rooms 0 and 1 are identical and quite separate. The only difference is that in room 0 there is a big 0 drawn on the wall and in room 1 there is a big 1 drawn on the wall. So as this is repeated, there are 2, then 4, then 8, etc., copies, and each of them remembers a different sequence of 0's and 1's. You are asked to bet on your immediate and less immediate future feeling. Precisely: we ask you to choose among the following bets: Immediate: A. I will see 0 on the wall. B. I will see 1 on the wall. C. I will see 0 on the wall and I will see 1 on the wall. D. I will see 0 on the wall or I will see 1 on the wall. Less immediate: A'. I will always see 0 on the wall. B'. I will always see 1 on the wall C'. I will see as many 0 and 1 on the wall D'. I will see an incompressible sequence of 0 and 1 on the wall And there are three versions of the experiences. In a first version you are always reconstituted in the two rooms. Okay, let's handle just that for now. We suppose obviously that you want maximize your benefit(s). Well, since you are asking *me*, then naturally I'll want a global maximum for me, and so a maximal sum for each instance. Each Lee-i is offered 5$ each time his bet is confirmed, but loses 5$ if he makes a wrong bet. And yes, it would be possible to emphasize to each instance that he is to attempt to maximize his own instance's earnings. What will be your strategy in each version? Will your strategy differ? Now if the Lees know all these facts, then they'll anticipate being in both rooms upon each iteration. Therefore, they'll anticipate losing $5 in one room and gaining $5 in the other. They'll also realize that all bit sequences are being carried out. Therefore, it doesn't make any difference whatsoever. The expectation of each sequence is exactly the same number of dollars: zero. I don't get the significance of this. Lee Note that I have purposefully avoided the use of first person in the question, and so C can be considered as a little ambiguous. My point will be to make you accept there is indeed an ambiguity in C. In the second version we tell you in advance that once on 2 iterations, you are reconstituted in one room only, and this one is chosen by random with a coin. In the third version we don't tell you if we choose the first version or the second
