Independence conglomerability

Conglomerability says that if you have an event E and a partition {R_i : i ∈ I} of the probability space, then if P(E∣R_i) ≥ λ for all i, we likewise have P(E) ≥ λ. Absence of conglomerability leads to a variety of paradoxes, but in various infinitary contexts, it is necessary to abandon conglomerability.

I want to consider a variant on conglomerability, which I will call independence conglomerability. Suppose we have a collection of events {E_i : i ∈ I}, and suppose that J is a randomly chosen member of I, with J independent of all the E_i taken together. Independence conglomerability requires that if P(E_i) ≥ λ for all i, then P(E_J) ≥ λ, where ω ∈ E_J if and only if ω ∈ E_J(ω) for ω in our underlying probability space Ω.

Independence conglomerability follows from conglomerability if we suppose that P(E_J∣J=i) = P(E_i) for all i.

However, note that independence conglomerability differs from conglomerability in two ways. First, it can make sense to talk of independence conglomerability even in cases where one cannot meaningfully conditionalize on J = i (e.g., because P(J=i) = 0 and we don’t have a way of conditionalizing on zero probability events). Second, and this seems like it could be significant, independence conglomerability seems a little more intuitive. We have a bunch of events, each of which has probability at least λ. We independently randomly choose one of these events. We should expect the probability that our randomly chosen event happens to be at least λ.

Imagine that independence conglomerability fails. Then you can have the following scenario. For each i ∈ I there is a game available for you to play, where you win provided that E_i happens. You get to choose which game to play. Suppose that for each game, the probability of victory is at most λ. But, paradoxically, there is a random way to choose which game to play, independent of the events underlying all the games, where your probability of victory is strictly bigger than λ. (Here I reversed the inequalities defining independence conglomerability, by replacing events with their complements as needed.) Thus you can do better by randomly choosing which game to play than by choosing a specific game to play.

Example: I am going to uniformly randomly choose a positive integer (using a countably infinite fair lottery, assuming for the sake of argument such is possible). For each positive integer n, you have a game available to you: the game is one you win if n is no less than the number I am going to pick. You despair: there is no way for you to have any chance to win, because whatever positive integer n you choose, I am infinitely more likely to get a number bigger than n than a number less than or equal to n, so the chance of you winning is zero or infinitesimal regardless which game you pick. But then you have a brilliant idea. If instead of you choosing a specific number, you independently uniformly choose a positive integer n, the probability of you winning will be at least 1/2 by symmetry. Thus a situation with two independent countably infinite fair lotteries and a symmetry constraint that probabilities don’t change when you swap the lotteries with each other violates independence conglomerability.

Is this violation somehow more problematic than the much discussed violations of plain conglomerability that happen with countably infinite fair lotteries? I don’t know, but maybe it is. There is something particularly odd about the idea that you can noticeably increase your chance of winning by randomly choosing which game to play.

Another simple way to see a problem with infinitesimal probabilities

Suppose I independently randomly and uniformly choose X and Y between 0 and 1, not including 1 but possibly including 0. Now in the diagram above, let the blue event B be that the point (X, Y) lies one one of the two blue line segments, and let the red event R be that it lies on one of the two red line segments. (The red event is the graph of the fractional part of 2x; the blue event is the reflection of this in the line y = x.) As usual, a filled circle indicates a point included and an unfilled circle indicates a point not included; the purple point at (0, 0) is in both the red and blue events.

It seems that B is twice as likely as R. For, given any value of X—see the dotted line in the diagram—there are two possible values of Y that put one in B but only one possible value of X that puts one in R.

But of course the situation is completely symmetric between X and Y, and the above reasoning can be repeated with X and Y swapped to conclude that R is about twice as likely as B.

Hmm.

Of course, there is no paradox in classical probability theory where we just say that the red and blue events have zero probability, and twice zero equals zero.

But if we have any probability theory that distinguishes different events that are classically of zero-probability and says things like “it’s more likely that Y is 0.2 or 0.8 than that Y is 0.2” (say because both events have infinitesimal probability, with one of these infinitesimals being twice as big as the other), then the above reasoning should yield the absurd conclusion that B is more likely than R and R is more likely than B.

Technically, there is nothing new in the above. It just shows that when we have a probability theory that distinguishes classically zero-probability events, that probability theory will fail conglomerability. I.e., we have to reject the reasoning that just because conditionally on any value of X it’s twice as likely that we’re in B as in R, therefore it’s twice as likely that we’re in B as in R. We already knew that conglomerability reasoning had to be rejected in such probability theories. But I think this is a really vivid way of showing the point, as this instance of conglomerability reasoning seems super plausible. And I think the vividness of it makes it clear that the problem doesn’t depend on any kind of weird trickery with strange sets, and that no mere technical tweak (such as moving to qualitative or comparative probabilities) is likely to get us out of it.

Classical probability theory is not enough

Here’s a quick argument that classical probability cannot capture all probabilistic phenomena even if we restrict our attention to phenomena where numbers should be assigned. Consider a nonmeasurable event E, maybe a dart hitting a nonmeasurable subset of the target, and consider a fair coin flip that is causally isolated from E. Let H and T be the heads and tails results of the flip. Then let A be this disjunctive event:

• (E and H) or (not-E and T).

Intuitively, event A clearly has probability 1. If E happens, the probability of A is 1/2 (heads) and if E doesn’t happen, it’s also 1/2 (tails). (The argument uses finite conglomerability, but it is also highly intuitive.)

So a precise number should be assigned to A, namely 1/2. And ditto to H. But we cannot have these assignments in classical probability theory. For if we did that, then we would also have to assign a probability to the conjunction of H and A, which is equivalent to the conjunction of E and H. But we cannot assign a probability to the conjunction of E and H, because E and H are independent, and so we would have a precise probability for E, namely P(E)P(H)/P(H)=P(E&H)/P(H), contrary to the nonmeasurability of E.

Shuffling an infinite deck of cards

Suppose I have an infinitely deep deck of cards, numbered with the positive integers. Can I shuffle it?

Given an infinite past, here is a procedure: n days ago, I perfectly fairly shuffle the top n cards in the deck.

When one reshuffles a portion of an already perfectly shuffled finite deck of cards, the full deck remains perfectly shuffled. So, the top n cards in the infinitely deep deck are perfectly shuffled for every finite n.

Can we argue that the thus-shuffled deck generates a countably infinite fair lottery, i.e., that if we pick cards off the top of the deck, all card numbers will be equally likely? At the moment I don’t know how to argue for that. But I can say that we get what I have called a countably infinite paradoxical lottery, i.e., one when any particular outcome has zero or infinitesimal probability.

For simplicity, let’s just consider picking the top card off the deck and consider a particular card number, say 100. For card 100 to be at the top of the deck, it had to be in the top n cards prior to the shuffling on day −n for each n. For instance, on day −1000, it had to to be in the top 1000 cards prior to the shuffling. The subsequent 1000 shufflings together perfectly shuffle the top 1000 cards. Thus, the probability that card 100 would end up at the top is 1/1000, given that it was in the top 1000 cards on day −1000. But it may not have been. So, all in all, the probability that card 100 would end up at the top is at most 1/1000. But the argument generalizes: for any n, the probability that card 100 would end up at the top is at most 1/n. Hence, the probability that card 100 would end up at the top is zero or infinitesimal.

If taking an infinite amount of time to shuffle is too boring, you can also do this with a supertask: one minute ago you shuffle the top card, 1.5 minutes ago you shuffle the top two cards, 1.75 minutes ago you shuffled the top three cards, and so on. Then you did the whole process in two minutes.

All the paradoxes of fair countably infinite lotteries reappear for any paradoxical countably infinite lottery. So, the above simple procedure is guaranteed to generate lots of fun paradoxes.

Here is a fun one. Carl shuffles the infinite deck. He now offers to pay Alice and Bob $20 each to play this game: they each take a card off the top of the deck, and the one with the smaller number has to pay $100 to the one with the bigger number. Alice and Bob happily agree to play the game. After all, they know the top two cards of the deck are perfectly shuffled, so they think it’s equally likely that each will win, and hence each calculates their expected payoff at 0.5×$100 − 0.5×$100 + $20 = $20. He puts them in separate rooms. As soon as each sees their own card (but not the other's), he now offers a new deal to them: if they each agree to pay him $80, he’ll broker a deal letting them swap their cards before determining who is the winner. Alice sees her card, and knows there are only finitely many cards with a smaller number, so she estimates her probability of being a winner at zero or infinitesimal. So she is nearly sure that if she doesn’t swap, she’ll be out $100, and hence it’s obviously worth swapping, even if it costs $80 to swap. Bob reasons the same way. So they each pay Carl $80 to swap. As a result, Carl makes $80+$80−$20−$20=$120 in each round of the game.

Causal Finitism, of course, says that you can’t have an infinite causal history, so you can’t have done the infinite number of shufflings.

Another dice game for infinitely many people

Consider:

• Case 1: There are two countably infinite sets, A and B, of strangers. The people are all alike in morally relevant ways. I get to choose which set of people gets a lifetime supply of healthy and tasty food.

Clearly, it doesn't matter how I choose. And if someone offers me a cookie to choose set A, no harm in taking it and choosing set A, it seems.

Next:

• Case 2: Countably infinitely many strangers have each rolled a die, whose outcome I do not see. Set S6 is the set of people who rolled a six and set S12345 is the set of people who rolled something other than a six. The people are all alike in morally relevant ways. I get to choose which set of people gets a lifetime supply of healthy and tasty food.

Almost surely, S6 and S12345 are two countably infinite sets. So it seems like this is just like Case 1. It makes no difference. And if you offer me a cookie to choose S6 to be the winners, no harm done if I take it.

But now suppose I focus in on one particular person, say you. If I choose S6, you have a 1/6 chance of getting a significant good. If I choose S12345, you have a 5/6 chance. Clearly, just thinking about you alone, I should disregard any cookie offered to me and go for S12345. But the same goes when I focus on anybody else. So it seems that Case 2 differs from Case 1. If Case 1 is the whole story--i.e., if there is no backstory about how the two sets are chosen--then it really doesn't matter what I choose. But in Case 2, it does. The backstory matters, because when I focus in on one individual, I am choosing what that individual's chance of a good is.

But now finally:

• Case 3: Just like Case 2, except that you get to see who rolled what number, and hence you know which people are in which set.

In this case, I can't mentally focus in on one individual and figure out what is to be done. For if I focus in on someone who rolled six, I am inclined to choose S6 and if I focus in on someone who rolled non-six, I am inclined to choose S12345, and the numbers of people in both sets are equal. So I don't know what to do in this case?

Maybe, though, even in Case 3, I should go for S12345. For maybe instead of deciding on the basis of the particular situation, I should decide on the basis of the right rule. And a rule of favoring the non-six rollers in circumstances like this is better for everyone as a rule, because every individual will have had a better chance at the good outcome then?

Or maybe we just shouldn't worry about the case where you see all the dice, because that's an impossible case according to causal finitism? Interestingly, though Cases 1 and 2 only require an infinite future, something that's clearly possible.

Non-conglomerability

This result is probably known, and probably not optimal. A conditional probability function P is conglomerable provided that for any partition {H_i} (perhaps infinite and maybe even uncountable) of the state space if P(A|H_i)≥r for all i, then P(A)≥r.

Theorem. Assume the Axiom of Choice. Suppose P is a full conditional probability function (i.e., Popper function) P on an uncountable space such that:

1. all singletons are measurable
2. the function satisfies this regularity condition for all elements x and y: P({x}|{x,y})>0
3. there is a partition of the probability space into two disjoint subsets A and B with the same cardinality such that P(A)>0 and P(B)>0

Then P is not conglomerable.

Conditions (2) and (3) are going to be intuitively satisfied for plausible continuous probabilities, like uniform and Gaussian ones. So in those cases there is no hope for a conglomerable conditional probability.

Sketch of proof: Let Q be a hyperreal-valued unconditional probability corresponding to P, so that P(X|Y)=Q(X∩Y)/Q(Y). The regularity condition (2) implies that that there is a hyperreal α such that Q(F)/α is finite, non-zero and non-infinitesimal for each finite set F. (Just let α=Q({x_0}) for any fixed x₀.) Let R(F) be the standard part of Q(F)/α for any finite set α. Then P(F|G)=R(F∩G)/R(G) for any finite sets F and G. Moreover, R is finitely additive and non-zero on every singleton.

Since A² has the same cardinality as A, there is a function f from B to the subsets of A with the property that f(b) and f(c) are disjoint if b and c are distinct and every f(b) is uncountable. Choose a finite number c such that P(A)<c/(1+c). For each b in B, choose a finite subset F_b of f(b) such that R(F_b)>cR({b}). Such a finite subset exists as R is finitely additive and the sum of an uncountable number of non-zero positive numbers is always infinity. Let H be the union of the F_b as b ranges over B. Then A−H has at most the cardinality of B. Let h be a one-to-one function from A−H to B. For each b in B, let G_b=F_b if there is no a in A−H such that h(a)=b; otherwise, let G_b=F_b∪{a} for such an a. Let H_b={b}∪G_b. Then R(G_b)>cR({b}) and so R(G_b)/R(H_b)>c/(1+c). Then P(A|H_b)=P(G_b|H_b)=R(G_b)/R(H_b)>c/(1+c). But the H_b are a partition of our probability space, and P(A)<c/(1+c), so we have a violation of conglomerability.