probability in ensemble theories
[I am starting a new thread on this question because I feel I may have confused matters with increasingly elaborate thought experiments in my posts to Eric Cavalcanti's thread on observation selection effects.] The occurence of events which at first glance seem to be very unlikely should not necessarily surprise us. For example, the probability that a particular person wins first prize in a lottery may be one in a million, but the probability that SOMEONE wins first prize is usually greater than 50% for most lotteries. We can avoid confusion by being clear on what the desired outcome is before the event takes place: Pr(someone wins)=1/2, Pr(John Smith wins)=1/1,000,000. Consider the situation in a generic ensemble theory in which every possible outcome occurs. Clearly, Pr(some version of John Smith wins)=1. The outcome (some version of John Smith wins) is analogous to the outcome (someone wins) in the above example. Now, if some version of John Smith wins, it may be argued that this is surprising, because there is only a 1/1,000,000 probability that that PARTICULAR version wins. But for this to be surprising, we would have to specify the desired outcome before the event takes place: just as in the single world example above we have to specify that a particular individual will be the winner, in the ensemble theory we would have to further specify that a particular version of that individual will be the winner. The difficulty is that it is impossible to specify a particular version of a particular individual in an ensemble theory. The different worlds do not come labelled in any way, and it is not possible in general to pick out any attribute which will single out a particular world or individual, because the only necessary difference is the outcome of interest. In other words, the only way to specify which version of John Smith will win is to specify the outcome (that version of John Smith who will win, will win), which is obviously not going to lead to any surprises. Stathis Papaioannou _ Smart Saving with ING Direct earn 5.25% p.a. variable rate: http://ad.au.doubleclick.net/clk;7249209;8842331;n?http://www.ingdirect.com.au/burst6offer.asp?id=8
Ambjørn et al.
Of possible general interest - J. Ambjørn J. Jurkiewicz and R. Loll (also a writeup in Nature news, at http://www.nature.com/news/2004/041004/full/041004-17.html) Emergence of a 4D World from Causal Quantum Gravity http://dx.doi.org/10.1103%2FPhysRevLett.93.131301
Re: S, B, and a puzzle by Boolos, Smullyan, McCarthy
Bruno Marchal wrote: You made a relevant decomposition of the problem, and you are on the right track. Actually I'm not sure the ja da McCarthy's amelioration adds anything deep to the problem. It will be enough to take into account that a double negation gives an affirmation. I've thought about McCarthy's version for a while and I can't figure it out, even if I make the simplifying assumption of two truth-tellers and one random-answerer. It seems like the 3 bits of information you get from their answers just shouldn't be enough to tell the meaning of Ja vs. Da *and* tell you the identity of all three gods. 3 bits of information should only allow you to choose from 8 possibilities, but there seem to 12 possibilities here: (Ja=yes, god #1=knight, god#2=knave) (Ja=yes, god#1=knight, god#2=knife) (Ja=yes, god#1=knave, god#2=knight) (Ja=yes, god#1=knave, god#2=knife) (Ja=yes, god#1=knife, god#2=knight) (Ja=yes, god#1=knife, god#2=knave) (Da=yes, god #1=knight, god#2=knave) (Da=yes, god#1=knight, god#2=knife) (Da=yes, god#1=knave, god#2=knight) (Da=yes, god#1=knave, god#2=knife) (Da=yes, god#1=knife, god#2=knight) (Da=yes, god#1=knife, god#2=knave) Then I thought there might be a clever solution where you can figure out the identity of the gods without ever knowing the meaning of Ja and Da. But if you don't figure out the meaning of Ja and Da, it seems to me you're really only getting 2 bits of information from their answers, since it doesn't actually matter what the first answer is, only whether the second and third answer are the same as or different from the first--you can't distinguish between the answers Ja-Da-Ja and Da-Ja-Da, for example. 2 bits should only be enough to choose from 4 possibilities, but there are 6 possibilities you need to choose from to find the identity of all three gods: (god #1=knight, god#2=knave) (god#1=knight, god#2=knife) (god#1=knave, god#2=knight) (god#1=knave, god#2=knife) (god#1=knife, god#2=knight) (god#1=knife, god#2=knave) So either way I'm stumped...can you give me the solution? If you want to let other people keep trying to figure it out you can just email it to me. Jesse
Re: Ambjørn et al.
Download the article free of charge here: http://arxiv.org/abs/hep-th/0404156 - Oorspronkelijk bericht - Van: Pete Carlton [EMAIL PROTECTED] Aan: [EMAIL PROTECTED] Verzonden: Tuesday, October 12, 2004 07:09 PM Onderwerp: Ambjørn et al. Of possible general interest - J. Ambjørn J. Jurkiewicz and R. Loll (also a writeup in Nature news, at http://www.nature.com/news/2004/041004/full/041004-17.html) Emergence of a 4D World from Causal Quantum Gravity http://dx.doi.org/10.1103%2FPhysRevLett.93.131301
Re: S, B, and a puzzle by Boolos, Smullyan, McCarthy
On Mon, 2004-10-11 at 22:51, Bruno Marchal wrote: As a Price, I give you the (known?) Smullyan McCarthy As a Price, or a Prize? :) puzzle. You are in front of three Gods: the God of Knights, the God of Knaves, and the God of Knives. The God of Knight always tells the truth. The God of Knaves always lies, and the God of Knives always answers by yes or no randomly. You must find which is which, through some questions. You can ask no more than three yes-no (answerable) questions. (Each question must be asked to one God, but you can ask more than one question to a God; only then there will be a God you can no more ask a question). And (added McCarthy) I let you know that all the Gods, although they understand English, will answer the yes-know question by either JA or DA, and you are not supposed to know which means yes and which means no. Wow, that was a hard one! I have been thinking about it all day, but I think I got to the solution. Let's see... I developed a whole method for solving this kind of problem. :) Labelling the Knights' God as T, the Knaves' God as F, and the Knives' God as R, we have 6 possible 'states' we want to distinguish, which are all the permutations of them. I'll make 3 questions, which can have two possible outcomes each, JA(J) or DA(D). This means that in the end my experiment has 8 possible outcomes. From this initial analysis it is clear that I cannot aim to ask questions which would give me information about the statements JA=YES or JA=NO, because for that I would need to distinguish between 12 states. But since I have 2 outcomes more than states, it is possible that I might in some cases get this last answer, but only as a bonus. So I make a table where I try to fit a set of 3 questions whose outcomes are going to distinguish these states. One possible table is the following: State 1st J DJJ JD DJ DD Final --- TRF J J J JJJ TFR D JJ DJJ FRT J D J JDJ FTR D DJ DDJ RTF J/D J DD D JJD/DDD RFT J/D D JD D JDD/DJD --- The blank spaces mean that we don't care what would be the answer in those cases, because they have been already ruled out. It's important to leave blank spaces in at least two lines which correspond to 'R's in the same column, because we can always make the question to the the God in that column so that we can arrange the random answers to fall in those places. The columns 'J', 'JJ' and so on indicate the outcomes that I wish the next question to have given outcome 'J', 'JJ' and so on for the previous questions. Some more thought shows that you can always come up with *some* question that fits whatever choice of outcomes you want, given the constraints above. So we have a lot of freedom to choose the outcomes in the above table, the only problem will be to think of the corresponding verbal questions. So the first problem is to come up with some question which would be answered as in the '1st' column. This question could be, using the idea of Jesse Mazer, but with an extra trick: If I asked you Is the 2nd God the 'R', would you say 'JA'? The interesting thing is that this question does not tell us if JA is YES or NO, but a minute of thought shows that the outcomes are the same whatever JA means. After some thought, the other questions can be: Column'J': 3rd god, if I asked you 'Are you the 'F'', would you say 'JA'?; Column'D': 2nd god, if I asked you 'Are you the 'F'', would you say 'JA'?; Columns 'JJ' and 'JD': 3rd god, if I asked you 'Is the 2nd god 'R'', would you say 'JA'?; Columns 'DJ' and 'DD': 2nd god, if I asked you 'Is the 3rd god 'R'', would you say 'JA'?. So I think these would solve the problem. It was a lot of fun! Let me see what you guys think. As an extra challenge, can you think of a way to find out if JA=YES in some of the outcomes? It would involve different questions, but I think it should be possible in principle. Eric.
Re: S, B, and a puzzle by Boolos, Smullyan, McCarthy
Nice work, Eric! Your solution looks right to me. I now realize my mistake, I was thinking that if the gods are in a particular order (say, TRF) and Ja has a particular meaning (say, Ja=yes) and you get a particular series of answers (say, JJJ) then if you reverse the meaning of Ja and ask the same questions, that means you'll also get the reverse answers (in this case DDD). But if the question is of the form If I asked X, would you say 'Ja'? this isn't actually the case. The key seems to be that if you ask a question of the form If I asked X, would you say 'Ja'? then if X is true, both the knight and the knave will answer Ja regardless of whether Ja means yes or no, and if X is false then both the knight and the knave will answer Da regardless of the meaning of Ja. So with that trick in mind, the solution to this problem will be exactly like the solution to the problem where the gods actually say yes and no...for example, I could take my previous solution to that problem: your first question should be to ask the first God If I asked you 'is the second God the God of Knives', would you say 'yes'? If the first God answers yes, you know the God of Knives is either the first or the second God, so you can ask the third God, If I asked you 'are you the God of Knights', would you say 'yes'? and after that you can ask the third God If I asked you 'is the first God the God of Knives', would you say 'yes'? and this will be enough to tell you the identity of all three Gods. On the other hand, if the answer to your first question was no, then you know the God of Knives is either the first or the third God, so you would ask the *second* God the same two subsequent questions as above. ...and simply replace every yes with Ja and every no with Da: your first question should be to ask the first God If I asked you 'is the second God the God of Knives', would you say 'Ja'? If the first God answers Ja, you know the God of Knives is either the first or the second God, so you can ask the third God, If I asked you 'are you the God of Knights', would you say 'Ja'? and after that you can ask the third God If I asked you 'is the first God the God of Knives', would you say 'Ja'? and this will be enough to tell you the identity of all three Gods. On the other hand, if the answer to your first question was Da, then you know the God of Knives is either the first or the third God, so you would ask the *second* God the same two subsequent questions as above. ...keeping in mind, again, that if you ask a god If I asked you X, would you say 'Ja'?, then if he says Ja that means X must be true if the god was a knight or a knave, and if he says Da X must be false if the god was a knight or a knave. Jesse
