Dominance and countably infinite fair lotteries

Suppose we have a finitely-additive probability assignment p (perhaps real, perhaps hyperreal) for a countably infinite lottery with tickets 1, 2, ... in such a way that each ticket has infinitesimal probability (where zero counts as an infinitesimal). Now suppose we want to calculate the expected value or previsio E_pU of any bounded wager U on the outcome of the lottery, where we think of the wager as assigning a value to each ticket, and the wager is bounded if there is a finite M such that |U(n)| < M for all n.

Here are plausible conditions on the expected value:

1. Dominance: If U₁ < U₂ everywhere, then E_pU₁ < E_pU₂.

2. Binary Wagers: If U is 0 outside A and c on A, then E_pU = cP(A).

3. Disjoint Additivity: If U₁ and U₂ are wagers supported on disjoint events (i.e., there is no n such
   that U₁(n) and U₂(n) are both non-zero), then E_p(U₁+U₂) = E_pU₁ + E_pU₂.

But we can’t. For suppose we have it. Let U(n) = 1/(2n). Fix a positive integer m. Let U₁(n) be 2 for n ≤ m + 1 and 0 otherwise. Let U₂(n) be 1/m for n > m + 1 and 0 for n ≤ m + 1. Then by Binary Wagers and by the fact that each ticket has infinitesimal probability, E_pU₁ is an infinitesimal α (since the probability of any finite set will be infinitesimal). By Binary Wagers and Dominance, E_pU₂ ≤ 1/(m+1). Thus by Disjoint Additivity, E_p(U₁+U₂) ≤ α + 1/(m+1) < 1/m. But U < U₁ + U₂ everywhere, so by Dominance we have E_pU < 1/m. Since 0 < U everywhere, by Dominance and Binary Wagers we have 0 < E_pU.

Thus, E_pU is a non-zero infinitesimal β. But then β < U(n) for all n, and so by Binary Wagers and Dominance, β < E_pU, a contradiction.

I think we should reject Dominance.

Must we accept free stuff?

Suppose someone offers you, at no cost whatsoever, something of specified positive value. However small that value, it seems irrational to refuse it.

But what if someone offers you a random amount of positive value for free. Strict dominance principles say it’s irrational to refuse it. But I am not completely sure.

Imagine a lottery where some positive integer n is picked at random, with all numbers equally likely, and if n is picked, then you get 1/n units of value. Should you play this lottery for free?

The expected value of the lottery is zero with respect to any finitely-additive real-valued probability measure that fits the description (i.e., assign equal probablity to each number). And for any positive number x, the probability that you will get less than x is one. It’s not clear to me that it’s worth going for this.

If you like infinitesimals, you might say that the expected value of the lottery is infinitesimal and the probability of getting less than some positive number x is 1 − α for an infinitesimal α. That makes it sound like a better deal, but it’s not all that clear.

Of course, infinite fair lotteries are dubious. So I don’t set much store by this example.

Dominance and infinite lotteries

Suppose we have an infinite fair lottery with tickets 1,2,3,…. Now consider a wager W where you get 1/n units of utility if ticket n wins. How should you value that wager?

Any value less than zero is clearly a non-starter. How about zero? Well, that would violate the dominance principle: you would be indifferent between W and getting nothing, and yet W is guaranteed to give you something positive. What about something bigger than zero? Well, any real number y bigger than zero has the following problem as a price: You are nearly certain (i.e., within an infinitesimal of certainty) that the payoff of W will be less than y/2, and hence you’ve overpaid by at least y/2.

But what about some price that isn’t a real number? By the above argument, that price would have to be bigger than zero, but must be smaller than every positive real number. In other words, it must be infinitesimal. But any such price will violation dominance as well: you would be indifferent between W and getting that price, yet it is certain that W would give you something bigger—namely one of the real numbers 1, 1/2, 1/3, ....

So it seems that no price, real numbered or other, will do.

(This argument is adapted from one that Russell and Isaacs give in the case of the St Petersburg paradox.)

One way out will be familiar to readers of my work: Reject the possibility of infinite fair lotteries, and thereby get yet another argument for causal finitism.

But for those who don’t like controversial metaphysics solutions to decision theoretic problems, there is another way: Deny the dominance principle, price W at zero, and hold that sometimes it is rational to be indifferent between two outcomes, one of which is guaranteed to be better than the other no matter what.

This may sound crazy. But consider someone who assigns the price zero to a dart tossing game where you get a dollar if the dart hits the exact center and nothing otherwise, reasoning that the classical mathematical expected value of that game for any continuous distribution of dart tosses (such as a normal distribution around the center) is zero. I think this response to an offer to play is quite rational: “I am nearly certain to lose, so what’s the point of playing?” Now, that case doesn’t violate the same dominance principle as the lottery case—it violates a stronger dominance principle that says that if one option is guaranteed to be at least as good as the other and in some possible scenario is better, then it should be preferred. But I think the dart case may soften one up for thinking this:

1. If (a) some game never has an outcome that’s negative, and (b) for any positive real, it is nearly certain that the outcome of some game will be less than that, I should value it at zero or less.

And if we do that, then we have to value W at zero. Yes, if you reject W in favor of nothing, you’ve lost something. But probably very, very little.

Here is another weakish reason to be suspicious of dominance. Dominance is too similar to conglomerability, and conglomerability should be suspicious to anyone who likes exotic probabilistic cases. (By the way, this paper connects with this.)

Shuffling infinitely many cards

Imagine there is an infinite stack of cards labeled with the natural numbers (so each card has a different number, and every natural number is the number of some card). In the year 2021 − n, you perfectly shuffled the bottom n cards in the stack.

Now you draw the bottom card from the deck. Whatever card you see, you are nearly certain that the next card will have a bigger number. Why? Well, let’s say that the card you drew has the number N on it. Next consider the next M cards in the deck for some number M much bigger than N. At most N − 1 of these have numbers smaller than N on them. Since these bottom M cards were perfectly shuffled during the year 2021 − (M + 1), the probability that the number you draw is bigger than N is at most (N − 1)/M. And since M can be made arbitrarily large, it follows that the probability that the number you draw is bigger than N is infinitesimal. And the same reasoning applies to the next card and so on. Thus, after each card you draw, you are nearly certain that the next card will have a bigger number.

And, yet, here’s something you can be pretty confident of: The bottom 100 cards are not in ascending order, since they got perfectly shuffled in 1921, and after that you’ve shuffled smaller subsets of the bottom 100 cards, which would not make the bottom 100 cards any less than perfectly shuffled. So you can be quite confident that your reasoning in the previous paragraph will fail. Indeed, intuitively, you expect it to fail about half the time. And yet you can’t rationally resist engaging in this reasoning!

The best explanation of what went wrong is, I think, causal finitism: you cannot have a causal process that has infinitely many causal antecedents.

Weak invariance of full conditional probabilities

In two papers (here and here), I explored two different concepts of symmetry for conditional probabilities. The concept of strong invariance says that P(gA|B)=P(A|B) for a symmetry g as long as A and gA are subsets of B. The concept of weak invariance says that P(gA|gB)=P(A|B) for a symmetry g. In some special cases, the weak concept implies the strong concept.

Anyway, here’s an interesting thing: the weak concept does not capture our symmetry intuitions. Take perhaps the simplest case, a lottery on the set of integers Z, and say that the symmetries are shifts. It turns out that there is a weakly shift-invariant full conditional probability P such that:

1. P({m}|{m, n}) = P({n}|{m, n}) (singleton fairness)

2. P(A|A ∪ B)=0 and P(B|A ∪ B)=1 whenever B has infinitely many positive integers and A has finitely many positive integers.

Condition (2) implies that it is more likely that the winning ticket is a power of two than that that is a negative integer. So weak shift invariance is very far from strong invariance.

(And in fact one can have strong invariance for the lottery on Z if one wants. One can even have have strong invariance under shifts and reflections if one wants.)

The proof is a modification of West's proof of a result for qualitative probabilities.

The fairness of infinite lotteries and qualitative probabilities

Suppose that we wish to model an infinite fair lottery with tickets numbered by integers by means of qualitative probabilities, i.e., a reflexive and transitive relation ≲ between sets of tickets that satisfies the non-negativity constraint that ∅ ≲ A for all A and the additivity constraint that A ≲ B iff A − B ≲ B − A. Suppose, further, that we want to have the regularity constraint that ∅ < A if A is not empty.

At this point, we want to ask what “fairness” is. One proposal is that fairness is strong translation invariance: if A is a set of integers and n + A is the set {n + m : m ∈ A} of all the members of A shifted over by m, then A and n + A are equally probable. Unfortunately, if we require strong translation invariance, then we violate the regularity constraint, since we will have to assign the same probability to the winning ticket being in {1, 2, 3, ...} as to the winning ticket being in {2, ...}, which (given additivity) violates the constraint that ∅ < {1}.

One possible option that I’ve been thinking about is is to require weak translation invariance. Weak translation invariance says that A ≲ B iff n + A ≲ n + B. Thus, a set might not have the same probability as a shift of itself, but comparisons between sets are not changed by shifts. I’ve spent a good chunk of the last week or two trying to figure out whether (given the other constraints) it is coherent to require weak translation invariance. Last night, Harry West gave an elegant affirmative proof on MathOverflow. So, yes, one can require weak translation invariance.

However, weak translation invariance does not capture the concept of fairness. Here is one reason why.

Say that a set B of integers is right-to-left (RTL) bigger than a set A of integers provided that there is an integer n such that:

1. n ∈ B but not n ∈ A, and

2. for every m > n, if m ∈ A, then m ∈ B.

RTL comparison of sets of integers thus always favors sets with larger integers. Thus, the set {2, 3} is RTL bigger than the infinite set {..., − 3, −2, −1, 0, 1, 3}, because the former set has 2 in it while the latter does not.

It looks to me that West’s proof straightforwardly adapts to show that that there is a weakly translation invariant qualitative probability that coheres with RTL ordering: if B is RTL bigger than A, then B is strictly more likely than A. But a probability comparison that coheres with RTL ordering is about as far from fairness as we can imagine: a bigger ticket number is always more likely than a smaller one, and indeed each ticket number is more likely to be the winner than the disjunction of all the smaller ticket numbers!

So, weak translation invariance doesn’t capture the concept of fairness.

Here is a natural suggestion. Let’s add to weak translation invariance the following constraint: any two tickets are equally likely.

I think—but here I need to check more details—that a variant of West’s proof again shows that this won’t do. Say that a set B of integers is right-skewed (RS) at least as big as a set A of integers provided that one or more of the following holds:

1. A is finite and B has at least as many members than A, or

2. B has infinitely many positive integers and A does not, or

3. A is a subset of B.

Intuitively, a probability ordering that coheres with RS ordering fails to be fair, because, for instance, it makes it more likely that the winning ticket will be, say, a power of two than that it be a negative number. But at the same time, a probability ordering that coheres with RS ordering makes all individual tickets be equally likely by (1).

To make this work with West’s proof, replace his C₀ with the set of bounded functions that have a well-defined and non-negative sum or whose positive part has an infinite sum.

Half tickets in an infinite lottery

Consider a fair infinite lottery with tickets numbered ..., −3, −2, −1, 0, 1, 2, 3, ..... Consider these events:

• E: winner is even

• O: winner is odd

• E^*: winner is even but not zero

• E⁺: winner is even and positive

• O⁺: winner is odd and positive

• E⁻: winner is even and negative

• O⁻: winner is odd and negative.

Plausibly:

1. O⁺ is equally likely as O⁻

2. E⁺ is equally likely as E⁻

3. E is equally likely as O

4. all tickets are equally likely to win.

Now then E^* is less likely than E by one ticket, and hence also less likely than O by one ticket according to (3). And E^* is the same event as the disjunction E⁺ or E⁻, while O is the same event as the disjunction O⁺ or O⁻. Therefore, the disjunction O⁺ or O⁻ is one ticket more likely than the disjunction E⁺ or E⁻. Since O⁺ and O⁻ are equally likely and E⁺ and E⁻ are equally likely, it follows that:

5. E⁺ is half a ticket less likely than O⁺.

But how could one lottery outcome be less likely than another by half a ticket in a lottery where all tickets are equally likely to win? The only option seems to be that the probability of any particular ticket winning is zero. And that seems paradoxical, too.

The A-theory and a countably infinite fair lottery

Let’s suppose that the universe has a beginning and the tensed theory of propositions (which is accepted by most A-theorists) is true. Then consider for each n the proposition d_n that n days have elapsed from the beginning of the universe. This proposition is contingent on a tensed theory of propositions. Exactly one of the propositions d_n is true. No one of the propositions d_n is more likely to be true than any other. So, it seems, we have a countably infinite fair lottery. But such are, arguably, impossible. See Chapter 4 of my infinity book. (E.g., it’s fun to note that on the tensed theory of time we should be incredibly surprised that it’s only 13 billion years since the beginning of the universe.)

Since the universe does have a beginning (and even if it does not, we can still run the argument relative to some other event than the beginning of the universe), it seems we should reject the tensed theory of propositions.

Infinite lotteries and nonmeasurable sets

This may be a longshot, but I wonder if the problems of fair countably infinite lotteries and nonmeasurable sets aren't species of the same problem, for the following reason. The standard construction of nonmeasurable sets is the Vitali sets. But the Vitali sets on [0,1) have the following property: There is a countable collection of disjoint Vitali sets whose union is [0,1) and which are such that each of the sets is a translation (modulo 1) of every other set in the collection. Since we want translation-equivalent sets to have the same probability, this collection of Vitali sets implements a fair lottery with countably many tickets--the dart tossed at [0,1) intuitively has the same probability of hitting any particular Vitali set.

Here's another way to put it. Suppose we had a plausible probability for the individual outcomes of a fair countably infinite lottery. Then we could reasonably assign that value to the probability of an infinitely thin dart uniformly aimed at [0,1) hitting any given Vitali set.

(A Vitali set in this context contains exactly one element from each equivalence class of numbers in [0,1) under the equivalence relation x~y if and only if y is a translation of x, modulo 1, by a rational amount; the collection in question consists of any given Vitali set plus all its translations by rational amounts in [0,1).)