## Wednesday, March 6, 2013

### A variant on my argument against probability in multiverse scenarios

This is a variant of the argument here. I am not sure which version I like more. This version is lacking one distracting feature of the original, but I kind of like the original more.

You're suddenly informed by an angel that a countably infinite number of people have just each rolled a fair and independent indeterministic die. In the case of each of these people, you should surely assign probability 1/6 that that person rolled six. The angel then adds that infinitely many people rolled six and infinitely many didn't. This doesn't surprise you—after all, that was just what you expected from the Law of Large Numbers.

The angel then adds that he divided up all the die rollers into pairs, one member of each pair having rolled six and the other not. This doesn't seem to be a big deal. After all, you knew that they could be thus divided up as soon as you heard that infinitely many rolled six and infinitely many didn't. The mere fact that they were so divided seems irrelevant.

Finally, the angel transports you to meet each pair of paired die rollers, pair by pair. In each pair, you now know for certain that one person got six and the other didn't. Let's say that at some point you meet Jennifer and Patricia. What probabilities do you assign to each having rolled six? You can't stick to your old assignment of 1/6 to each roller. For if you did that, you'd be violating the probability calculus, since you know for certain that exactly one of the two got a six, so you better have P(Jennifer got six)+P(Patricia got six)=1. But you can't go for any asymmetric assignment either, since your evidence regarding Jennifer and Patricia is exactly symmetric. That leaves you with two options: you must refuse to assign any probability to the probability of Jennifer getting six and to the probability of Patricia getting six, or you must assign probability 1/2 in both cases. And since this reasoning applies to everybody, you'll be assigning 1/2 to everybody or no-probability to everybody.

Assigning 1/2 to everybody is untenable, because you could then re-run the scenario with a different way of partitioning (say, into groups of three, two of whom rolled six, and one who didn't).

So it seems that you just need to refuse to assign probabilities. But from which point on? As soon as you learned that there were infinitely many people rolling independent fair dice, you already knew that some partitioning like the above was possible. Intuitively, you didn't learn anything important at any subsequent step. (You did genuinely learn something when you learned there were infinitely many sixes and infinitely many non-sixes, but since that information had probability 1 before you learned it, it shouldn't have affected your probabilities significantly.)

One way out is to deny the possibility of engaging in probabilistic reasoning in a scenario where there are infinitely many independent copies of an experiment. This way out insists that from the beginning, as soon as we knew about the infinitary aspects of the experiment, we should have refused to assign probabilities. (But I worry: What if the experiments are sequential? Shouldn't it be possible to engage in probabilistic reasoning then? And yet one can make a similar story work in sequential contexts.)

Maybe, though, one can switch from assignments of 1/6 to a no-probability at the last step. When I meet Patricia and Jennifer, I meet a highly biased sample of a pair of die rollers: after all, in an unbiased sample, I would have no guarantee that exactly one of the two rolled six. The bias takes a form that cannot be probabilistically handled by me: I have no probability distribution on ways of pairing individuals that assigns a probability to Patricia and Jennifer being paired together. And where there is bias that cannot be probabilistically handled, I must suspend judgment. This is not a very attractive way out. But maybe it's still the right one? (Notice, though, that if you are told ahead of time that you will be meeting people in pairs like this, then you will violate a plausible generalization of van Fraassen's reflection principle if you initially assign 1/6 and then switch to no-probability.)

Alexander R Pruss said...

Before anybody complains that I neglect interval-valued probabilities, let me say that I count those as no-probability options. You could, I suppose, assign [1/6,5/6] to the Jennifer and Patricia rolling six hypotheses. Those don't count as probabilities here.

But in any case, the argument still applies in those cases: there doesn't seem to be a good point where the probabilities would change from sharp 1/6 to unsharp [1/6,5/6].

Bones said...

"you now know for certain that one person got six and the other didn't"

The event already happened, so what does it mean to assign a probability?

Alexander R Pruss said...

These probabilities are degrees of rational belief or something like that.

I'm about roll a die. I assign 1/6 to getting six. I roll the die, but it rolls under the sofa. Do I stop assigning 1/6 to six?

Bones said...

Rolling two dice is a different event though and in this case you're given information about at least one of them. That seems like a totally different event than losing the dice under the couch. Also, under D&D rules that'd be a re-roll.

William said...

Does the fallacy of the biased sample apply here? Aren't the pairings a biased sample of an infinite number of die rolls, making the probability change due to the angel pairing causing a sample bias?

Alexander R Pruss said...

William:

I think there is something like that here, but this still doesn't answer what I should do when I see the pair. I can't keep on assigning 1/6 to Jennifer getting six and 1/6 to Patricia getting six, since I know that one of them got a six.

William said...

You can change your mind from 1/6 to 1/2 and justify the change due to sampling bias?

Huume said...

"Also, under D&D rules that'd be a re-roll. " -- I nearly spit my rice out in a burst of laughter

Mark said...
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Mark said...

Suppose a million people roll a die and I tell you that at least one person rolled a 6 and at least two people didn't. I then present you with a 6er/non-6er pair (X, Y) at random. Seems like P(X rolled a 6) should be 1/2, since P(X rolled a 6) + P(Y rolled a 6) = 1. Then I present you with a pair (A, B) randomly such that exactly one of (X, A, B) rolled a 6. Now it seems like P(X rolled a 6) should be 1/3.

How is this different from the infinite case?

Alexander R Pruss said...

Let's say there are N 6ers and M non-6ers. Then there will be NM possible pairs. And you choose one at random for my benefit.

If X is a 6er, then X appears in exactly M possible pairs. If X is not a 6er, then X appears in exactly N possible pairs. But M is approximately equal to 5N. So, if X is a 6er, it is about five times more likely that X would end up being in the random pair than if X is not a 6er. So, the fact that X appears in the pair you showed me is indeed evidence in favor of X. (This reminds me of the Monty Hall problem.)

So, so far so good. No paradox, but just ordinary Bayesian stuff.

Now, you find a pair A,B such that exactly one of X,A,B rolled six. This gives me no new information about X, however. If X,A,B were a randomly chosen triple with the property that exactly one rolled a six, then I should indeed assign 1/3 to X rolling six. But it's not. Instead, X is a random member of a random 6er-non6er pair, and then A and B are chosen to match.

Mark said...

I agree with everything you wrote, but I'm still not yet seeing how it differs from the infinite case. You put Patricia in a room with Jennifer. It seems like they both have a probability of 1/2 of having rolled a 6. You put Patricia in a room with two other people such that exactly one of them rolled a 6. Why would that make the probability go to 1/3?

William said...
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Bones said...

Mmmm. When you pair the dice rolls you're doing a totally different thing. You're asking what's the probability that you pick the person with the six. That's a totally different event than asking what's the probability that one of those roll a six. I'm just not sure why you don't think it's relevant that/how you do these pairings.

William said...

With regard to the 3/2/13 scenario, the angel has a 25/36 chance of getting two people with a non-six if they choose both simultaneously, so they cannot do it that way unless they lied to you about how they did the choosing.

They either chose you first and than chose your partner in the room after they checked your die roll, or chose your partner first then chose you based on your die roll after the fact.

If they chose you first, the chance is 1/6 of you having rolled a 6, but if they chose your partner first then you were chosen with a selection bias, and your chance of a 6 is 5/6. So the net is also 1/2 in that scenario, unless they tell you they chose you first, in which case it goes back to 1/6.

ah25 said...

These are two different paradoxes with different resolutions. They are all being driven by the fact that there is no well-defined way to have an countably infinite set of events each with equal probability (no uniform distribution possible over a countably infinite set, in the more formal language used in some comments), since the probabilities cannot sum to one. As a result, if you were to pick two people 'at random' (loosely speaking, since random sampling does not make sense with no well-defined probability distribution), and put them in a room, knowing nothing about them except that they are members of the set of people who rolled dice, we could not sensibly assign probabilities, because there is no well defined distribution over the population (note that I everywhere use 'population' to signify 'set of people who rolled dice', following the convention in statistics). In all cases proposed here, however, we have further information that makes everything discussed above irrelevant.

In the variant where an outside observer is put in a room with one of the matched pairs, their relevant information boils down to "exactly one of these people rolled a six, and exactly one didn't." Once they know this, and assuming symmetry, the clear assignment is 1/2. This does not depend at all on any other part of the story. The only relevant information for assigning probability to people in your room is the probability distribution over people in your room. The distribution in the population is only relevant in so far as you can use it to infer information about the distribution in your room. Hence, since we are given the exact distribution of people in the room here, the population distribution is completely irrelevant, well-defined or not. As an example, if you blindly picked one of two balls, knowing one was red and one blue, and were also told these had in turn been drawn from a hat containing 20 blue balls and 10 red, it would not be sensible to refuse to claim the ball you picked was blue with probability 1/2, since it is just as reasonable to claim it is 2/3. This is, however, essentially identical to the line of reasoning present in the dice-rolling story, just obscured by the use of infinite sets. By the same logic, your relevant information in the case with three people is 'one of these people rolled sixes, two didn't' and you can assign (1/3,2/3) to everyone without fear of contradiction, since further information about people not in your room is again irrelevant.

In the original case, your initial belief is, quite sensibly, "I rolled a six with probability 1/6." When the angel then tells you the rest, this changes to "with probability 1/6, I rolled a six and will be put into a room with a twin who didn't, and with probability 5/6 I didn't roll a six and will be put in a room with a twin who did," which, while containing new information, tells you nothing new about whether you rolled a six. In this case, the only relevant information is what you know about yourself. Continuing the example above, if you were to draw a ball out of the hat of 30, you should not alter your belief that you drew a red ball with probability 1/3 if you are also told that you will be put in a room with someone who drew the opposite color. The fact that not everyone can be matched with an alternate color partner shouldn't matter for your assessment of what you drew as long as you know only that you will be so matched regardless of what you draw, and thus cannot infer anything new about your own draw by the fact that you are so assigned. The same is true in the infinite case, even though such matching is possible there.

In the 'sequential case' mentioned, the only difference is that a population distribution exists. This is still irrelevant if you know the distribution in your room.

William said...

Can the angels put a given pair in the room without exercising selection bias?

William said...

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.

I know that of 36 possibilities, 26 without at least one 6 are excluded and one with 2 sixes is also excluded.

What remains that the angels could have picked are:

1-6, 6-1, 2-6, 6-2, 3-6, 6-3, 4-6, 6-4, 5-6, 6-5.

May this changes my information to support a 1/2 odds?

Alexander R Pruss said...

ah25:
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.
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.
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.

ah25 said...

Alexander Pruss:

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.

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.

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.

ah25 said...

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.

ah25 said...

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).

Alexander R Pruss said...

ah25:

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.

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?

3. In the second version, the observer foresees that every 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.

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.

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.

ah25 said...

In these stories, we are inclined to assert two seemingly conflicting things:

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

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

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:

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.

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.

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).

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.

ah25 said...

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.

ah25 said...

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.

Alexander R Pruss said...

"If you are then told the draw was in {11,12}, this changes nothing"

I am confused. It changes a lot of things. It sets the probability its being 13 to zero, for instance.

ah25 said...

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.

ah25 said...

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).

ah25 said...

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).

Alexander R Pruss said...

So, in the first case, you stick to 1/6, and in the second, you switch to 1/2?

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.

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.

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.

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.

So according to this argument, you should answer both cases in the same way.

William said...

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.

This suggests that there would be an infinite number of possible worlds where the angel never is able to get started telling the story!

If the probability for you cannot change, the angel should not be able to show up at all.

William said...

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.

Am I entitled to change my mind about the significance of the study's findings? (something like this actually happened btw).

Alexander R Pruss said...

William:

Yes, this may be a case where there is biased sampling that it's impossible to correct for.

But that's not a complete solution. For we still have to say when, if ever, we change probabilities, and why then.

It is a neat and simple solution to say that as soon as you learn about the infinity, you then abandon all probabilistic hope.

ah25 said...

Alexander Pruss:

Alexander R Pruss said...

But surely the probabilities shouldn't depend on whom we focus on first.

ah25 said...

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.

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.

William said...

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 this book.

ah25 said...

William:

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.

ah25 said...

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:

1) If I pick a friend at random, they rolled a 6 with probability 1/6.

2) If I pick a friend at random, their roommate rolled 6 with probability 5/6

3) If I pick a room at random, it will contain one of each type

4) If I pick a room and then one of its occupants, that person rolled 6 with probability 1/2

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.

ah25 said...

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).