tag:blogger.com,1999:blog-3891434218564545511.post8046769836173587827..comments2024-03-27T20:37:09.185-05:00Comments on Alexander Pruss's Blog: A variant on my argument against probability in multiverse scenariosAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger41125tag:blogger.com,1999:blog-3891434218564545511.post-49942812349870678442013-03-09T15:20:51.785-06:002013-03-09T15:20:51.785-06:00The crux of the issue might be this: since the die...The crux of the issue might be this: since the die roll has already happened, "Jones rolled a six with probability x" is shorthand for "if Jones were to do this arbitrarily many times, he would roll a 6 in x of them" (this assumes you use the 'frequentist' interpretation of probability, but I think this must coincide with the 'rational' one in cases like the ones we've considered where you know enough to use it--it is, for example, the one you won't lose money on in a fair bet). In scenarios where we can use this interpretation with x unique, as in the 4 statements above, if we name the random draw 'Jones,' this probability is well defined. If we try to apply both (1) and (4) to Jones, though, as our universal friend would try to do, the conditional is vacuously true for any probability (even inadmissable ones such as -1) since the premise (x=1/6 and x=1/2) is contradictory, so the quantity 'probability Jones rolled a six' is no longer well-defined. Theoretically, of course, you can still have any subjective probability you want, but we can no longer find a uniquely 'rational' one (although if this only happens if you have infinite friends or some similar condition, this would only be a problem for rational agents with both the ability to have infinite acquaintances and the ability to not observe the die rolls--some type of angel, perhaps).ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-8307464726358850162013-03-09T14:44:32.422-06:002013-03-09T14:44:32.422-06:00We are only able to reason as above in all our cur...We are only able to reason as above in all our current cases, though, because the cases can all be rephrased to focus on only finite sets, and ignore the infinite nature of the problem. If we consider a case where, by some magic, you are friends with the entire set, you can still rationally believe:<br /><br />1) If I pick a friend at random, they rolled a 6 with probability 1/6.<br /><br />2) If I pick a friend at random, their roommate rolled 6 with probability 5/6<br /><br />3) If I pick a room at random, it will contain one of each type<br /><br />4) If I pick a room and then one of its occupants, that person rolled 6 with probability 1/2<br /><br />Now, however, if you inquire about their particular friend Jones1, there is no clear way out of the paradox, because we cannot conclude that (1) and (2) apply but (4) doesn't as we could when you have 1 friend.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-30080124401571570312013-03-09T14:37:09.082-06:002013-03-09T14:37:09.082-06:00William:
I agree. In fact, if we let A='Jones...William:<br /><br />I agree. In fact, if we let A='Jones1 rolled a 6' and B='Jones1090 rolled a six' it is almost identical to our two-person cases, except that we are considering P(A|AUB and not 'A intersect B') rather than P(A|B,AUB). Since 'A intersect B' is a subset of AUB, P(AUB and not 'A intersect B')=P(AUB)-P('A intersect B')=1-(5/6)^2-(1/6)^2=10/36. Thus, we have P(A|AUB and not 'A intersect B')=P(A and AUB and not 'A intersect B')/P(AUB and not 'A intersect B')=P(A and not B)/(10/36)=((1/6)*(5/6))/(10/36)=1/2, as we have been claiming.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-46859033581310379662013-03-09T02:20:25.176-06:002013-03-09T02:20:25.176-06:00The room is like Berkson's paradox except for ...The room is like Berkson's paradox except for the infinite sample pool and the fact that rooms with both rolling a 6 are excluded. See <a href="http://books.google.com/books?id=DoG8QjF5q58C&pg=PA151&dq=berkson's+paradox+statement&hl=en&sa=X&ei=5O86UdmYE9D_yQGW2YCQBw&ved=0CDAQ6AEwAA#v=onepage&q=berkson's%20paradox%20statement&f=false" rel="nofollow"> this book.</a>Williamhttps://www.blogger.com/profile/12533263841520213358noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-69641888417477290512013-03-08T23:11:55.030-06:002013-03-08T23:11:55.030-06:00Not unless the order gives you additional informat...Not unless the order gives you additional information. If we step into a random room, and happen to notice one of its members first, there is no reason for us to favor that person. If we are friends with John1, and later discover his roommate, though, the fact that he is paired with John gives us evidence in favor of his having rolled a 6--the crucial factor is not the order but the relative likelihood of this person having been selected dependiing on type. If you were John9010's best friend the opposite would be true, but since you cannot both be friends with John1 and first encounter John 9010 as a random stranger paired with him you cannot do the reverse in the same scenario. If you were friends with both, they will be paired with probability zero, but we could argue that 1/2 is reasonable because the 'pairing revelation' gives you new information about both.<br /><br />There are other cases where recieving a random draw from an ex ante symmetric group of objects gives you assymetric information about those objects ex post: If we consider a variant of the Monty Hall problem where you are randomly assigned a door and the prize is behind, say, the first one, you will rationally assign 1/3 to each ex ante, but you will assign 1/3 to door 1 ex post if assigned it and 2/3 if you don't. The probabilities of each door having been assigned the prize are identical, but in any given scenario, you end up with different information and different rational beliefs.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-7712105190267828102013-03-08T20:14:35.446-06:002013-03-08T20:14:35.446-06:00But surely the probabilities shouldn't depend ...But surely the probabilities shouldn't depend on whom we focus on first.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-23635773434665594262013-03-08T20:12:57.166-06:002013-03-08T20:12:57.166-06:00Alexander Pruss:
This case has a crucial differen...Alexander Pruss:<br /><br />This case has a crucial difference from your second case--namely that we start by fixing our focus on Jones1 and later introduce Jones1090 as his match. This falls into the case of 'drawing from the set of individuals,' where we know each individual is in a room with someone who rolled a 6 with probability 5/6. In your original second case, we pick a room, and find it contains Jones 1090 and Jones1, who we haven't met before. This falls into the '[drawing from the set of pairs' case, with 1/2 probability for each. Our information on the people is different in the two cases--in the first, we start with Jones1 (who we are assuming is a random draw from the population). When we meet Jones1090, we know something further about him--he was picked to be the opposite of Jones1, and Jones1 rolled a 6 with probability 1/6. Thus, Jones1090 is not a random draw from the population, because his type is not independent of Jones1's, so we cannot use the same reasoning to evaluate his type we used for Jones1. In the original second case, we just know that we have a room with one of each type, and our information about the two people is symmetric. The sequential case we've been worried about is distinct from both of these. In this case, you know Jones1, you know you've been assigned a room, and you later find your room contains Jones1 (which is not a foregone conclusion as it was when you follow Jones1 to his room). In this case, you receive further relevant information about Jones1, namely that he was selected by a sampling process that favors people who rolled a six, and as a result, your subjective probability that he rolled a six increases.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-72936184091377180322013-03-08T19:18:27.951-06:002013-03-08T19:18:27.951-06:00William:
Yes, this may be a case where there is b...William:<br /><br />Yes, this may be a case where there is biased sampling that it's impossible to correct for.<br /><br />But that's not a complete solution. For we still have to say when, if ever, we change probabilities, and why then.<br /><br />It is a neat and simple solution to say that as soon as you learn about the infinity, you then abandon all probabilistic hope.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-35347033053849920112013-03-08T17:42:41.612-06:002013-03-08T17:42:41.612-06:00Not to drag this out too long, but this has curren...Not to drag this out too long, but this has current world implications. Say a study is done that shows that among a group of Republicans paired with age and sex-matched Democrats, the brains of the Republicans are different. I accept these results, Later, an angel tells me that the Republicans were selected from a group of private college students, and the age-matched controls were chosen from a group of homeless persons in a nearby city.<br /><br />Am I entitled to change my mind about the significance of the study's findings? (something like this actually happened btw). Williamhttps://www.blogger.com/profile/12533263841520213358noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-5488880742566618682013-03-08T15:36:33.116-06:002013-03-08T15:36:33.116-06:00I don't think the angel can show up and truthf...I don't think the angel can show up and truthfully tell you the story and really demonstrate the pairs unless the sampling has already been biased. So whenever you believe the angel's information about the pairing is when your prior probability beliefs should change.<br /><br />This suggests that there would be an infinite number of possible worlds where the angel never is able to get started telling the story!<br /><br />If the probability for you cannot change, the angel should not be able to show up at all.<br />Williamhttps://www.blogger.com/profile/12533263841520213358noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-42576486568873537912013-03-08T14:58:40.060-06:002013-03-08T14:58:40.060-06:00So, in the first case, you stick to 1/6, and in th...So, in the first case, you stick to 1/6, and in the second, you switch to 1/2?<br /><br />I am not sure this is a defensible combination. Now modify the first case. You're not actually rolling, but the person rolling, call him Jones1, is your best friend. You start in the same room as your best friend, watching him roll. And then the angel announces he's made a pairing and will show you all the pairs. Moreover, the angel offers to first show you the pair containing Jones1.<br /><br />So, you now see Jones1 paired with somebody, say Jones1090. By what you said about the first case, it seems that Jones1 should stick to assigning 1/6 to Jones1 rolling six. But why would you disagree with him? Indeed, what does it matter who actually rolled the die, whether it was you or Jones1? What matters is just the die. So you should assign whatever Jones1 should assign, namely assign 1/6 to Jones1 rolling six, according to what you said about the first case. And of course 5/6 to Jones1090.<br /><br />And then you see all the other pairs. And presumably you assign 1/2 to everybody in them, by your reasoning in the second case.<br /><br />But, wait, this isn't any different from my second case, except for three features (a) you started in the same room as Jones1, (b) he was your best friend and (c) the angel showed you the Jones1 pair first. But none of these features seem significant vis-a-vis the probability of Jones1 rolling six.<br /><br />So according to this argument, you should answer both cases in the same way.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-43199031595197194312013-03-08T14:37:46.482-06:002013-03-08T14:37:46.482-06:00Given this, let's go through cases: In the ini...Given this, let's go through cases: In the initial story, an individuals type is (from their perspective) a random draw from the original iid distribution, so they rolled a six with probability 1/6. Once his roll is determined, though, any mapping drawn next will put him in a room with the opposite type. If he rolls a six 1/6 of the time, this implies his roommate won't 1/6 of the time and vice versa. In our second case, if our observer is assigned a room, he is essentially picking one of the pairs in an iid fashion. The odds of a given person in his pair rolling a six is 1/2. Furthermore, if he chooses n pairs and then picks a person from their union, the probability is still 1/2, and remains so as n becomes arbitrarily large. Intuitively, he is using a different sampling procedure, one that is biased in favor of giving him people who rolled sixes. In the case where he is shown Jennifer in the population, and then draws a room in which she happens to turn up, something similar is going on. When considering Jennifer initially, his information is just that she is a member of the set of die rollers (hence the assignment of 1/6). Once she shows up in his room, this increases to 'member of the population who was drawn in a sampling process biased in favor of sixes.' The fact that she was drawn here is new evidence in favor of her rolling a 6, hence the change. Conversely, if we fix a room first and then draw one of its members from the original population, that is evidence in favor of that member not rolling a six, although this may not make sense given our story. Our ability to do this in both cases hinges crucially on being able to treat the identification of the individual or room we are considering as fixed, so that we only get information specific to the individual from the second process--in a situation where we draw a room and a person simulteneously and the person is in the room, it is not clear what to deduce. I suspect that all of our current worries about the reflection principle in our story fit into the first case, though, and in this case believing "Jennifer rolled a 6 with probability 1/2 if she is in my room and 1/2 otherwise' is consistent with the reflection principle (from the point you can foresee whether she's in your room or not on, you have a fixed belief).ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-80417192117084512062013-03-08T14:37:30.943-06:002013-03-08T14:37:30.943-06:00Here's a more precise (probably better) way to...Here's a more precise (probably better) way to state my argument. We can formalize this scenario in a way that preserves its exact structure as follows: Let our set of die rollers be a countably infinite number of draws from a uniform distribution on the interval [0,1]. We will say they rolled a six if their draw was in the interval (5/6,1] (which has probability 1/6) and we can imagine they all have names (Jennifer, etc.). In this case, for any individual, they rolled a six with probability 1/6, and the outcome of their roll is statistically independent of that of any other individual, so the outcomes for each individual are independent and identically distributed (iid). From this we can infer that any subset of our individuals that is also iid will be such that each individual rolls a six with probability 1/6 and the expected proportion of sixes rolled in the subset is 1/6 (assuming the subset is finite). We can construct the process of pairing into rooms as a mapping that transforms the set of all individuals into a set of pairs of individuals such that each pair has 1 of each type and each individual is in exactly one pair. If we are given one of these pairs, we cannot infer that each individual rolled a 6 with probability 1/6 because independence is violated. The probability I rolled a 6 given my roommate didn't is one, independence would require P(I rolled x|roommate rolled y)=P(I rolled x) for all x and y. Instead, they have a joint distribution with P(I rolled 6 and you didn't)=P(I didn't and you did)=1/2, with all other outcomes zero, implying that the probability each person in the room rolled a 6 is 1/2. Note that we can without loss of generality assume that both the assignment of sixes and nonsixes and the mapping of individuals to rooms are fixed for the duration of each scenario, but unknown to participants (i.e. we randomly draw an assignment of sixes and then a compatible mapping before anything else happens). ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-3575974144281978562013-03-08T14:36:13.638-06:002013-03-08T14:36:13.638-06:00You're right--I withdraw the numbers analogy. ...You're right--I withdraw the numbers analogy. The problem, though, is that a number's 'name' implies its status as even or prime, while a persons' doesn't, so the same shouldn't hold in the other case. We can fix it by making our set {a,b,c,d,e} and saying 2 of these are blue and 3 green (let's say a and b are blue and c, d and e are green). Say the procedure is to draw a letter, pair it with one of the opposite color (all equally likely), and then show you the pair. Let {x,y} be the pair, and adopt the convention that x is the number drawn and y is matched to it (implying that you cannot distinguish between {x,y} and {y,x}). Then the set of all outcomes is {a,b}X{c,d,e}U{c,d,e,}X{a,b} (where 'X' is the Cartesian product and 'U' the set union). If you knew how the letters are colored, as with the numbers, then seeing a and c means the outcome is either {a,c} or {c,a}, and since both are equally likely initially, you assign each probability 1/2 by Bayes' rule. Since you don't know the coloring, seeing a and c tells you 'either a was drawn, a is blue and c is green; a was drawn, a is green and c is blue; c was drawn, a is blue and c is green or c was drawn, a is green and c is blue,' in other words, you know only that a and c are of different colors and one or the other was chosen. Both 'a is blue and c is green' and 'c is blue and a is green' happen in three possible colorings, so the probability that a is blue given a and c are different is 1/2. The probability a is drawn given a is blue, however, is 2/5, as is P(c drawn|c blue), while P(a drawn|a green)=P(c drawn|c green)=3/5. Thus, P(blue letter drawn| a or c was drawn and a and c are different colors)=P(a drawn|a blue)P(a blue)+P(c drawn|c blue)P(c blue)=1/2*2/5+1/2*2/5=2/5. Thus, seeing a and c does not change your assessment of this probability as long as you do not know the coloring.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-86966464558122928102013-03-08T11:01:45.232-06:002013-03-08T11:01:45.232-06:00"If you are then told the draw was in {11,12}..."If you are then told the draw was in {11,12}, this changes nothing"<br /><br />I am confused. It changes a lot of things. It sets the probability its being 13 to zero, for instance. Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-23904813230469192922013-03-08T10:31:25.988-06:002013-03-08T10:31:25.988-06:00Regarding the betting example: If you are told a n...Regarding the betting example: If you are told a number has been drawn (with equal probabilities) from the set {11,12,13,14,16,17,18,19} (which has 5 evens and 4 primes) you should bet based on the draw being even with probability 4/5. If you are then told the draw was in {11,12}, this changes nothing, since you could have been told it was in some such pairing regardless of the outcome. If instead, you are told that numbers will be drawn repeatedly until either 11 or 12 is drawn (picking rooms unitl you find Jennifer's) with the first 11 or 12 retained you should bet on 1/2, as the probability of an even is just the probability 12 is drawn first and vice versa. Hence, the odds are genuinely different in the case where we know which pair your draw will be in as well as that it will be in such a pair.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-49759468158740612362013-03-08T10:30:31.141-06:002013-03-08T10:30:31.141-06:00The fact that we can no longer rely on the fact us...The fact that we can no longer rely on the fact used in the above finite cases that the set of people who did not roll sixes is five times as large as the set who did (because the two sets now have the same cardinality) prevent us from using the reasoning we used for Jennifer for all individuals simultaneously--we avoided paradox above because we could not fix more than one room or person within the same setup. However, among any finite subset of die rollers, we can say that about 1/6 of them will be the person in their room who rolled a six, whereas we can say for any room that exactly 1/2 of its occupants did. This is counterintuitive, but no more paradoxical than similar results involving countable sets. By analogy, the set of even numbers and the set of primes have the same cardinality (countably infinite). If we are given an arbitrary finite set of consecutive integers, the number of evens will almost surely exceed the number of primes. If, however, we are given a set of two numbers, one drawn from the primes and one from the evens, there will always be the same number of each. If we are handed some group of numbers drawn in some arbitrary fashion from the union of the two, we cannot assign a well-defined probability (we could do this in the other two cases only because the subsets we consider have a special form). There is no 'relative frequency of primes' in the union of evens and primes in the strong sense that would allow us to predict frequencies in random draws from a union of finite sets. This was the basis of my initial argument that anything we can say about probabilities must be based on the specific subset we consider, not the entire setup.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-56390024097194661392013-03-08T10:30:00.569-06:002013-03-08T10:30:00.569-06:00In these stories, we are inclined to assert two se...In these stories, we are inclined to assert two seemingly conflicting things:<br /><br />1) Each person has probability 1/6 of rolling a six, and thus has probability 1/6 of being the person in their room who rolled a six<br /><br />2) There is one of each type in every room, and thus the probability of someone in a given room having rolled a six is 1/2<br /><br />In our case with countably many rollers, these are not contradictory, but in a finite set they would be. However, we can consider partial finite-set analogues that retain most features of each story:<br /><br />Imagine considering a fixed individual, who you know will be placed in a room with someone of the opposite roll, ignoring all other rollers (since we cannot do this for all of them in the finite case). Then we can reasonably assert this person is the one in the room who rolled a six with probability 1/6, and could do the same if we changed the setup to focus on any other given individual.<br /><br />Alternatively, if we fix a room and learn that one person of each type will be put in it, but know nothing about other rooms, we can reasonably assign 1/2 probabilities.<br /><br />From the point of view of our observer in the more recent version, provided he will be assigned a single, specific room, we can partition the set of die rollers into four groups: the person in his room who rolled 6, the person in his room who didn't, the people not in his room who rolled 6, and the people not in his room who didn't. If the finite case, he will initially believe Jennifer rolled a 6 with probability 1/6. If also told Jennifer is in his room, this will rationally change to 1/2 by Bayes' rule, as he will learn that she cannot be in the latter 2 sets, and eliminating these sets changes the relative frequency of sixes and non-sixes in the remainder to 1/2 (since the set of non-sixes not in his room is much larger).<br /><br />On the other hand, if we are told as in the first case that Jennifer will be placed in some such room, then (again assuming finitely many rollers), anyone aware of the setup should believe Jennifer rolled a 6 with probability 1/6--both her roommate and our observer should also believe this in this case--since her roommate will be chosen from the set of others who rolled sixes with probability 5/6 and from the set who didn't with probability 1/6, her roommate is more likely to have rolled a six given that she was chosen. Her view is not symmetric with Jennifer's since she was chosen to match Jennifer, but not vice versa.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-4947150140549803382013-03-08T07:46:15.665-06:002013-03-08T07:46:15.665-06:00ah25:
1. We disagree when we assign significantly...ah25:<br /><br />1. We disagree when we assign significantly different credences to the same proposition. If you assign credence 1/6 to string theory and I assign credence 1/2 to string theory, we disagree. <br /><br />2. You might say that in the second version there is no well-defined Bayesian probability because the agent observes something she doesn't have a probability for. She observes the order in which the pairs are shown to her, and she presumably has no probability for the order. But that doesn't eliminate something analogous to the Reflection Principle. If you foresee that upon being given more evidence, you will cease to assign a probability to some piece of evidence, shouldn't you cease right away? <br /><br />3. In the second version, the observer foresees that <em>every</em> person will eventually be in a room. So, if she will set her credence to 1/2 for a particular person, then by Reflection, she should already have set her credence to 1/2. Which is absurd, because the angel can then change your credences simply by making a list, telling you that he's made it, and announcing his intention of showing it to you.<br /><br />4. In the second version, suppose you switch to 1/2 as you suggested. Imagine a sequence of independent runs of the second version, but each time the angel tells you that he will stop showing you pairs once he gets to a pair containing Jennifer (to ensure the run doesn't take an infinite number of steps; one could also suppose a super-task, but that raises its own problems). And then you're offered a bet: you get $1.01 if Jennifer rolled six, and you pay $1.00 if she didn't. If you changed to 1/2, you rationally accept. But if you accept this, then on average on five runs of the experiment out of six, you will be paying $1.00 and one run out of six you will be getting $1.01. In other words, you'll on average be losing. And, moreover, since you're still just as smart as you were before, you will be able to see this point, and yet you will still accept the bet rationally.<br /><br />Now, it may be true that betting odds can come apart from credences in cases like Sleeping Beauty. But in cases like that, the number of times a bet is offered to you is dependent on the events you're betting on. In such cases, the two might come apart. But this isn't a case like that.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-68612048238740471402013-03-08T00:46:41.876-06:002013-03-08T00:46:41.876-06:00On second thought, I'm not sure the reflection...On second thought, I'm not sure the reflection principle is violated in this case either (this reflection principle is new to me, although it looks a lot like Bayes' rule). In any case where our observer can foresee whether a person will be in the room or not, they will believe the 'right' thing from the start. Am I right in thinking the reflection principle only applies to such cases? (My friend draws the ace of clubs from a deck and will show it to me shortly. I will come to believe it is a spade with probability 1 then. It would be strange for my believing this with probability 1/4 now to be irrational).ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-24830891089368263132013-03-08T00:34:25.688-06:002013-03-08T00:34:25.688-06:00It seems based on the last exchange I misunderstoo...It seems based on the last exchange I misunderstood slightly. In any case where we only care about the probabilities in the room there is no paradox, but there is in the before/after case in the variant. Bayesian reasoning is only well-defined in cases where the agent starts with a well-defined subjective probability and never observes zero probability events. One way out of this is to claim that things like Bayes' Rule and the reflection principle are only necessary conditions for rational belief in such 'well-behaved' cases.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-85971356141450601372013-03-08T00:30:15.156-06:002013-03-08T00:30:15.156-06:00Alexander Pruss:
In the first case, I am not conv...Alexander Pruss:<br /><br />In the first case, I am not convinced there is a disagreement, because we are dealing with subjective probabilities: each person is thinking 'based on my information, my subjective probability that I rolled a 6 is 1/6,' but since these are claims about two different objects (person a's subjective probability distribution and person b's subjective probability distribution), these beliefs do not conflict.<br /><br />In the second case: Regarding the reflection principle, there are two types of people in the population--ones who will end up in your room and ones who won't. You should rationally assign probability 1/6 to 'someone not in my room rolled a 6' and probability 1/2 to 'someone in my room rolled a 6.' Since the probability a given person will end up in your room is, loosely speaking, zero (2/infinity=0), you will rationally assign probability 1/6 to any given person in the population, if you don't know if they will end up in your room. If you later find out that person is in your room, we do have a problem, but this is an instance of a very general problem involving Bayesian updating in response to 0-probability events, and doesn't depend on any of the specifics. Our observer in this case will never lose money betting on 1/2 if the betting is restricted to the 2 people in his room. If betting on a random die roller, he will not rationally bet 1/2, since that person is not in his room with probability 1. If the angel truthfully tells him x out of the n people in his room rolled sixes, he should rationally assign x/n to people in his room, and 1/6 to people whose status is unknown, and as long as n is finite, we still only have a problem if he is asked about someone and later finds out they are in his room.<br /><br />The point about subjective probabilities in the first case also applies to the disagreement between cases--the observer's subjective probability is a separate object from either die roller's.ah25https://www.blogger.com/profile/12742408346616343363noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-85189505567949492502013-03-07T23:41:53.858-06:002013-03-07T23:41:53.858-06:00ah25:
This is plausible and helpful, except for th...ah25:<br />This is plausible and helpful, except for the "No paradox" part.<br />In the first case, you end up with disagreement between two people who agree on all the evidence and on the methods for evaluating it. That's paradoxical.<br /> In the second case, you get a violation of the Reflection Principle, plus if you bet in situations like this according to 1/2, you will on average lose. Moreover, the angel can get you to sign any rational number probability he wishes, strictly between zero and one, just by telling you truths. These things are paradoxical.<br />Plus it's paradoxical to think the two cases should disagree. For in the first case you're assigning a different probability from the one that is being assigned in the second by an agent who has strictly more information than you do. <br />These are not contradictions, of course, but they are paradoxical.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-47859384797934752602013-03-07T23:24:58.721-06:002013-03-07T23:24:58.721-06:00When I arrive in the room, I know something new ab...When I arrive in the room, I know something new about my roll. I know that certain possibilities about the combination of my roll with the other guy's roll are excluded, or I would not be in the room. <br /><br />I know that of 36 possibilities, 26 without at least one 6 are excluded and one with 2 sixes is also excluded.<br /><br />What remains that the angels could have picked are:<br /><br />1-6, 6-1, 2-6, 6-2, 3-6, 6-3, 4-6, 6-4, 5-6, 6-5.<br /><br />May this changes my information to support a 1/2 odds?<br />Williamhttps://www.blogger.com/profile/12533263841520213358noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-26580511183788874042013-03-07T23:07:57.009-06:002013-03-07T23:07:57.009-06:00Can the angels put a given pair in the room withou...Can the angels put a given pair in the room without exercising selection bias?Williamhttps://www.blogger.com/profile/12533263841520213358noreply@blogger.com