A double lottery and non-normalized probabilities

Suppose a positive integer N is generated by a fair lottery.

Then, a random integer K is chosen between 1 and N (inclusive).

What information does this give you about N?

Obviously you now know that N ≥ K. Anything else?

Consider some specific pair of numbers n ≥ k, and suppose we’ve found out that K = k. What’s the probability that N = n? Of course P(N=n|K=k) = 0/0. But what if we do this as a limiting procedure. Suppose first that N is randomly chosen between 1 and M where M ≥ n, and let P_M be the probabilities for this case. Then

• P_M(N=n|K=k) = (1/M)(1/n)/[(1/M)Σ_j=k^Mj⁻¹] = (1/n)/Σ_j=k^Mj⁻¹.

Take the limit as M goes to infinity. Since Σ_n=k^∞j⁻¹ = ∞, the limit is zero, so we don’t have a meaningful distribution for N.

On the other hand, what if we independently choose two random integers K₁ and K₂ between 1 and N? Suppose n ≥ k_i for i = 1, 2. Let k^* = max (k₁,k₂). Then:

• PM(N=n|K1=k1,K2=k2) = (1/M)(1/n2)/[(1/M)Σj=k*Mj−2] = (1/n2)/Σj=k*Mj−2.

Take the limit as M → ∞ and call that P(N=n|K₁=k₁,K₂=k₂). The limit behaves like ck^*/n², for a constant c > 0, and generates a well-defined probability for N = n.

With zero samples, we don’t have a well-defined probability for N. With one sample, we still don’t. But with two samples (or more), now we do. This is a rummy thing: how is it that sampling turns probabilistic nonsense into sense?

This is making me more friendly to using non-normalized probabilities. After all, the fair lottery for N is easily modeled by the constant probability p₀(n) = 1. With one sample N = k, we have p₁(n) = 1/n for n ≥ k and p₁(n) = 0 for n < k. With two samples k₁, k₂, we have p₂(n) = 1/n² for n ≥ max (k₁,k₂) and p₂(n) otherwise. All this makes perfect sense. And there is a lovely mathematical feature of non-normalized probabilities: conditionalization is conjunction. The conditional probability of an event A on event B is just the probability of A ∩ B.

Non-normalized probabilities aren’t going to solve all problems with infinite fair lotteries. For instance, I toss a fair coin and generate a number N with the following rule. On heads, I choose N with my fair lottery on the positive integers. On tails, I choose N such that the probability of N = n is 2⁻ⁿ (e.g., I toss an independent fair coin and let N be the number of the first toss that gives heads). What’s my non-normalized probability p(x,n), where x is heads or tails and n is a positive integer? We surely want ∑_np(H,n) = ∑_np(T,n): the total probability of the heads options equals the total probability of the tails options. But clearly p(T,n) has to exponentially decrease so ∑_np(T,n) is finite and non-zero. On the other hand, p(H,n) is constant, so ∑_np(H,n) is zero or infinity. So they can’t be equal.

But I wonder if one could say something like this: Non-normalized probabilities make sense in certain cases, and in those cases it’s reasonable to use them?

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.

Probability for truly fair infinite lotteries

Long ago, in correspondence with Plantinga and me, Peter van Inwagen suggested that the only way to model a countably infinite fair lottery is by assigning probability zero to every finite set of tickets, probability one to every co-finite set of tickets (a subset A of a set B is co-finite [relative to B] provided that the set of members of B that are not in A is finite), and an undefined probability to every other subset.

Van Inwagen’s proposal has been growing on me. In a truly fair lottery, the ordering of ticket numbers is irrelevant. Therefore, if Ω is the set of tickets and π is any permutation of Ω, the probability of A should be the same as that of πA, with each defined if and only if the other is. In other words the probability function should be permutation-invariant.

Proposition. There are only two finitely-additive real-valued probabilities invariant under all permutations of a countably infinite set Ω: the trivial probability that assigns 0 to the empty set, 1 to Ω and is undefined for all other subsets, and van Inwagen’s probability that assigns 0 to every finite set, 1 to every co-finite set and is undefined for all other subsets.

There is something very appealing about van Inwagen’s proposal: it’s the only finitely-additive real-valued probability that really captures the idea of a countably infinite fair lottery. I can't remember if van Inwagen had the above proposition in the correspondence, but he might have.

Proof of Proposition: For any two subsets X and Y that are neither finite nor co-finite, there is a permutation of Ω mapping X onto Y. Thus, by permutation invariance, either all sets that are neither finite nor co-finite have a probability or none do. Suppose first that all do. In that case, they all have equal probability. Let A be the evens, B be the odds, C the numbers equal to 0 modulo 4 and D the numbers equal to 2 modulo 4. They all must have equal probability. But A is the disjoint union of C and D, so by finite additivity, if all three have equal probability, all three must have probability zero. And so does B. Thus, P(Ω) = P(A) + P(B) = 0, a contradiction.

So, only sets that are neither finite nor co-finite have a probability. If the only subsets that have a probability are ⌀ and Ω, we are done. Suppose some other subset has a probability. If that subset is co-finite, its complement will have to have a probability too, so in either case there is a finite non-empty subset A that has a probability. Let A′ be a finite non-empty subset that has the same cardinality as A but intersects A in only one element. By permutation invariance, A′ has a probability. Thus, so does the intersection of A and A′. Hence, at least one singleton has a probability. Hence by permutation invariance all singletons have a probability. By finite additivity, that probability must be zero. It follows that all finite sets have probability, and that probability is zero, and all co-finite sets have probability, and that probability is one.

Remark 1: Suppose that we allow the probabilities to take values in some non-Archimedean ordered field. Then there are more possibilities. Specifically, for any positive infinitesimal α, we can define a probability that assigns to every finite set the probability nα where n is the set’s cardinality and to every co-finite set the probability 1 − nα where n is the cardinality of the set’s complement. And these are the only extra possibilities.

Remark 2: If we drop the countability condition on Ω, and assume the Axiom of Choice, then in the setting of the Proposition we can prove that P(A) is 0 or 1 for every subset A for which P(A) is defined.

Label independence and lotteries

Suppose we have a countably infinite fair lottery, in John Norton’s sense of label independence: in other words, probabilities are not changed by any relabeling—i.e., any permutation—of tickets. In classical probability, it’s easy to generate a contradiction from the above assumptions, given the simple assumption that there is at least one set A of tickets that has a well-defined probability (i.e., that the probability that the winning ticket is from A is well-defined) and that has the property that both A and its complement are infinite. John Norton rejects classical probability in such cases, however.

So, here’s an interesting question: How weak are the probability theory assumptions we need to generate a contradiction from a label independent countably infinite lottery? Here is a collection that works:

1. The tickets are numbered with the set N of natural numbers.

2. If A and B are easily describable subsets of the tickets that differ by an easily describable permutation of N, then they are equally probable.

3. For every easily describable set A of tickets, either A or its complement is (or both are) more likely than the empty set.

4. If A and B are disjoint and each is more likely than the empty set, then A ∪ B is more likely than A or is more likely than B.

5. Being at least as likely as is reflexive and transitive.

Here, Axioms 3 and 4 are my rather weak replacement for finite additivity (together with an implicit assumption that easily describable sets have a well-defined probability). Axiom 2 is a weak version of label independence, restricted to easily describable relabeling. Axiom 5 is obvious, and the noteworthy thing is that totality is not assumed.

What do I mean by “easily describable”? I shall assume that sets are “easily describable” provided that they can be described by a modulo 4 condition: i.e., by saying what value(s) the members of the set have to have modulo 4 (e.g., “the evens”, “the odds” and “the evens not divisible by four” are all “easily describable”). And I shall assume that a permutation of N is “easily describable” provided that it can be described by giving a formula f_i(x) using integer addition, subtraction, multiplication and division for i = 0, 1, 2, 3 that specifies what happens to an input x that is equal to i modulo 4. (E.g., the permutation that swaps the evens and the odds is given by the formulas f₂(0)=f₀(x)=x + 1 and f₃(x)=f₁(x)=x − 1.)

Proof: Let A be the set of even numbers. By (3), A or N − A is more likely than the empty set. But A and N − A differ by an easily describable permutation (swap the evens with the odds). So, by (2) they are equally likely. So they are both more likely than the empty set. Let B be the subset of A consisting of the even numbers divisible by four and let C = A − B be the even numbers not divisible by 4. Then B and C differ by an easily describable permutation (leave the odd numbers unchanged; add two to the evens divisible by four; subtract two from the evens not divisible by four). Moreover, A and B differ by an easily (but less easily!) describable permutation. (Exercise!) So, A, B and C are all equally likely by (2). So they are all more likely than the empty set. So, A = B ∪ C is more likely than either B or C (or both) by (4). But this contradicts the fact that A is equally likely as B and C.

Uncountably infinite fair lottery

A fair lottery is going to be held. There are uncountably infinitely many players and the prize is infinitely good. Specifically, countably infinitely many fair coins will be tossed, and corresponding to each infinite sequence of heads and tails there is a ticket that exactly one person has bought.

Along comes Truthful Alice. She offers you a deal: she’ll take your ticket and give you two tickets. Of course, you go for the deal since it doubles your chances of winning, and Alice gives the same deal to everyone else, and everyone else goes for it. Alice then has everyone’s tickets. She now proceeds as follows. If you had a ticket with the sequence X₁X₂X₃..., she gives you the tickets HHHHHX₁X₂X₃... and HHHHTX₁X₂X₃.... And she keeps for herself all the tickets that start with something other than HHHH.

So, everyone has gone for the deal, and Alice has a 15/16 chance of winning (since that’s the chance that the coin sequence won’t start with HHHH). That’s paradoxical!

This paradox suggests that there may be something wrong with the concept of a fair infinite lottery even when the number of tickets is uncountable.

Here is one way to soften the paradox. If you reason with classical probability theory, without any infinitesimals, you will agree that the deal offered you by Alice doubles your chances of winning, but you will also note that the chance it doubles is zero, and doubling zero is no increase. So if you reason with classical probability theory, you will be indifferent to Alice’s deal. There is still something strange in thinking that Alice is able to likely enrich herself at the expense of a bunch of people doing something they are rationally indifferent about. But it’s less surprising than if she can do so at the expense of people doing what they rationally ought.

There is another thought which I find myself attracted to. The very concept of a fair lottery breaks down in infinite cases. If the lottery were fair, exchanging a ticket for two tickets would be a good deal. But the lottery isn’t fair, because there are no infinite fair lotteries.

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.

Yet another infinite hat-guessing story

Suppose first a countably infinite line of blindfolded people standing on tiles numbered 0,1,2,…, with the ones on a tile whose number is divisible by 10 having a red hat, and the others having blue hats. Suppose you’re in the line, with no idea where, but apprised of the above. It seems you should reasonably think: “Probably my hat is blue.”

But then the blindfolded people are shuffled, without any changes of hats, so that now it is the tiles with numbers divisible by 10 that have the blue hatters and the others have the red hatters. Such mere shuffling shouldn’t change what you think. So after being informed of the shuffle, it seems you should still think: “Probably my hat is blue.” It is already puzzling, though, why the first arrangement defined the probabilities and not the second. (What does temporal order have to do with these probabilities?)

Now suppose you gather the nine people after you (in the tile order—even though you are blindfolded, I suppose you can tell which direction the tile number numbers increase) along with yourself into a group of ten. In any group of ten successive people on the line, there is exactly one blue hat and nine red hats. Yet each of the ten of you thinks: “Probably my hat is blue.” And by a reasonable closure, you each also think: “Probably the other nine all have red hats.” You talk about it. You argue about it. “No, I am probably the one with the blue hat!” “No, my hat is probably the blue one.” “No, you’re probably both wrong: It’s probably mine.” I submit there is no rational room for any resolution to the disagreement, and indeed no budging of probabilities, no matter how much you pool your data, no matter how completely you recognize your epistemic peerhood, no matter how you apply exactly the same reasonable principles of reasoning. For nothing you learn from the other people is evidentially relevant. This is paradoxical.

Yet another infinite lottery machine

In a number of posts over the past several years, I’ve explored various ways to make a countably infinite fair lottery machine (assuming causal finitism is false), typically using supertasks in some way.

Here’s another, slightly simplified from a construction in Norton. Suppose we toss a countably infinite number of fair coins to make an array with infinitely many infinite rows that could look like this:

HTHTHHHHHHHTTT...
THTHTHTHTHHHHH...
HHHHHTHTHTHTHT...
...

Make sure that nobody looks at the coins after they are tossed. Here’s something that could happen: each row of the array contains one and only one tails. This is unlikely (probability zero; Norton originally said it's nonmeasurable, but that was a mistake, and we're coauthoring a correction to his paper) but possible. Have a robot scan the array—a supertask will be needed—to verify whether this unlikely event has happened. If not, we have failed to make the machine. But if yes, our array will look relevantly like:

HHTHHHHHHHHHHH...
HHHHHTHHHHHHHH...
HHTHHHHHHHHHHH...
...

Continue making sure nobody looks at the coins. Put a robot at the beginning of the first row. Now, you have an countably infinite fair lottery machine that you can use over and over. To use it, just tell the robot to scan the row it’s at, announce the position of the lone tails, and move to the beginning of the next row. Applied to the above array, you will get the sequence of results 3,6,3,….

Of course, it’s very unlikely that we will succeed in making the machine (the probability is zero). But we might. And once we do, we can run as many paradoxes of infinity as we like. And we might even find ourselves lucky enough to be in a universe where some natural random process has already generated such a lucky array, in which case we don’t even have to flip the coins.

Once we have the machine, we can have lots of fun with it. For instance, it seems antecedently really unlikely that the first hundred times you run the machine, the numbers you get will be in increasing order. But no matter how many numbers you've pulled from the machine, you are all but certain that the next number will be bigger than any of them.

From a certain A-theory of time to a countably infinite fair lottery

Suppose:

1. The past has to be finite.
2. The future has to be infinite.
3. The A-theory of time is true.

Then contingent reality appears to generate a countably infinite fair lottery: Simply let N be the number of days since the beginning of time, rounded down. Surely no one day is more likely to be objectively present than another, so N is the outcome of a fair lottery with tickets numbered 0,1,2,.... But such lotteries are well known to lead to many paradoxes (e..g, see chapter 4 of Infinity, Causation and Paradox). Thus, one shouldn't hold all of (1)-(3).

Not every A-theorist has this problem: only those who accept (1) and (2) as well.

Countably infinite fair lotteries and sorting

There is nothing essential new here, but it is a particularly vivid way to put an observation by Paul Bartha.

You are going to receive a sequence of a hundred tickets from an countably infinite fair lottery. When you get the first ticket, you will be nearly certain (your probability will be 1 or 1 minus an infinitesimal) that the next ticket will have a bigger number. When you get the second, you will be nearly certain that the third will be bigger than it. And so on. Thus, throughout the sequence you will be nearly certain that the next ticket will be bigger.

But surely at some point you will be wrong. After all, it's incredibly unlikely that a hundred tickets from a lottery will be sorted in ascending order. To make the point clear, suppose that the way the sequence of tickets is picked is as follows. First, a hundred tickets are picked via a countably infinite fair lottery, either the same lottery, in which case they are guaranteed to be different, or independent lotteries, in which case they are nearly certain to be all different. Then the hundred tickets are shuffled, and you're given them one by one. Nonetheless, the above argument is unaffected by the shuffling: at each point you will be nearly certain that the next ticket you get will have a bigger number, there being only finitely many options for that to fail and infinitely many for it to succeed, and with all the options being equally likely.

Yet if you take a hundred numbers and shuffle them, it's extremely unlikely that they will be in ascending order. So you will be nearly certain of something, and yet very likely wrong in a number of the cases. And even while you are nearly certain of it, you will be able to go through this argument, see that in many of the judgments that the next number is bigger you will be wrong, and yet this won't affect your near certainty that the next number is bigger.

From a past-infinite causal sequence to a paradoxical lottery: A cosmological argument

Infinite fair lotteries are well-known to be paradoxical. Let's say that an infinite fair lottery is played twice with tickets 1,2,3,.... Then whatever number wins first, you can be all but perhaps certain that in the next run of the lottery a bigger number will win (since the probability of any particular number winning is zero or infinitesimal, so the probability that the winner is a member of the finite set of numbers smaller than or equal to the first picked number is zero or infinitesimal). So as you keep on playing, you can be completely confident that the next number picked will be bigger than the one you just picked. But intuitively that's not what's going to happen. Or consider this neat paradox. Given the infinite fair lottery, there is a way to change the lottery that makes each ticket infinitely more likely to win. Just run a lottery where the probability of ticket n is 2^-n (which is infinitely bigger than the zero or infinitesimal probability in the paradoxical lottery)

What makes the infinite fair lottery paradoxical is that

1. there is a countable infinity of tickets

and

2. each ticket has zero or infinitesimal chance of winning.

Let's stipulate that a lottery is "paradoxical" if and only if it satisfies (1) and (2).

Suppose now that a past-infinite causal sequence is possible (e.g., my being caused by my parents, their being caused by theirs, and so on ad infinitum). Then the following past-infinite causal sequence is surely possible as well. There is a machine that has always been on an infinite line with positions marked with integers: ...,-3,-2,-1,0,1,2,3,.... Each day, the machine has tossed a fair coin. If the coin was heads, it moved one position to the right on the line (e.g., from 2 to 3) and if it was tails, one position to the left (e.g., from 0 to -1). The machine moved in no other way.

We can think of today's position of the machine as picking out a ticket from a countably infinite lottery. Moreover, this countably infinite lottery is paradoxical. It satisfies (1) by stipulation. And it's not hard to argue that it satisfies (2), because of how random walks thin out probability distributions. (And all we need is finite additivity for the argument.)

So if past-infinite causal sequences are possible, paradoxical lotteries are as well. But paradoxical lotteries are not possible, I say. So past-infinite causal sequences are not possible. So there is an uncaused cause.

Another absurdity about infinite fair lotteries

It is easy to generate a method for choosing a natural number in such a way that the probability of choosing n is 2⁻ⁿ. For instance, toss a fair coin and let n be the number of the first toss on which you get heads. Thus, n=1 if you get heads on your first toss, n=2 if you get tails on the first and heads on the second, n=3 if you get tails on the first two and heads on the third, and so on. (And if you never get heads, count that as just another way of choosing 1—after all, the probability of never getting heads is zero.)

Could one have a way of choosing a natural number, guaranteed to return some natural number, but such that the probability of choosing n is between 4⁻ⁿ and 3⁻ⁿ? Surely not! That would be absurd: the probabilities would be too small.

But now suppose you can have an infinite fair lottery. Follow the following procedure. Toss a fair coin until you get heads. If it took an even number m of tosses to get to heads, let your chosen number be n=m/2. If it took an odd number of tosses to get to heads, or if you never got to heads, then choose the natural number n via an infinite fair lottery.

What's the probability of getting n in this new process? Well, the probability of getting heads for the first time on the 2nth toss is 2⁻²ⁿ=4⁻ⁿ. But if you didn't get heads for the first time on an even-numbered toss, you still have a chance of getting n, namely via the infinite fair lottery. The latter chance is either zero or infinitesimal. So, the probability of getting n is 4⁻ⁿ or 4⁻ⁿ plus an infinitesimal. But it's absurd to have a random choice of natural number where the probability of getting n is 4⁻ⁿ. And if it's 4⁻ⁿ plus an infinitesimal, then it's between 4⁻ⁿ and 3⁻ⁿ, which we agreed was absurd.

So infinite fair lotteries lead to absurdity.