Shuffling an infinite deck

Suppose infinitely many blindfolded people, including yourself, are uniformly randomly arranged on positions one meter apart numbered 1, 2, 3, 4, ….

Intuition: The probability that you’re on an even-numbered position is 1/2 and that you’re on a position divisible by four is 1/4.

But then, while asleep, the people are rearranged according to the following rule. The people on each even-numbered position 2n are moved to position 4n. The people on the odd numbered positions are then shifted leftward as needed to fill up the positions not divisible by 4. Thus, we have the following movements:

• 1 → 1

• 2 → 4

• 3 → 2

• 4 → 8

• 5 → 3

• 6 → 12

• 7 → 5

• 8 → 16

• 9 → 6

• and so on.

If the initial intuition was correct, then the probability that now you’re on a position that’s divisible by four is 1/2, since you’re now on a position divisible by four if and only if initially you were on a position divisible by two. Thus it seems that now people are no longer uniformly randomly arranged, since for a uniform arrangement you’d expect your probability of being in a position divisible by four to be 1/4.

This shows an interesting difference between shuffling a finite and an infinite deck of cards. If you shuffle a finite deck of cards that’s already uniformly distributed, it remains uniformly distributed no matter what algorithm you use to shuffle it, as long as you do so in a content-agnostic way (i.e., you don’t look at the faces of the cards). But if you shuffle an infinite deck of distinct cards that’s uniformly distributed in a content-agnostic way, you can destroy the uniform distribution, for instance by doubling the probability that a specific card is in a position divisible by four.

I am inclined to take this as evidence that the whole concept of a “uniformly shuffled” infinite deck of cards is confused.

Does my existence by itself confirm a multiverse?

Suppose I am considering two hypotheses, H₁ and H₂, and according to H₂ there are more people. Does the fact that I exist give me reason to prefer H₂, all other things being equal? If so, then my existence is apt to confirm the existence of a multiverse over a single universe.

Here is one reason to think this works. The probability that I exist in a given world, all other things being equal, seems proportional to the number of people in that world. Each person in that world corresponds to another opportunity for me to exist.

While this is tempting, here is a toy model that should give us pause. Suppose that I am defined by a real number parameter between 0 (inclusive) and 1 (not inclusive). According to hypothesis H₁, a single real number is picked uniformly at random in the range, and the person with that parameter is created. According to hypothesis H₂, two real numbers are picked uniformly and independently in the range, and persons corresponding to these are created. Learning that a person with my parameter is created seems to provide me with evidence for H₂, since it’s twice as likely on H₂ as on H₁.

But this is tricky. In classical probability theory, it is correct to say that my parameter is twice as likely to be generated on H₂ as on H₁, but that’s only because both probabilities are zero, and zero is twice zero, so while H₂ is twice as likely as H₁, it is also true that H₁ is twice as likely as H₂!

Perhaps, though, we want to depart from classical probability theory in some way, say by allowing non-zero infinitesimal probabilities or by an intuitive handwavy “this is twice as likely as that”. However, it is then no longer clear that on H₂ there is twice as big a chance of hitting my parameter. For there are (infinitely) many ways of picking a number between 0 and 1 uniformly randomly.

Here’s one way:

1. You write down “0.”, then roll a fair ten-sided die infinitely many times, writing down the results as the digits after the decimal point, thereby generating a decimal representation of a number. If the number ends with infinitely many nines, try again.

(The final proviso is to ensure that intuitively each number is equally likely. Without that proviso, 1/10 would be more likely than 1/3, as there would be two ways of getting 1/10, namely 0.1000... and 0.0999..., but only one way to get 1/3, namely 0.3333.....)

Here is another way:

2. You write down “0.”, then roll a fair ten-sided die infinitely many times, omitting the results of the first die throw, but writing down the results as the digits after the decimal point, thereby generating a decimal representation of a number. If the number ends with infinitely many nines, try again.

Intuitively, method B has ten times the probability of generating any given number than method A has, as long as literally the numerically same die throws occur in the two cases. For consider the number 1/3 = 0.3333.... By method A to generate it you need every die to show a three. By method B, to generate 1/3, all you need is for all the die throws other than the first one to be threes, and so there are ten times as many ways to generate the number.

Now, if the single selection of a parameter on H₁ uses method B while the double selection of a parameter on H₂ uses method A, then intuitively we are five times as likely to generate my parameter on H₁ than on H₂. Thus merely saying that on both hypotheses the parameters are generated uniformly is insufficient to determine how the comparison between the probabilities of generating my parameter goes.

We might insist that in both hypotheses the same method for generating parameters is used. But notice that in cosmological applications, this is implausible. If H₂ is some multiverse hypothesis and H₁ is a single universe hypothesis, we are unlikely to be able to count on the two hypotheses involving even the same laws of nature, much less the same selection process for the parameters of the persons. (Besides all this, it is really unclear what it even counts to say that there are two different runs of method A.)

So, here’s what I am thinking. On classical probability theory, there is no difference in the probability of my parameter getting generated on H₂ than on H₁, because both probabilities are zero. On non-classical probability theory, we can perhaps make sense of a difference between the probabilities, but cannot count on the hypothesis with more people being more likely to generate my parameter.

Given all this, there does not seem to be a way of making sense of comparing the evidential impact of my existence on the two hypotheses using probabilistic methods. Maybe all we have is intuition.

Domination and uniform spinners

About a decade ago, I offered a counterexample to the following domination principle:

1. Given two wagers A and B, if in every state B is at least as good as A and in at least one state B is better than A, then one should choose B over A.

But perhaps (1) is not so compelling anyway. For it might be that it’s reasonable to completely ignore zero probability outcomes. If a uniform spinner is spun, and on A you get a dollar as long as the spinner doesn’t land at 90^∘ and on B you get a dollar no matter what, then (1) requires you to go for B, but it doesn’t seem crazy to say “It’s almost surely not going to land at 90^∘, so I’ll be indifferent between A and B.”

But now consider the following domination principle:

2. Given two wagers A and B, if in every state B is better than A, then one should choose B over A.

This seems way more reasonable. But here is a potential counterexaple. Consider a spinner which uniformly selects a point on the circumference of a circle. Assume x is any irrational number. Consider a function u such that u(z) is a real number for any z on the circumference of the circle. Imagine two wagers:

• A: After the spinner is spun and lands at z, you get u(z) units of utility

• B: After the spinner is spun, the spinner is moved exactly x degrees counterclockwise to yield a new landing point z′, and you get u(z′) units of utility.

Intuitively, it seems absurd to think that B could be preferable to A. But it turns out that given the Axiom of Choice, we can define a function u such that:

3. For any z on the circumference of the circle, if z′ is the result of rotating z by x degrees counterclockwise around the circle, then u(z′) > u(z).

And then if we take the states to be the initial landing points of the spinner, B always pays strictly better than A, and so by the domination principle (2), we should (seemingly absurdly) choose B.

Remarks:

• The proof of the existence of u requires the Axiom of Choice for collections of countable sets of reals). In my Infinity book, I argued that this version of the Axiom of Choice is true. However, arguments similar to those in the book’s Axiom of Choice chapter suggest that the causal finitist has a good way out of the paradox by denying the implementability of the function u.

• Some people don’t like unbounded utilities. But we can make sure that u is bounded if we want (if the original function u is not bounded, then replace u(z) by arctan u(z)).

• Of course the function u is Lebesgue non-measurable. To see this, replacing u by its arctangent if necessary, we may assume u is bounded. If u were measurable and bounded, it would be integrable, and its Lebesgue integral around the circle would be rotation invariant, which is impossible given (3).

It remains to prove the existence of u. Let ∼ be the relation for points on the (circumference of the circle) defined by z ∼ z′ if the angle between z and z′ is an integer multiple of x degrees. This is an equivalence relation, and hence it partitions the circle into equivalence classes. Let A be a choice set that contains exactly one element from each of the equivalence classes. For any z on the circle, let z₀ be the point in A such that z₀ ∼ z. Let u(z) be the (unique!) integer n such that rotating z₀ counterclockwise around the circle by an angle of nx degrees yields z. Then for any z, if z′ is the result of rotating z′ by x degrees around the circle, then u(z′) = u(z) + 1 > u(z) and so we have (3).

Independence, uniformity and infinitesimals

Suppose that a random variable X is uniformly distributed (in some intuitive sense) over some space. Then :

1. P(X = y)=P(X = z) for any y and z in that space.

But I think something stronger should also be true:

2. Let Y and Z be any random variables taking values in the same space as X, and suppose each variable is independent of X. Then P(X = Y)=P(X = Z).

Fixed constants are independent of X, so (1) follows from (2).

But if we have (2), and the plausible assumption:

3. If X and Y are independent, then X and f(Y) are independent for any function f,

we cannot have infinitesimal probabilities. Here’s why. Suppose X and Y are independent random variables uniformly distributed over the interval [0, 1). Assume P(X = a) is infinitesimal for a in [0, 1). Then, so is P(X = Y).

Let f(x)=2x for x < 1/2 and f(x)=2x − 1 for 1/2 ≤ x. Then if X and Y are independent, so are X and f(Y). Thus:

4. P(X = Y)=P(X = f(Y)).

Let g(x)=x/2 and let h(x)=(1 + x)/2. Then:

5. P(Y = g(X)) = P(Y = X)

and

6. P(Y = h(X)) = P(Y = X).

But now notice that:

7. Y = g(X) if and only if X = f(Y) and Y < 1/2

and

8. Y = h(X) if and only if X = f(Y) and 1/2 ≤ Y.

Thus:

9. (Y = g(X) or Y = h(X)) if and only if X = f(Y)

and note that we cannot have both Y = g(X) and Y = h(X). Hence:

10. P(X = Y)=P(X = f(Y)) = P(Y = g(X)) + P(Y = h(X)) = P(Y = X)+P(Y = X)=2P(X = Y).

Therefore:

11. P(X = Y)=0,

which contradicts the infinitesimality of P(X = Y).

This argument works for any uniform distribution on an infinite set U. Just let A and B be a partition of U into two subsets of the same cardinality as U (this uses the Axiom of Choice). Let g be a bijection from U onto A and h a bijection from U onto B. Let f(x)=g⁻¹(x) for x ∈ A and f(x)=h⁻¹(x) for x ∈ B.

Note: We may wish to restrict (3) to intuitively “nice” functions, ones that don’t introduce non-measurability. The functions in the initial argument are “nice”.

More on comparing zero probability sets

There are two devices, A and B, each of which generates an independent uniformly distributed number between 0 and 1. You have a choice between two games.

Game 1: You win if B generates the number 1/2.

Game 2: You win if B generates the number generated by A.

Perhaps you say: "I don't care. I have infinitesimal or zero probability of winning." To make you care, suppose the game is free but the payoff is infinite.

Intuitively, you're equally likely to win either game. There is no reason to choose one over the other, given that the two devices are independent. It's no easier or harder to get 1/2 than to get whatever number A generates.

But there is another way of seeing the situation. Graph the state space of the game with the x-coordinate corresponding to A and the y-coordinate corresponding to B. The state space is then the unit square. But on Game 1, the victory region is a horizontal line y=1/2 while on Game 2, it is the diagonal line y=x. But the diagonal line has the square root of two, approximately 1.414, as its length, while the horizontal line has unit length. So you should choose Game 2.

Really?  I still think it makes no difference.  (And that it makes no difference shows that rotation invariance does not apply to all cases of uniform distribution in the square, since the horizontal line when rotated becomes obviously shorter than the diagonal one.)