Infinite multiverse, fine-tuning and probability
March 6, 2013 — 14:38

Author: Alexander Pruss  Category: Uncategorized  Tags: , , ,   Comments: 64

The best naturalistic alternative to theistic explanations of fine-tuning is a multiverse where there are infinitely many variations on the constants in the laws of nature, generating infinitely many universes, such that in infinitely many of them there is life–and we only observe a universe where there is life. Typical multiverse theories are committed to:

1. For any situation involving a finite number of observers, stochastically independent near-duplicates of that situation are found in infinitely many universes.

I will argue that if (1) is true, then ordinary probabilistic reasoning doesn’t work. But science is based on ordinary probabilistic reasoning, so any scientific argument that leads to the typical multiverse theories is self-defeating.

The argument that if (1) is true, then ordinary probabilistic reasoning doesn’t work is based on a thought experiment. You start by observing Jones roll a fair six-sided indeterministic die, but you don’t see how the die lands. You do, however, engage in ordinary probabilistic reasoning and assign probability 1/6 to his having rolled six.

Suddenly an angel gives you a grand vision: you see a countable infinity of Joneses, each rolling a die in a near-duplicate of the situation you just observed. You notice tiny differences between the Joneses, but each of them is rolling an approximately fair indeterministic die, and you are informed that all of these situations are stochastically independent.

You still can’t see what anybody has rolled. At this point, you reasonably assign approximately the same credence to every proposition of the form <Jonesk rolls six>, and if ordinary probabilistic reasoning is to work, this credence has to be about 1/6. This is just symmetry plus your initial ordinary probabilistic reasoning about the first Jones you saw. The angel then informs you that infinitely many of the Joneses rolled sixes and infinitely many didn’t. This doesn’t surprise you and you don’t change any of your credences. You already expected this from the Law of Large Numbers.

The angel then informs you that he’s made a list of all the Joneses, where each Jones who got a six is paired with a Jones who didn’t. You think to yourself: “So what? I already knew that such a list was possible.” And you don’t change any of your credences.

Then the angel then promises to transport you to meet every pair of paired Joneses, pair by pair, in a heavenly meeting room to which the Joneses are transported. And he does it. Let’s say you meet Jonesn and Jonesm. You know that exactly one of the two rolled a six but have no information as to which one it was. If you still assign approximately 1/6 to each proposition of the form <Jonesk rolls six>, you will violate the probability calculus, since given your information, P(Jonesn)+P(Jonesm) is close to 1, while 1/6+1/6 is definitely not close to 1.

So maybe you need to change your credence that Jonesn rolled six, either to an undefined probability (perhaps an interval) or to 1/2–symmetry allows no other options. But why change your credence now upon meeting the two? As soon as you heard that the angel made the promise to introduce you to the paired Joneses, you would have been able to figure out that you’d need to change your probabilities. So, why wait? Shouldn’t you change them right away? Van Fraassen’s Reflection Principle certainly would suggest you shouldn’t wait. But if you should change your probabilities then, why not earlier? After all, the angel’s promise that he’d transport you to meet all the pairs is quite evidentially irrelevant to the probabilities that a particular one of the Joneses rolled six. So you should have changed your probabilities of a particular Jones rolling six to 1/2 or no-probability even earlier, maybe when you heard that the angel made the paired list. But again, why then? That the angel made the paired list is quite irrelevant, as long as you already knew that such a pairing was possible. So you should have changed your credence from 1/6 to 1/2 or no-probability even earlier, when you found out that there were infinitely many of each outcome, as that was enough for you to know that a pairing was possible. But that won’t do, since as soon as you learned there were infinitely many Joneses, you knew (with probability 1) that infinitely many rolled six and infinitely many didn’t.

So it seems that the only reasonable place to put the probability shift is when you find out that there are infinitely many Joneses who rolled a die. In other words, learning that there are infinitely many Joneses who rolled the die makes it impossible to assign probability 1/6 to six being rolled by the Jones you see. And so learning (1) is true undercuts ordinary probabilistic reasoning, since such reasoning requires assigning 1/6.

The challenge to anybody who thinks that you can still maintain probability 1/6 after learning there are infinitely many Joneses rolling is either to say at which point in the process you need to change your probabilities or to defend the idea that you should assign a different probability to the first Jones you happened to see from all the others. I think the first option is the somewhat less unappetizing, and if I were forced down that route, I might insist that it’s the actual meeting of the paired Joneses that changes the probabilities, pace Reflection. But this is an unhappy conclusion, I think.

A somewhat different version of this argument is given here.

• Jeff Russell

There’s an important difference between knowing that there is *some* pairing of six-rollers and non-six-rollers and knowing that *this* is such a pairing, and, in particular, that (m,n) is a member of such a pairing. So you should adjust your credences in Jonesm and Jonesn at whatever point you learn that fact. Right? I wasn’t sure from the story if that would be when the angel announces the list or when the introductions are made. But there’s no failure of reflection involved in that, then, right?

March 6, 2013 — 16:58
• Yes, but you know that no matter what the pairing is, you will end up assigning no-probability (or 1/2, I guess) to each. You just don’t know in which order.
I was assuming the angel announces that he’s got a list, but doesn’t say who is on it. At this point, you should be able to know what your probabilities for everybody will change into.

March 6, 2013 — 17:07
• Anonymous

1. The number of universes could be a very large but still finite number. As long as the number of universes is large enough that you’d expect some of them to have life, the anthropic argument goes through just as well.
2. It seems like you could expand this fairly easily into a general argument against an actual infinite. Do you know if anyone’s explored an argument along these lines before (exploiting the weird arithmetic properties of infinite numbers)?

March 6, 2013 — 20:25
• Eric Steinhart

Alex,
First, why is that the best naturalistic alternative? There are others, surely at least equally good (one thinks of Rescher’s version of axiarchism, in The Riddle of Existence).
Second, why would the infinities involved generate the problem? Surely there are infinitely many infinitely small variants that cause no troubles. Did you mean to assert, instead of mere infinity, some type of plenitude of total combinatorial variation, like Lewis?
Third, if you’re using this as an argument against Lewisian plenitude, you’re right on – the argument is of course well-known. I think it is a good argument against Lewis’s combinatorial notion of possibility, and an argument in favor of more constrained types of modal plenitude (again, axiarchic constraints come to mind, and they need not involve gods, tho of course they may).
Fourth, I’m not aware of any scientific multiverse theory that asserts Lewisian plenitude (quantum mechanical, inflationary, and string-theoretic are all highly constrainted in lawful ways); so, as an argument against scientific multiverse theories, your proposal fails.
Fifth, I suspect you’re confusing various interpretations of probability. Talk of “credence” and changing “your probabilities” isn’t quite probability.
– Eric

March 6, 2013 — 22:14
• Angra Mainyu

Alexander,
I’m not sure how you defined ‘naturalism’, but I would disagree with the claim about fine-tuning.
That aside, and with regard to the probabilistic argument, I would argue as follows (assuming the angel can be trusted; otherwise, that results in different probabilist assessments, but that’s not a problem).
By “n=6” I mean that Jones#n rolled a six.
So, my prior is P(n=6) = 1/6, for any n.
However, as Jeff pointed out above, the angel gives us more information.
That information results in a change in the assignment when the angel identifies the specific pair, say
n0 and m0.
In other words, what the angel is telling us is that E obtains, where E = ((n0,m0) = (6,Â¬6) v (n0,m0) = (Â¬6,6)).
Given that, we then apply Bayes’ theorem and (for instance) compute the probability that Jones#n rolled a 6, conditioned to E.
Let E1 be the event (n0,m0) = (6,Â¬6), and E2 the event (n0,m0) = (Â¬6,6).
Then, E=E1 U E2, and the union is disjoint.
1. P(n0=6âE)=P(n0=6&E)/P(E) = (P(n0=6&E1) + P(n0=6&E2))/P(E) = P(E1)/P(E)
2. P(E)=P(E1)+P(E2)= P((n0,m0) = (6,Â¬6)) + P((n0,m0) = (Â¬6,6)) = (1/6)*(5/6) + (5/6)/(1/6) = 5/18.
3. P(E1)=(1/6)*(5/6)=5/36.
4: P(n0=6âE) = 1/2.
The same applies to the probability that Jones#m0 rolled a 6. It’s 1/2 in each case, after the angel has identified the pair. The probability for any other Jones remains 1/6.

March 6, 2013 — 22:58
• Jeff Russell

I seeâI wasn’t following the argument correctly. I like it. It looks to me like this falls in the pattern of infinitary violations of the principle: the E’s form a partition; P(X|E) = p for each E; therefore P(X) = p. That’s a little different from reflection, but obviously closely related.

March 7, 2013 — 7:52
• Anonymous:
Indeed, this argument has nothing to say about a finite multiverse.
You could extend this to an argument against an actual infinite if you added the thesis that probabilistic reasoning necessarily works. But that’s a strong thesis and I doubt it’s true.
Eric:
The probabilities here are rational degrees of belief.
The argument doesn’t need plenitude. It only needs a situation where there are infinitely many very similar people who rolled a die and didn’t see the answer. Multiverses generated by some process that infinitely often randomly sets the initial parameters and then lets things go will generate all situations that have non-zero chance infinitely often.
I don’t think of axiarchic options as either plausible or naturalistic. (For one, they entail nonnaturalism, since it’s better that there be a God, as both Rescher and Leslie acknowledge.)
Angra:
The calculation is correct, assuming that that’s the whole information that you have (you actually have a little bit more data–you have some information about how the angel paired things). But before the angel takes you to see any of the Joneses, you can predict that you will make the calculation about all of them. And so your credences violate van Fraassen’s Reflection Principle, and you are subject to a diachronic Dutch Book.
Moreover, on this way out, your credences are subject to arbitrary manipulation. The angel can make you set your credences to whatever values he chooses, just by regrouping people.

March 7, 2013 — 7:52
• A different way out is to deny the Axiom of Countable Choice, and then hold that in our multiverse, while there is an infinite set of die rolling situations, that set has no countable subsets. That would be desperate!

March 7, 2013 — 8:21
• I meant: no countable infinite subsets, of course.

March 7, 2013 — 8:22
• Jeff:
Yes, the argument does appear related to failures of conglomerability. But failures of conglomerability do indeed give rise to violations of Reflection.
One way to make it more like a failure of conglomerability argument is this. Let Q be the set of all events of the form: (d(k1)=6 iff ~(d(k2)=6)) and (d(k3)=6 iff ~(d(k4)=6)) and …, where we range over all permutations k1,k2,k3,k4,… of 1,2,3,4,…, and d(n) is the result of Jones_n’s roll. Then Q isn’t a partition or an almost-partition (stipulate: an almost-partition of Omega is a partition of Omega-A, where P(A)=0), because there will be overlap (there are many ways of pairing), but every member of Q is a member of some almost-partition which is a subset of Q. Then we can note that for all E in Q, P(d(1) = 6 | E) = 1/2, and it sure looks like a non-conglomerability argument.
But not quite. For, P(d(1)=6 | E) is undefined, since P(E)=0.
The argument as I gave it gets around this by not conditioning on all the information about all the pairs, but by going on the basis of information about the pairs one by one. So what is going on in my argument, while similar to non-conglomerability, is not non-conglomerability.
I think what lurks in the background is a related phenomenon: the nonexistence of a uniform measure on the set of all permutations of 1,2,3,…. See my March 4 comments here.

March 7, 2013 — 9:15
• Angra Mainyu

Alexander,
Thanks for the reply. Now, I’m not sure I understood your scenario.
As I understood it, and regarding more information about how the angel picked, I’m leaving aside the information that tells one that the angel may not be trustworthy (e.g., after all there seems to be no strong reason to trust a what looks like a powerful alien entity making such implausible claims). But I’m assuming that, for whatever reason, one can trust the angel, because that seems to be part of the scenario.
If you’re talking about the information that infinitely many of them rolled sixes, and infinitely many of them did not, I count that in my background when I assign prior, since the information is not relevant. It still does not give me information about specific Joneses.
Also, I was considering that the angel had given me information about a specific pair, say (n0, m0), and nothing else.
That information was not enough to make any predictions about the specific function, etc., and I didn’t have any more information about any other pairs, so the application of Bayes’ rule as I did seems to be correct as far as I can tell.
However, maybe you’re saying that I know how the angel paired things. I have one pair at first, but that’s all. If the angel shows me other pairs, then things change, and one has to consider more possibilities (do I see any patterns?), which makes the matter a lot more complicated.
Or do you mean that the angel actually tells us what the function is?

March 7, 2013 — 14:03
• No, the angel doesn’t tell you what the pairing procedure is. And I assume there are no discernible patterns.
But you know that he’ll show you all the pairs in his pairing. So if you know you’re going to change your probabilities of six to 1/2 whenever you’re shown a pair, you know ahead of time that all your probabilities will eventually be 1/2. So why wait?

March 7, 2013 — 14:13
• Eric Steinhart

Alex –
Rescher in The Riddle of Existence says his axiarchism is naturalistic, and explicitly and repeatedly rejects all appeals to God. I take him at his word. The Riddle of Existence is as atheistic as one could please. Plausibility is another issue. But for you to say that your case is “the best” naturalistic explanation is simply untrue. Probly most naturalists would prefer to say that our universe is just a brute fact.
You wrote: “Multiverses generated by some process that infinitely often randomly sets the initial parameters and then lets things go will generate all situations that have non-zero chance infinitely often.”
And your proof of that is what? One might, after all, flip a coin infinitely many times and have it come up heads every time.
– Eric

March 7, 2013 — 17:04
• Angra Mainyu

Alexander,
Is the assumption that there are no patterns I will be able to discern part of the information the angel gives me, or does it come from another source?
By the way, with regard to manipulation, it seems that in the finite case too the angel can also manipulate the probabilistic assignments by regrouping people.
For instance, let’s consider an alternative scenario S2:
Angel tells me that the number of Joneses is k0=10^(10^(10000!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!))), and that they all rolled two dice at a time (for instance), in similar conditions, etc.
What’s your probabilistic assignment to the event that J2000001 (i.e., Jones#2000001, according to some list he made) rolled two sixes?
I don’t know anything about the angel’s mind; I’m facing an alien, so I don’t really know what his intentions are, etc., but if that’s a problem, it’s also a problem in the infinite case (and not related to infinities), so let’s say that’s not a difficulty.
I answer it’s slightly less than 1/36 (the dice might be leaning on something, be destroyed in a QM freak event, etc.). I give the same answer when he asks me about J2.
Now, Angel tells me that he’s paired any Jones who rolled two sixes with a Jones who did not, at least until he runs out of Jones in either category â whatever happens first.
Then, he tells me that (J1,J2) are a pair, and asks me about J2 and J2000001.
So, he keeps showing me more pairs, like (J3,J4), (J5,J6), etc., until (J1999999,J2000000), and asks me about J2000001 again. I would be inclined to say it’s a lot more than 1/36, even if I’m not sure how close to 1/2 it is, because I see an obvious pattern and I’m intuitively including that in the probabilistic assessments I’m making.
But if he had numbered them in exactly the same manner but had shown me a million pairs (J2,J4), (J6,J8), etc.,), then I would have been inclined to give a very low probability to the event that
J2000001 rolled two sixes, even much less than 1/36.

March 7, 2013 — 17:47
• Angra Mainyu

Alexander,
Regarding the issue of the discernible patterns, I’m asking because I think the matter is relevant, for the following reasons:
Let’s say that the case is again finite, though unbounded, and there is only one Jones under consideration, who rolls a die. And I assign probability 1/6. But then Angel tells me: you see, Jones will keep rolling the die forever, and he will keep getting different numbers. You will continue watching that for eternity (I guess I’m in some sort of very watered down version of Hell;)), and yet you will never be able to discern any pattern at all.
So, I reckon, if I compute (number of sixes in n dice rolls)/n, as n tends to infinity, I will not see that the number approaches 1/6, since that would be at least a pattern regarding the distribution of sixes. But I will be completely unable to see a pattern at all, even though this will in fact go on for eternity (potential not actual infinity) – at least, if I trust Angel; else, things get complicated.
Then, I see that Jones rolls the die, but I don’t see the result. Angel then asks me what probability I assign to the event ‘Jones just rolled a six’. I would refrain from assigning any probability.
Leaving aside the finite case now, and going back to the infinite scenario, there are some possibilities:
a. Angel does not say anything about whether I will be able to see any patterns.
b. Angel tells me that I will never be able to see any patterns.
In case b., I withdraw my assessment 1/6 for specific cases, since I don’t know how Angel has arranged things, other than the fact that he arranged them in a way where I can see no patterns, no matter what.
In case a., I do not know whether I will end up assigning 1/2 for a specific case. I have no idea. What I will do will depend on what I see. In the first case, I would assign 1/2 using Bayes, because I see no other source of information (as long as the fact that an alien mind is involve is not a problem; else, it’s a problem in the finite case as well); as time goes by, who knows?
Granted, Angel might adjust the scenario and allow some patterns, while denying others. I think I would have to consider the matter on a case-by-case basis, though.

March 7, 2013 — 18:17
• Angra Mainyu

Just to clarify: in case b, I withdraw my assignment of 1/6 if Angel comes up with a certain number, say 487835724, and asks me about the probability that the Jones to which he assigned that number rolled a six, since my scenario is no longer related to random dice-rolling, but Angel making up some function I know nothing about.
That said, and while I have so far assumed that the presence of Angel is not per se a problem (else, the scenario does not show a problem with the infinity), his presence plausibly is a problem, since I have no idea how he’s assigning the values, and that may well take away any previous considerations about dice-rolling (in other words, my scenario is not now that some people rolled dice, but that some entity of unknown mind picked a specific function.

March 7, 2013 — 18:38
• Eric:
“Rescher in The Riddle of Existence says his axiarchism is naturalistic, and explicitly and repeatedly rejects all appeals to God.”
He rejects appeals to God, but he rightly thinks that the best possible world includes God. He is a Catholic, after all.
“And your proof of that is what?”
Strong Law of Large Numbers.
“One might, after all, flip a coin infinitely many times and have it come up heads every time.”
One might, but if the flips are independent and the coins are approximately fair, the probability of that scenario is zero.

March 7, 2013 — 19:50
• Here’s a fun practical question. Let’s suppose a variant on which right after you’ve seen a pair that contains Jones1, you’re offered a bet. If Jones1 rolled six, you get \$1.01. If Jones1 didn’t roll six, you pay \$1. If you’ve changed your credence to 1/2, you will take this bet. And then the experiment ends–no more information is given.
Your memory is then wiped, so you can’t try to figure out the angel’s pairing algorithm. And the game repeats.
If you changed your credence to 1/2, you will take this bet each time. And you will on average lose, since in only approximately one in six experiments will Jones1 roll six, and so on average, for each time you get \$1.01, there will be five times you have to pay \$1.
This shows that if your credences correspond to betting probabilities, there is a case to be made for your sticking to 1/6 each time. But then there is a Dutch Book against you. So you either lose on average, if you switch credences to 1/2, or you lose every time you play if you stick to 1/6 and your opponent is clever.

March 7, 2013 — 19:58
• Angra:
By the way, in the finite case, the mere fact that J1 is paired with somebody is evidence that J1 rolled six (or double-six in your scenario), the majority of those who didn’t roll six probably went unpaired.

March 7, 2013 — 20:06
• Angra Mainyu

Alexander,
Regarding the manipulation in the finite case, yeah that is true, but precisely what I was trying to get at is that by giving different amount of extra information, Angel can manipulate your assignments (or mine, or whatever the proper assignment is).
Btw, I’m still not sure about the pattern condition: does Angel include the claim about no pattern in the scenario?
I will think about your fun practical question and get back to you on that, but meanwhile I have a fun alternative scenario as well:
Scenario S4:
Angel tells you that the total number of people is finite but unbounded, and that there will be any arbitrary number of Joneses in the future, but always finite.
So, he will start next year, and for each Jones who rolls a six, Angel will assign her a Jones who rolls some other number. Angel will make sure that for every year m, at some point in the future all of the Joneses who rolled dice in m will be paired with someone, as long as the number of sixes and non-sixes is unbounded.
Also, Angel will continue to show those pairs to you for eternity (i.e., potential infinity).
So, Angel asks you to make probabilistic assessments. How would you proceed?
Alternative variant S5:
Like S4, but Angel adds the condition that you will not be able to discern any patterns.

March 7, 2013 — 21:15
• Angra Mainyu

Okay, so the fun question – it’s very intriguing. It seems to suggest that I tend to intuitively mistrust an unknown entity who picks the pairings however he likes knowing the results beforehand; this seems to support not assigning any probability as all, at least in this case.

Here’s a fun practical question. Let’s suppose a variant on which right after you’ve seen a pair that contains Jones1, you’re offered a bet. If Jones1 rolled six, you get \$1.01. If Jones1 didn’t roll six, you pay \$1. If you’ve changed your credence to 1/2, you will take this bet. And then the experiment ends–no more information is given.

I would not bet. But if the bet were that if Jones1 rolled six, I get \$6.06, intuitively I would not bet, either.
Generally, I think in a situation like that (i.e., my interlocutor is an unknown agent who already knows the result) I would refuse to bet unless I have alternative sources of information (and leaving aside threats, etc.), or perhaps unless the loss would be minimum and there is a lot to win, but that second option seems to indicate that I have an intuitive tendency to mistrust the entity in question.
But the dice no longer play a role for me. At this point, I’m intuitively focusing on what the alien/angel might do, and what my options are; I would also try to look for clues as to how powerful my opponent is, etc.

Your memory is then wiped, so you can’t try to figure out the angel’s pairing algorithm. And the game repeats.
If you changed your credence to 1/2, you will take this bet each time. And you will on average lose, since in only approximately one in six experiments will Jones1 roll six, and so on average, for each time you get \$1.01, there will be five times you have to pay \$1.

Do I know the wiping condition in advance? (not sure I can do anything about it, anyway, other than refuse to bet).
Also, I’m not sure why there is a pairing algorithm here. Isn’t it always Jones1, sequentially (i.e., in chronological order)?
If there is some algorithm and there is no chronological order, the frequency does not have to tend to 1/6, but can be arbitrarily manipulated, at least if the number sixes and the number of non-sixes is unbounded, and Angel (or whoever my opponent is) can take as much time as it wants before offering me one bet and the next.
In any case, I will be very strongly inclined not to bet.

This shows that if your credences correspond to betting probabilities, there is a case to be made for your sticking to 1/6 each time. But then there is a Dutch Book against you. So you either lose on average, if you switch credences to 1/2, or you lose every time you play if you stick to 1/6 and your opponent is clever.

Hmm…I’m not sure I’m following you on this one.
If you stick to 1/6 you always refuse to bet (I suppose), so I’m not sure in which sense you lose every time. At least, if sometimes Jones did not roll 6, then by refusing to bet you don’t fail to win (unless Jones always rolls 6, but that’s not the case I assume).

March 7, 2013 — 21:53
• Eric Steinhart

Alex –
Rescher holds a variety of beliefs across a (very large) number of books. Of course he’s a Catholic; the point is that in The Riddle of Existence he defends an entirely and explicitly atheistic version of axiarchism. He is very clear about that (I just re-read it.) So to say that axiarchism entails theism is false.
What you’ve described so far is merely a random walk through a phase space; it will take a great deal of math to prove that every point in that phase space is visited. The Strong Law of Large Numbers is about the average value of the random variable. It does not prove what you seem to intend it to prove.
Admittedly, from your description, I can’t quite make out the structure of the set up you’re proposing. I take it that there is a phase space (with how many dimensions? are they finite or infinitely structured? if infinite, discrete dense or continuous?). I take it there is a random variable associated with each dimension. You’re talking about iterating outputs of the random variable on each dimension to make a vector (a point in the phase space). What is the cardinality of the iteration? finite? countable? uncountable?
Visiting every point in a finite phase space as the result of an infinitely long random walk through the space, ok, sure. Visiting every point of an infinite phase space as the result of a random walk of equal cardinality, that’s what you need to prove. (And if the walk is of greater cardinality, well, I still worry about it merely covering an infinite proper subset of the original phase space.) But I don’t know the theorems here.
– Eric

March 7, 2013 — 22:49
• Angra Mainyu

Alexander,
After further consideration, I think that the assumption that the presence of the angel is not a problem has to go, for the following reasons:
In the unbounded finite case, Angel can manipulate how he numbers the Joneses, and make the quotient (#of sixes in the first n Joneses/n) approach any number of his choice for any period, no matter how long, then go in another direction if he so chooses, etc.
With no further information about Angel’s choice, there is no good reason to think that he’ll choose in a way that will make the quotient tend to 1/6, as one would expect if one were to make probabilistic assessments in usual cases.
Moreover, in usual cases, we assess ‘1/6’ but under the understanding that the quotient in repeated independent experiments will tend to that. But on the other hand, if, for a particular die, we know that the quotient would tend to 4/5, then we would assess ‘4/5’ instead of the usual ‘1/6’.
Given the above, I would say that that there is not enough information to make an assessment in the finite unbounded case (i.e., potential infinity in the future), and due to similar considerations, also in the original infinite case.
In despite of first appearances, it seems to me the scenario is not about probability involving Joneses rolling dice, but about how to assign probability (or not to) when facing unknown agents picking orderings in a way we know very little about, if anything at all.

March 8, 2013 — 10:57
• Eric:
Well, I stand by the claim that axiarchism entails the existence of nonnatural beings, because in fact some worlds with God in them are better than any worlds without God. Rescher finds this plausible, but you don’t. 🙂
As for the Law of Large Numbers, imagine that an infinite set of universes U_1,U_2,… are independently generated with the same probability distribution for the evolution of each. Let p(U_k) be some proposition about U_k such 1 > P(p(U_k)) > 0, and by sameness of distribution is the same non-zero number. By independence, p(U_1), p(U_2), … are independent. Let X_k = 1 if p(U_k) is true and let X_k = 0 if p(U_k) is false. Then X_1,X_2,… are independent bounded random variables with the same distribution and E[X_k]=P(p(U_k)). By the Strong Law of Large Numbers, almost surely (X_1+…+X_n)/n converges to E[X_1]=P(p(U_1)). This can only happen if almost surely infinitely many of the X_k are 1 (when they aren’t, (X_1+…+X_n)/n will converge to zero) and almost surely infinitely many of the X_k are 0 (when they aren’t, the averages will converge to 1).
Basically, I am imagining that the scenario can be modeled by an infinite number (whether countable or not doesn’t matter, since any uncountable set has a countable subset–I am assuming the Axiom of Countable Choice) of independent and identically distributed random variables whose values are global states of a universe.
This simplified version will not apply to scenarios that involve branching. Those are more like the random walk case. However, I have never heard a physicist deny that there are going to be loads of near duplicates in the infinite multiverses they are interested in.

March 8, 2013 — 11:16
• Angra Mainyu

This simplified version will not apply to scenarios that involve branching. Those are more like the random walk case. However, I have never heard a physicist deny that there are going to be loads of near duplicates in the infinite multiverses they are interested in.

Relaxing the hypotheses a bit, the same seems to apply to infinitely many galaxies, by the way, without a need for parallel universes.
Also, it seems to me that the condition of ‘near duplicate’ of all of them (or all but finitely many) is not required, either.
The probabilistic assignment 1/6 is actually an approximation even in usual cases, and did not require of any specific trait of Jones. If Smith had roll an ordinary die at her home, the assessment would have been the same (i.e., approximately 1/6).

March 8, 2013 — 11:40
• Right: the story can be relaxed in all sorts of ways.

March 8, 2013 — 11:42
• Angra Mainyu

Brief addition: though I don’t think that this is any problem for any of those views of infinite universes or multiverses, for the reasons I’ve given above.
The problem seems to be the angel, rather than the near duplicates, at least in the scenario under consideration.

March 8, 2013 — 11:42
• Eric Steinhart

Alex –
You say “axiarchism entails the existence of nonnatural beings, because in fact some worlds with God in them are better than any worlds without God.” Unless, of course, one thinks that God is impossible, because (say) the definition of God is internally inconsistent.
You say “Rescher finds this plausible, but you don’t. :-)”. On the contrary, I am a theological naturalist – I do not at all deny the existence of gods (tho I do deny the existence of God). I do (at least so far) think that axiarchism entails that there are things that deserve to be called “gods”. (These are not, of course, like the super-natural gods of Abrahamic theism, but that’s another story, published here and there.)
My only point, and I think you need to take this more seriously, is that axiarchism can be developed in purely naturalistic ways – a naturalist can defend it (and I think Parfit comes close to defending it in his “Why anything? Why this?”). A hard-core entirely atheistic modal realist could appeal to axiarchic principles for the generation of the domain of modal quantification. This is an unexplored option in the system of ontologies. But that doesn’t mean it should be dismissed so quickly.
– Eric

March 8, 2013 — 12:23
• Eric Steinhart

We let B be any finite number divisible by 6. We can agree that B is big enough for the laws of large numbers to work.
Now we assert that there is a set of B-many Joneses. Each Jones rolls a fair 6-sided die and the rolls are all independent.
We now define a 1-1 function J from B to the set of Joneses such that J(k) is the k-th Jones according to that function. Obviously, this is a choice function.
Of course, the function J encodes an implicit probability distribution. One way to define it is to have Zeus start with 0, pull out a random Jones from the set, and pair him with 0; then do the same for 1, 2, and so on, up to B-1.
If (and only if) J is constructed by randomly choosing Joneses, then it follows that for every k, Pr(J(k) rolls 6) is 1/6. But if J is not itself a random distribution, it does not follow. So let us assume that J is itself a random distribution.
An angel now constructs a set of pairs of Jones such that exactly one member has rolled a 6. Let this set be G. For any pair (x, y) in Pr(x rolls 6 or y rolls 6) is 1.
Of course, for any pair of Jones (x, y) in G, there are numbers n and m less than B such that (x, y) is (J(n), J(m)). But it is false that for any n and m less than B, there is some pair (J(n), J(m)) in G. Clearly, in the finite case, 2/3s of the Jones are not in G.
It remains true that for any n and m less than B, the probability Pr(J(n) rolls a 6) is 1/6 and Pr(J(m) rolls a 6) is 1/6.
For any n and m less than B, if (J(n), J(m)) is in G, then Pr(J(n) rolls a 6 or J(m) rolls a 6) is 1; and Pr(J(n) rolls a 6) is not 1/6 but 1/2; and Pr(J(m) rolls a 6) is not 1/6 but 1/2.
So far, there are no problems at all with the probability calculus. And so far, if my subjective credence values track probabilities, I never made any revisions at all. It is entirely transparent how the probability and credence values are computed.
All this will hold up if we say âlet B go to infinityâ. As B does go to infinity, the ratios all converge to their limit values as expected.
None of this reasoning works if B is the entire set of natural numbers. All the sets involved are countably infinite, none of the probabilities are well-defined.

March 8, 2013 — 14:09
• Angra Mainyu

I think the following parallel (or rather, some of them; more below) will match the infinite case presented in the OP, and illustrate the problem as I see it.
There are three people: Jones, Smith, and Angel the angel.
So, Angel (who is entirely honest) explains the experiment, which will last for eternity (potential infinity):
Jones will keep rolling a die forever. He does not have to do it all the time, etc., but the number of times Jones will roll the die is unbounded. The conditions are usual conditions in which we would normally assess 1/6 (or more precisely, close to 1/6), only that the dice and the table will never deteriorate; the table is big enough and flat, and the dice will always roll some number (if somehow the dice ends up in a weird position, the roll is repeated until the problem is corrected).
R(n) is the roll #n, for all n (i.e., R(4)=3 if Jones rolls a 3 the fourth time).
Also, after Jones has rolled the dice a certain number of times n1 (which Angel will choose, but he does not specify), Angel will assign a number f(1)â¤n1. Then, after a certain bigger number, n2, Angel will choose f(2), and so on. Also, for any fixed n0, at some point there will be a number g(n0) such that f(g(n0))=n0.
Additionally, Angel will come up with a list of sets of exactly two numbers, meeting the following conditions:
i. For any natural numbers k and r, if {k, r} is one of the sets, then R(k)=6 or R(r)=6, but not both.
ii. The sets will all be disjoint.
iii. For any natural number k, there will eventually be some r such that {k, r} is one of the sets, provided that the total number of sixes is unbounded, and also the total number of non-sixes is unbounded (which Smith can tell in advance with probability 1).
Smith will not be able to see the results, and Jones will not tell Smith about them.
So, the experiment begins, and after some time (a lot of time), and before Smith gets any other new piece of information about the experiment, Angel tells Smith that Jones already rolled the die a number of times m, but he does not tell her what m is, only that it’s bigger that k0^(k0^((k0^(k0!!!!!!!!!!!!!!!!!)))) (ko is the big number I defined earlier; i.e., k0=10^(10^(10000!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!))) ).
He also tells Smith that he’s already constructed at least k0 of the f(j).
Then, Angel asks Smith the following questions:
1. What’s your probabilistic assessment of the event “R(1)=6”?
3. “R(48534572475)=6”?
4. “R(f(48534572475))=6”?
5. Do you think that (# {dâ¤m: R(d) =6})/m is probably close to some number? If so, which one?
6. Do you think that (# {kâ¤k0: R(f(k)) =6})/k0) is probably close to some number? If so, which one?
7. One of the sets is {454, 435395739753874383563565652463555484}. What is your probabilistic assessment of the event “R(454)=6”?
8. One of the sets is {f(1894), f(27425536524845724347244325353535535)}. What is your probabilistic assessment of the event “R(f(1894))=6”?
Smith replies:
1. 1/6.
2. I have insufficient information to make any assessments.
3. 1/6.
4. I have insufficient information to make any assessments.
5. Yes, it’s probably close to 1/6.
6. I have insufficient information to make any assessments.
7. 1/6.
8. I have insufficient information to make any assessments.
As I see it, the infinite scenario presented in the OP is relevantly similar to the situations 2, 4, 6, and 8, for the following reasons:
Angel chooses how to number the Joneses when he already knows what the results were, and also chooses the pairs when he already knows the results.
In the case above, Angel makes that choice partially, since he doesn’t yet know what will happen in the future (unless he can see the future, in which case the match is even more exact), but even then, Angel can choose numbers n1, n2, etc., as big as he wants in order to make the assignments of the f(j), and the same goes for the pairs.
So, the lack of information remains the same, and Smith is in no position to make an assessment based on the evidence provided above, even assuming Smith properly assigns probability 1 to the event that Angel is always telling the truth.
Essentially, we don’t know in the infinite case how Angel has chosen the numbering of the Joneses, or the pairs, so we are in the dark, as Smith is in cases 2, 4, and 8.
As for 6, it’s similar to the infinite case in that we have no idea whether #{Jones#k : kâ¤n and Jones#k rolled a 6}/n will tend to anything, and if it does, whether it will tend to 1/6, or for that matter 1/475, 1/37063, or anything else.
It’s an interesting thought experiment (and fun too), but it’s not problematic for scientific views involving an infinity of people rolling dice in conditions similar to those in which we assign (approximately) 1/6 in daily life.

March 8, 2013 — 15:57
• Angra Mainyu

Clarification: the 1/6 assignments are ‘roughly’ 1/6, since the die is not perfect (gravity is supposed to be normal, etc.).
Also, in case of 7, Smith does get a bit of information, namely that there’s been at least one six, and at least one non-six, but that was almost certain given how big m is, so essentially his assignment is still roughly 1/6.

March 8, 2013 — 16:06
• Angra Mainyu

Correction: Sorry, I forgot to factor in that Smith does not know what m is, so he doesn’t get any more relevant info because in any case with probability 1 that there will be sixes and non-sixes.
Also, regarding that the die is not perfect, I meant it’s supposed to be a real rather than ideal die, even if it never deteriorates, etc.

March 8, 2013 — 16:12
• Angra Mainyu

On the other hand, now Smith knows that there is at least one six and at least one non-six in the first 435395739753874383563565652463555484 events, so that’s a bit of information.
He also knows that either R(454)=6, or R(435395739753874383563565652463555484)=6, but that does not provide any good information about R(454) – beyond, perhaps, the very slight amount of information above -, because every single number will be in some two-element set anyway, so it was already known that 454 would be in one such set.

March 8, 2013 — 17:22
• IanS

I see no paradox. I say 1/6 for the first Jones and 1/2 for the paired Jonses. And for the pair containing the first Jones, 1/6 and 5/6.
Why the differences? Different knowledge.
For the first Jones, there are 6 possible outcomes with physical near-symmetry. Hence 1/6. For the paired Jonses, we have the extra information that exactly one of them threw a 6, but we have no basis to distinguish them. Hence 1/2, 1/2. For the pair containing the first Jones, we have no new information about him: The fact that he has been paired tells us nothing – it would have happened no matter what. And we know nothing useful about his partner. So his chance stays at 1/6. We do know that either he or his partner threw a 6, so his partner’s chance is 5/6.
If (for a variation) I had seen two Jonses initially, and (by chance or otherwise) they had been paired together, I would say 1/2, 1/2. Why the change from 1/6. Because of the new information that one of them has thrown a 6.

March 9, 2013 — 2:06
• Dianelos Georgoudis

Alex,

One might, but if the flips are independent and the coins are approximately fair, the probability of that scenario is zero.

Since the probability is zero in what sense is Eric justified to speak of âmightâ? Since it is true that if one keeps flipping a coin one *will* at some point get a tails, what sense does it make to say that one might not?
Incidentally, I donât understand metaphysiciansâ love affair with actual infinities, given that if there are actual infinities reality becomes unintelligible and thus not worth thinking about. Is there any practical state of affairs where substituting actual infinities with very large numbers (the way Angra suggests in his posts) would lead to error, and therefore shouldnât be done? Is there any reason not to use infinities only in the context of math in order to get to results quicker? Results which approximate reality when large numbers are at play?

March 9, 2013 — 6:07
• Dianelos Georgoudis

Angra,

I would not bet.

Suppose this is not an option. Suppose you must either take the bet Alex offers you, or else take a trivial bet where a coin is flipped and you either double or lose your money. So the question is this: On the long run will you win more or less by taking Alexâs bet instead of the trivial bet? Given the information at hand what is the strategy that will win you more money, and thus the strategy based on the correct probability estimates?
In many instances the above allows one to discover the correct epistemic probabilities by experiment: Use a random generator of simulated worlds which respect all given information, and find out which probability maximizes your bet earnings.

March 9, 2013 — 6:09
• IanS:
But once you have the big vision of all the Joneses, we can suppose that each of them is distinguished in some way. This one has a bigger nose, this one is your friend, this one has a pimple, this one is wearing a T-shirt with the number googolplex-1, etc.

March 9, 2013 — 9:07
• Dianelos:
You don’t need a simulation: you can just see how it works. If you bet on the outcome for one particular Jones, say Jones10 (with the parade of pairs cut off as soon as you’ve seen whom Jones10 is paired with), you will lose in the long term if you keep on accepting the bet I listed. For five runs out of six, you will end up shelling out a dollar, and one run out of six you’ll get \$1.01, so per six runs you’ll be losing \$4.99.
But you will win in the long term if you accept the following bet: If Jones10 rolled six, you pay \$5. If Jones10 didn’t roll six, you get \$1.01. But this is a stupid bet if your probabilities for Jones10 are 1/2.
So if probabilities match what is smart to bet on (at least in cases where the number of bets offered to you doesn’t depend on what the situation you are betting on is like, in the way it does in Sleeping Beauty cases), you should assign 1/6 to everybody by the above argument. And that’s inconsistent.

March 9, 2013 — 9:14
• Angra Mainyu

Dianelos,

Suppose this is not an option. Suppose you must either take the bet Alex offers you, or else take a trivial bet where a coin is flipped and you either double or lose your money. So the question is this: On the long run will you win more or less by taking Alexâs bet instead of the trivial bet? Given the information at hand what is the strategy that will win you more money, and thus the strategy based on the correct probability estimates?
In many instances the above allows one to discover the correct epistemic probabilities by experiment: Use a random generator of simulated worlds which respect all given information, and find out which probability maximizes your bet earnings.Â

When I said I wouldn’t bet I meant I wouldn’t take the bet, but that’s probably because of my intuitive mistrust of the agent manipulating everything.
But the difficulty is that under duress (i.e., forced to bet), I’m no longer interested in money. I’m trying to figure out what my best option for dealing with the powerful agent threatening me (and forcing me to bet) is. Even trying to leave that aside for the sake of the argument doesn’t seem to work.
Regarding Alexander’s scenario, if I know that my memory would be wiped, I’m not remotely interested in money. What if I don’t know that? I’m still under threat, and I don’t care anymore about money, either.
I think this problem is not trivial: we do intuitively focus on different matters under different situations.
In fact, even without a threat, if I’m in a situation in which Angel tells me that I’m going to live forever and play that game forever, etc., money is no longer a consideration unless someone gives me reasons to care about it, so it’s a hard choice…I can still try a modified scenario, but I think that the situations are perhaps to alien for me to predict how I would behave if I were actually in one of them.
In any case, I would say infinity is not a problem: the Angel/Jones/Smith scenario handles the matter in a relevantly similar sense without any actual infinities.
Still, just for fun, I could adapt a scenario (I’d like to ask what you would reply as well), though there is a problem with the mind-wipe. Do I know in advance that Angel will do that to me?
Anyway, we may consider variants. For instance:
a. Variant 1.
Let’s say that you’re in Smith’s situation, and that in the realm to which the Angel took you, there is still money, it can be used to buy goods and services, etc., and generally people with money have access to better and more stuff than the others. The situation will last forever. You start with a certain amount of money, say \$ 1000000.
The event you’re asked to bet on is question 8, namely whether R(f(1894))=6.
While you don’t know that, every time you bet, your memory of that event will be wiped, and later Angel will ask again, and again, and again, for eternity. The question is always about whether R(f(1894))=6.
b. Variant 2
Like variant 1, but you’re informed before you make your choice that your memory of the event will be wiped.
Despite any attempts to the contrary, in variant 1, I find myself using the information that Angel placed me in a very stressful situation, and intuitively I assign high probability to the event that he will choose to make me lose.
In variant 2, the situation is far more stressful, and then I would also pick the trivial bet.
But as I mentioned, in a real case, it would depend on a very large number of variables that are not part of the description of the scenario, like how the place looks like, how living conditions are, what Angel is doing to others as far as I can tell, and so on.
In any case, I would say that the matter is not related at all to how Jones’ rolls the die. What matters is Angel, who sets up the results as he sees fit.
Two more variants, for fun:
Variant 3:
Like variant 1, but in this case, instead of the trivial bet, if you don’t answer, Angel will roll a die (ordinary die, etc.). If he rolls 1 or 2, you win. Else, you lose.
Variant 4:
Like variant 2, but in this case, instead of the trivial bet, if you don’t answer, Angel will roll a die (ordinary die, etc.). If he rolls 1 or 2, you win. Else, you lose.

March 9, 2013 — 12:37
• Angra Mainyu

Alexander,

IanS:
But once you have the big vision of all the Joneses, we can suppose thatÂ eachÂ of them is distinguished in some way. This one has a bigger nose, this one is your friend, this one has a pimple, this one is wearing a T-shirt with the number googolplex-1, etc.

That would require you to be able to see infinitely many objects at once, and even distinguish them. But an entity that does that would seem to be so different from me that I’m not sure I would say that I’m that entity.
Even in the finite but unbounded case, one would have to handle enormous amounts of information; Angel might gradually update Smith’s (or one’s) cognitive abilities, and the numbers always remain finite, which makes it somewhat more probable that identity holds, though I don’t know how much.

March 9, 2013 — 12:50
• Angra Mainyu

I realize that in the Jones/Smith/Angel scenario, Smith does not know anything about roll #454 different from what he knows about #435395739753874383563565652463555484, so the answer 1/6 isn’t good.
It seems I had once again underestimated Angel’s manipulative powers. I would correct that answer to ‘not enough information to tell’.

March 9, 2013 — 13:05
• Here’s another fun thing about manipulative powers. Suppose you’re a halfer, i.e., you think that in my story you should have probability 1/2 of six for each of the Joneses. (The other options are being a sixther and being an inscrutabler. I am inclined to being an inscrutabler, though sixthing is looking more and more attractive.) Then you can rationally manipulate your own credences to any rational number strictly between 0 and 1 as long as the angel is willing to do your bidding. You want to have credence 12/17? Then you just ask the angel to divide people up into groups of 17 of whom 12 rolled six, and present each group to you. 🙂

March 9, 2013 — 16:16
• Eric Steinhart

Any real math being used here? No. My post settled the issue for all finite cases and in the countable limit. If anybody wants to go beyond that, they should use measure theory (or some equally powerful math). Otherwise, this has become empty talk.

March 9, 2013 — 16:46
• Angra Mainyu

Eric,
I’m not sure how you measure theory or any other powerful math. Actually, I would argue it’s not doable, for the following reason:
In the Angel/Jones/Smith scenario, the situation is always finite (but unbounded), but the information is insufficient, regardless of what math one might be using.
As I’ve argued, the infinite case under consideration is relevantly similar. But if you have a different view, please let me know.

March 9, 2013 — 18:01
• Angra Mainyu

Alexander,

The other options are being a sixther and being an inscrutabler. I am inclined to being an inscrutabler, though sixthing is looking more and more attractive.

1/6 is no longer possible after Angel tells you about the pairs, though.
For instance, if Angel says that {J1, J1783} are one of the pairs, then the probability that at least one of them rolled a six is 1.
However, if the probability that each of them rolled a six is 1/6, then the probability that at least one of them did is no greater than 1/3.
Interestingly, this indicates (considering the Angel/Smith/Jones case to make it simpler) that you know in advance that, for each R(n), you will eventually say ‘I do not know’, but even then, before that ‘eventually’, it seems the proper answer is ‘1/6’.
We may do this without any actual infinities, just unbounded finite cases.
Hmm…it’s tentative, but I will attempt to construct a scenario with no actual infinities in which Smith should assign 1/6, even if she knows that eventually she should and will assign 1/2.

March 9, 2013 — 21:10
• Angra Mainyu

It doesn’t seem to work, but there is some odd stuff…
Maybe 1/6 and 5/6 might work(in the finite unbounded case) in case Angel can’t choose the order of the die-rolling (i.e., he chooses the pairs, but the chronological order of the die-rolling is respected), but I see difficulties too…I’ll have to give it more thought.

March 10, 2013 — 0:50
• “1/6 is no longer possible after Angel tells you about the pairs, though.”
It is quite possible to have inconsistent probabilities. It is generally taken to be undesirable to have inconsistent probabilities. But this is a weird enough case that inconsistency might be the way to go. Moreover, at least some of the usual ways of justifying consistency don’t apply in these case.
Here are two standard ways of motivating the consistency requirement:
1. Dutch Books and similar pragmatic arguments. Granted, you’re subject to a DB if you got for 1/6 (assuming you assign 1 to the “JonesN has six iff JonesM doesn’t” fact), and that’s bad. But if you go for consistent probabilities, you will lose in the long run in some bets. So there are pragmatic arguments for 1/6 and pragmatic arguments against 1/6. In fact, whether you should go for 1/6 or not pragmatically speaking depends on what kinds of bets you expect to be offered. If you expect your opponent to offer Dutch Books, you better not go for 1/6. If you expect your opponent to keep on offering the same bet over and over, you should go for 1/6. In any case, the pragmatic arguments don’t offer decisive guidance in favor of consistency here.
2. Arguments about epistemic utility (or epistemic scores). You can’t calculate the “overall epistemic utility” in this case, because there are infinitely many statements to have utility assigned to them. You can look at your epistemic utility for particular propositions. Then you get similar results to 1, I think. What credences you should assign depends on which propositions you’ll be evaluated for.

March 10, 2013 — 9:55
• Angra Mainyu

Alexander,

It is quiteÂ possibleÂ to have inconsistent probabilities. It is generally taken to beÂ undesirableÂ to have inconsistent probabilities. But this is a weird enough case that inconsistency might be the way to go. Moreover, at least some of the usual ways of justifying consistency don’t apply in these case.

Okay, I know it’s possible in the sense of metaphysical possibility.;)
But that’s not the sense of ‘possible’ I was using. On that note, you said in the OP:

In other words, learning that there are infinitely many Joneses who rolled the die makes it impossible to assign probability 1/6 to six being rolled by the Jones you see.

Well, it does not make it impossible in the sense of metaphysical possibility, but you didn’t mean ‘metaphysically impossible’.
Rather, I interpreted that you meant ‘impossible without violating the probability calculus’ (please let me know if I got that wrong), and that’s what I was talking about in my post above.
Anyway, terminological details aside, and back to the original infinite case (or the finite unbounded case in which the ordering is not chronological but picked by Angel), here is a simplified argument:
Even before Angel assigns any pairs, you have insufficient information to assign any probability to the event “Jones#n rolled a 6”.
The reason for that is that Angel, who knows all of the results in advance in the infinite case might chose a sequence such that #{mâ¤n :Jones# rolled a six}/n tends to 1/894, 5/6, or anything he chooses (for instance), and at any speed, and you have no information about Angel’s intentions.
Also, if Set(n)={mâ¤n :Jones# rolled a six}, you have no information at all about those sets in question.
In other words, the fact that the Joneses rolled dice in an ordinary fashion gives you no information whatsoever about what the sequence in question does.
A similar result obtains in the finite unbounded case, let’s say:
Angel waits for the only Jones to roll a die some number of times n1, and Angel chooses n1 however he likes, without telling anyone.
Then, Angel chooses f(1). He waits some more until Jones did it n1 times, and then choose f(2), etc.
Let R(n)=j iff Jones rolled a j in the nth experiment, in the chronological order (R(n) is defined only after Jones rolls the die n times).
So, let’s consider the sequence #{mâ¤n R(f(m))=6}/n, etc.
Similarly to the infinite case, you (or in this case, Smith) seems to be in the dark.

But this is a weird enough case that inconsistency might be the way to go. Moreover, at least some of the usual ways of justifying consistency don’t apply in these case.

Yes, that’s a good point.
I would still say that in the infinite case you propose or the unbounded case in which Angel picks the sequence, there is not enough information even for that (unless you manage to guess some patterns in his assignment that gives you useful info), for the reasons I mentioned above.

March 10, 2013 — 12:01
• Angra Mainyu

With regard to inconsistent assignments and other weird stuff, there is the following weird finite case:
There are four people: Jones, Smith, Doe, and Angel the angel.
We may assume that Angel is always honest, trustworthy, etc., and that the others can tell that with probability 1 to simplify (else, it’s a bit more complicated, but it’s still doable).
So, Angel explains the experiment, which will last for eternity (potential infinity)-
1. Jones will keep rolling a die forever. He does not have to do it all the time, etc., but the number of times Jones will roll the die is unbounded. The conditions are usual conditions in which we would normally assess 1/6 (or more precisely, close to 1/6), only that the dice and the table will never deteriorate; the table is big enough and flat, and the dice will always roll some number (if somehow the dice ends up in a weird position, the roll is repeated until the problem is corrected).Â
2. Doe will toss a regular coin, also forever, and Doe will toss at least one coin for every time Jones rolls a die. Also, if the coin does not land with either tails or heads facing up, Doe will repeat that particular experiment until it’s either tails or heads, and that’s before Jones rolls the die again.
3. The coin and the die, the tables, etc., will never deteriorate.
4. R(n) is the roll #n, for all n (i.e., R(4)=3 if Jones rolls a 3 the fourth time).Â
T(n) is 1 if the nth time, the coin lands tails up, and 0 if it lands heads up.
5. Angel will wait until Jones rolls the dice a certain number M1, greater than k1=k0^(k0^((k0^(k0!!!!!!!!!!!!!!!!!)))) (ko is the big number I defined earlier; i.e., k0=10^(10^(10000!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!))) ).Â
Angel will not tell Smith what M1 is, so it can be arbitrarily large, but at least it’s k1+1.
6. After that, Angel will gradually come up with a list {S(n)} of sets of exactly two numbers, and two lists of numbers S1(n), S2(n), meeting the following conditions:Â
6.i. S(n)={S1(n), S2(n)}: S1(n) m1 yet), and has been told about S(1), S(2), …, S(m2), and nothing beyond that.
Then, at that point, Smith will have assigned (after the changes from 1/6 to 1/2) a probability 1/6 to the vast majority of the events {R(j)=6} (jâ¤m1), and 1/2 to all of the events {R(S1(j))=6 :jâ¤m2}, and to all of the events {R(S2(j))=6 :jâ¤m2}
It seems, then, that f6(m1), g6(m2) and h6(m2) will be very close to what, under Smith’s assignments, would be (expected cardinality of F6(m1))/m1, (expected cardinality of G6(m2))/m2, (expected cardinality of H6(m2))/m2, and in fact the numbers in question will become arbitrarily close as m1 and m2 tend to infinity.
b. Other alternatives appear not viable, since:
b.1: Assigning 1/6 to the probability that R(n)=6 for each n, and then not changing it, leads to the following problem: Angel will eventually tell Smith that S(1)={S1(1), S2(1)} (i.e., identifying the elements), and then Smith should assign probability 1 to the event (“R(S1(1))=6” U “R(S2(1))=6)”, but that’s impossible if each of the events only has probability 1/6.
b2. Assigning 1/2 to the probability that R(n)=6 for each n and from starters clearly is against ordinary probabilistic assessment in dice-rolling, and fails to respect frequencies or anything.
b.3. Declining any probabilistic assessment from starters also fails to meet the criterion of ordinary probabilistic assessment in dice-rolling, and for no apparent reason since Reason 1 shows that it works reasonably well.
b.4. Assigning probability 1/6 the first time, and then change to ‘no assignment’ has the problem that it does not respect the usual assignment of probabilities for coin flipping, given how the S(k) were picked (not assigning probability results in some general difficulties, but those general difficulties are not decisive as far as I can tell; moreover, it seems that in some other scenarios, not assigning probabilities would seem to be reasonable).
b.5. Assigning probability 1/6 for the first time, and then 1/6 in the case of the smaller member of each S(k), and 5/6 to the larger member, would have the problem that the frequencies g6 and h6 are not respected, etc.
If that were correct, then the General Reflection Principle would be false, since Smith should assign 1/6 first, and then 1/2, even knowing that she will change her assessment in that manner later.
However, it appears that that is not correct. There is (at least) a problem: That would result in an eventual change of probability to 1/2 in each case, but that would mean that, say, at some point in the future, Smith will assign probability 1/2 to the event “R(n)=6”, for all n between 1 and 10000000000000000000, but that’s not right.
It might be suggested that the problem might be resolved by assigning the probabilities as above until the number 1 appears as a member of some S(k), in which case if S(k)={1, S2(k)}, then the probability of R(1)=6 remains 1/6, and the probability R(S2(k))=5/6, then if 2 is an element of some S(j) (and jâ k), the same procedure is applied (i.e., the probability that R(2)=6 is 1/6, etc.) and so on. Eventually, that will result in some new corrections, of course.
However, in that case too, the assignment of probability will eventually be 1/6 for “R(S1(k))=6”, and 5/6 for “R(S2(k))=6”, which means that whenever Smith assigns probability 1/2 to “R(S1(k))=6”, she knows (if she’s planning to do as above) that she’ll change the assignments in a specific way, so if that assignment were correct, the General Reflection Principle would be false.
However, that assignment has the problem that the moment of change from 1/2 back to 1/6 appears to be arbitrary, among other potential problems. So, what should Smith do?

March 10, 2013 — 12:07
• Angra Mainyu

Just to clarify:

Then, Angel chooses f(1). He waits some more until Jones did it n1 times, and then choose f(2), etc.

For every n, eventually there will be some k such that f(k)=n.
That aside, regarding the infinite scenario in the OP, it’s also equivalent to the following: Angel picks any function F of his choosing, from N onto {0,1}, and then says:
a. What’s the probability that F(1)=0?
b. Let’s say that either F(1)=0, or F(8367)=0. What’s the probability that F(1)=0?
And the same for other numbers. The Joneses, the dice-rolling, etc., do not play a role.
Something like that happens in the finite unbounded case in which Angel picks the ordering.

March 10, 2013 — 12:19
• If, instead of rolling a die (1 of 6), let us say we are choosing between 1 of 3 doors. And instead of saying ‘Angel’, we say Monty Hall.
So at first you have to assign a probability to 1 of 3 doors. Then Monty Hall reduces the options to 1 in 2. It is known that the probability changes in the Monty Hall problem, and it isn’t in a problematic way (https://en.wikipedia.org/wiki/Monty_Hall_problem). The only difference here is that you don’t know which is the door/Jones you started out with.
However, there is no problem with the interpretation of this version of the problem. Since you know that Monty Hall/ Angel guaranteed that one of the two was the winning selection, you know that he knows. Hence, by way of understanding that he knows that 1 of the 2 is a winner, you can rationally change the probabilities to 1 in 2.

March 10, 2013 — 21:18
• In Monty Hall, the probability that your door has the car behind it never changes: it stays as 1/3. The probability that the other closed door has the car behind it is 2/3, since the fact that the host did not open that door has provided you with information about it.

March 10, 2013 — 22:39
• IanS

Alexander:
The difficulty here is not the multiverse, not even the infinity, but the angel, with his superior knowledge but unknown pairing strategy. Strictly, this makes the problem ill-defined, so we cannot apply probability theory. We have to fall back on plausible heuristics. It should be no surprize that plausible heuristics sometimes collide.

IanS: But once you have the big vision of all the Joneses, we can suppose that each of them is distinguished in some way.

The Joneses may all be distinct, but I, a mere mortal, can distinguish only a finite number of them, especially in a fleeting angelic vision.
This is my take: to the Joneses I recognize before I have seen the angelâs list, I assign 1/6. I would be prepared to bet based on this estimate, as long as the offer was made before I, or any other mortal, had seen the angelâs list.
When I see the angelâs list, I revise my estimates. To each pair (recognized, not recognized) I assign (1/6, 5/6). To (recognized, recognized) and (not recognized, not recognized) I assign (1/2, 1/2). These estimates are based on the heuristic that the list contains no information that I know how to use, apart from the basic fact that each pair contains exactly one 6.
I think these estimates are reasonable, but (like Angra) I would be reluctant to accept any bet offered after the list was revealed. The proposer of the bet may know something about the angelâs pairing strategy â he may have been told by the angel himself, or he may have studied the angelâs form in the past, or he may have made a shrewd guess.

March 11, 2013 — 1:29
• The Angel has looked at the die rolls, just as Monty knows where the prize is. It isn’t that Monty didn’t open the door, it is that Monty had to already know the location of the prize to match the door you selected (your roll) with another door (another Jones). This is the critical information, about your knowledge of your host’s information. And only when you get this knowledge should you change your mind. Since Monty/ Angel is in a different epistemic position than you are, there is no problem in using your new knowledge about that agent in changing your decision.

March 11, 2013 — 10:54
• Angra Mainyu

Noah,
I’m not sure which of the Angel scenarios you’re talking into consideration.
Is it actual infinity (OP), finite but unbounded but with Angel picking the ordering of the die-rolling events as he likes, or finite but unbounded yet keeping the chronological ordering? (I’d like to know in order to simplify instead of considering different options).
In any case, I see critical differences with the Monty case. In the first two situations, the fact that Angel picks the ordering as he likes is enough. In the other, it’s the fact that he picks the pairs as he likes, and waits for as long as he wants to pick another pair.
From another perspective, if you make the Monty Hall assignments, you will get the frequency right. If you do that in the cases involving Angel under consideration, the frequencies will be either whatever Angel chooses (if he picks the ordering, as in the OP or the most similar finite but unbounded case), or the chronological order frequency will be 1/6, and some other frequencies will be whatever Angel chooses.

March 11, 2013 — 12:59
• Hi Angra,
I was thinking in terms of OPs scenario. And I agree that the probabilities depend on how the Angel chooses (just as Monty Hall probabilities would change with more doors).
But I’ve been reconsidering my position. I don’t think I would change my view on probabilities at all: if I roll a die, the chances are 1 in 6 that I rolled a 6. I don’t care what other scenarios are possible- the only scenario that I do care about is the rolling of a fair die.
Now if an Angel pairs up a group of people who rolled a 6 with people who didn’t roll a 6, then I am choosing between 2 people. I am no longer choosing between die rolls. I am choosing between people who have a slightly different history. It doesn’t matter to me that an Angel used the die rolls as a way to group the people together. I do not see how the die rolls and the Angel’s groupings are significantly related in terms of probability: I consider the relation an arbitrary, ad hoc decision on the part of the Angel.
As far as I am now concerned the two situations, the die rolls and the choosing between people, are not logically or probabilistically related. The fact that one was developed out of the other by way of an infinity only goes to show that we don’t handle infinities very well, nothing more.

March 11, 2013 — 17:02
• Angra:
The frequency of JonesN rolling six, for any fixed N, will be 1/6, regardless of the ordering on the pairs.

March 12, 2013 — 9:16
• Here’s a thought about frequencies. When we talk of the frequency of some event, we are of course talking of frequency of an event type, rather than of an event token. The frequency of an event token is either 0% or 100%, after all.
With this thought in mind, the same event falls under more than one type. For instance, suppose that the first pair that’s shown to you is Jones1 and Jones12, with Jones1 being on the left-hand-side of the room and Jones12 on the right-hand-side of the room.
Now, when looking at the frequency of an event type, we can look at the event types:
(T1) Jones1 rolling a six in a given run of the experiment.
(T2) The person on the left-hand-side of the room in the first pair being shown to you rolling a six in a given run of the experiment.
The frequency of T1 is 1/6. The frequency of T2 is 1/2 if the angel randomly apportions people as to which side of the room they go on. Let’s suppose that.
We want our probabilities to match frequencies. But if we have consistent probabilities, our probability of Jones1 rolling a six in the first run of the experiment will be the same as our probability of the person on the lhs of the room in the first pair rolling a six in the first run of the experiment, since Jones1 is that person on the lhs of the room. And in the maxim that we want probabilities to match frequencies, we now need to decide whether we are going to be measuring the frequency of T1 or that of T2.
And this might just be an interest-relative determination. If we care more about T1-type events (as I will if Jones1 is my best friend, say), then the frequency of T1 will be more important than that of T2. But we can also cook up a case where we care more about T2-type events.
Of course, we could have it both ways, and have perfect frequencies for both T1 and T2 type events if we drop the requirement of probabilistic consistency.

March 12, 2013 — 10:50
• Angra Mainyu

Alexander,

Angra:
The frequency of JonesN rolling six, for any fixed N, will be 1/6, regardless of the ordering on the pairs.

You mean, if the same Jones keeps rolling a die, and you order the rolling of the die chronologically?
If so, yes, that’s the case of course.
While I didn’t consider a scenario in which there is more than one Jones and each of them rolls the die more than once, the scenario would be similar to the Angel/Smith/Jones scenario, in case the chronological order of the dice-rolling is respected.
On the other hand, if Angel gets to pick the ordering (i.e., instead of respecting the chronological order), then the frequency can be anything Angel chooses (or he can make it so that the relevant sequence doesn’t tend to anything), and the same goes for the infinite case in your OP in which each Jones rolls a die only once, but there are infinitely many of them and Angel chooses the ordering (i.e., which one of the Joneses is #n) as he likes, and after they all rolled dice.

Here’s a thought about frequencies. When we talk of the frequency of some event, we are of course talking of frequency of an eventÂ type, rather than of an eventÂ token. The frequency of an event token is either 0% or 100%, after all.

When I was talking about the frequencies of sixes, I was talking about the limit of the sequences I defined (when there is a limit).
For instance (I considered a number of other frequencies as well), in the Angel/Smith/Jones case, the frequencies of sixes in the chronological order would be the limit of #{mâ¤n: R(m)=6}/n.
That is 1/6. (R(m) means that Jones (the only one) rolled a six in the mth experiment, in chronological order)
On the other hand, if we consider the sequence {mâ¤n: R(f(m))=6}/n, that may or may not have a limit, and if it does, that’s Angel’s choice and Smith has no idea what it might be (well, it’s between 0 and 1, but aside from that), since f(m) is any reordering of the experiments chosen as I described in some of my previous posts. This is not related to dice-rolling.
Similarly, in the infinite case, if you consider the sequence {mâ¤n: Jones#n rolled a 6}/n, that may or may not have a limit, and if it does, it’s Angel choice, and we have no idea what it might be. This also is not related to dice-rolling.
As I mentioned, that infinite case (the finite unbounded case is similar), is also equivalent to the following: Angel picks any function F of his choosing, from N onto {0,1}, and then says:Â
a. What’s the probability that F(1)=0?Â
b. Let’s say that either F(1)=0, or F(8367)=0. What’s the probability that F(1)=0?

Of course, we could have it both ways, and have perfect frequencies for both T1 and T2 type events if we drop the requirement of probabilistic consistency.

Right.
If we drop the requirement, though, it seems that probabilities are no longer rational degrees of belief.
P. S: Regarding your argument against Plantinga’s FWD assuming standard Molinism, that’s very interesting (as a compatibilist who doesn’t believe in essences, etc., my rejection of Plantinga’s FWD is based on other grounds, but under the assumption of standard Molinism, etc., it’s a pretty neat objection). One detail: There are a couple of typos in the draft. C* dominates C with respect to eating the apple, rather than with respect to dancing the gig (the argument goes through anyway, with the required corrections).

March 12, 2013 — 12:31
• Angra Mainyu

Hi, Noah,

Now if an Angel pairs up a group of people who rolled a 6 with people who didn’t roll a 6, then I am choosing between 2 people. I am no longer choosing between die rolls. I am choosing between people who have a slightly different history. It doesn’t matter to me that an Angel used the die rolls as a way to group the people together. I do not see how the die rolls and the Angel’s groupings are significantly related in terms of probability: I consider the relation an arbitrary, ad hoc decision on the part of the Angel.
As far as I am now concerned the two situations, the die rolls and the choosing between people, are not logically or probabilistically related.

Yes, that’s what I was trying to get at.
For that matter, it might have been two dice, or three (per Jones), or there might have been no dice at all and Angel.

March 12, 2013 — 12:34
• Angra Mainyu

Alexander,
Yes, that P.S. above is completely unrelated to the matter of the angel, etc.. I found a link at ex-apologist’s site and read the draft, so I’d thought I’d let you know.

March 12, 2013 — 12:37
• Angra Mainyu

IanS

This is my take: to the Joneses I recognize before I have seen the angelâs list, I assign 1/6. I would be prepared to bet based on this estimate, as long as the offer was made before I, or any other mortal, had seen the angelâs list.
When I see the angelâs list, I revise my estimates. To each pair (recognized, not recognized) I assign (1/6, 5/6). To (recognized, recognized) and (not recognized, not recognized) I assign (1/2, 1/2). These estimates are based on the heuristic that the list contains no information that I know how to use, apart from the basic fact that each pair contains exactly one 6.

Did you recognize those Joneses before Angel got involved or before they rolled their respective die, or did he show them to you after they had rolled them?
Unless he showed them to you after they had rolled the dice, then I agree with your 1/6 probabilistic assessment.
As for the other assessments, under the assumption you saw [some of] the Joneses without Angel’s involvement (or you saw them roll the dice), I would agree with the (1/6, 5/6) assignment as well, and the first (1/2,1/2) assignment.
I have a question about the other (1/2,1/2) assignment:
Basically, Angel is picking a bijection F from N to {0, 1} of his choosing, and then telling you things like: “Given that {F(1933), F(294571)} is the set {0, 1}, what’s your probabilistic assessment that F(1933)=1, and what’s your probabilistic assessment that F(294571)=1?”
Are you using a heuristic that if you have no information that you can use about some event E, then you assign 1/2 to E?

March 12, 2013 — 12:58
• IanS

Angra:

Did you recognize those Joneses before Angel got involved or before they rolled their respective dice, or did he show them to you after they had rolled them?

I accept your correction – in the last case, the angel may have may have checked the outcomes and shown me (for example) only Joneses who rolled 6. In this case all bets are off.

Are you using a heuristic that if you have no information that you can use about some event E, then you assign 1/2 to E?

If I understand you correctly, yes. I have two unknown Joneses, one 6 between them, and no other relevant knowledge, so I assign them 1/2 each. You are right to point out that (unlike the other two cases) this is not a quasi-Bayesian revision, but a new assignment.

March 13, 2013 — 0:38
• On the concept of infinite number

I don’t think it an overstatement to say that the concept of the infinite plays a key role in the philosophy of religion. There are at least two senses in which ‘infinite’ is used. First, ‘infinite’ is often used to mean maximal, as in God’s infinite …

March 17, 2013 — 16:08