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My question relates to probabilities on countable infinite sets. For example, what is the probability of choosing an even number from the positive integers. Believe it or not I am interested in this question from a practical standpoint. I am writing a paper on Boltzmann's brains (Brains that occur spontaneously from the vacuum in De Sitter space). Specifically the problem is, is it more likely that I am a Boltzmann brain or a regular brain if space is infinite. There are of course no good calculations about what the odds of a Boltzmann brain identical to mine are vs. the odds of an identical brain evolving on a twin earth (my money would be that a copy on a twin earth is much more likely). But would the specific odds matter if space is infinite and there are an infinite number of both, regardless of which is more common? I know the answer would of course depend on your philosophy of probability (frequency, Bayesian, etc.) so I would be interested in the answer for each of the main theories in probability.

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In conventional countably additive probability, in order to assign a probability to the set of even positive integers, you would need to have a probability assigned to each integer, and then you add up the ones for the even integers.

You can't assign equal probabilities to all positive integers, since then either they're all $0$ and their sum is $0$, whereas the sum should be $1$, or they're all some positive number as the sum is $\infty$, whereas it should be $1$.

However, there is also the idea of letting the probability of each set $A\subseteq\{1,2,3,\ldots\}$ be the "density" of $A$, which is $$ \lim_{n\to\infty} \frac{|A\cap\{1,\ldots,n\}|} n. $$ If you do that, then probabilities are finitely additive but not countably additive, and then not all of the theorems of conventional probability hold.

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    $\begingroup$ Did you mean $A\cap\{1,2,3,\dots,n\}$ there? $\endgroup$ – Akiva Weinberger Dec 13 '15 at 18:24
  • $\begingroup$ Also, not all sets $A$ have a density, right? (For example, let $A$ be the set of numbers with an even number of digits.) $\endgroup$ – Akiva Weinberger Dec 13 '15 at 18:24
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    $\begingroup$ @AkivaWeinberger : Correct. Just as with probability distributions on the real line, where not all sets are measurable. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 13 '15 at 19:04
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    $\begingroup$ @MichaelCerullo : You shouldn't say there are countably infinite numers if you mean there are countably infinitely many numbers. The phrase "countably infinite numbers" properly means "numbers, each one of which, by itself, is countably infinite". If there are six numbers, each of which is countably infinite, then those are countably infinite numbers, but there are not countably infinitely many of them if there are only six. However, you can say a set of numbers is countably infinite without implying that each number in it is countably infinite. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 13 '15 at 19:06
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    $\begingroup$ Let's be clear that the word "density" as used here means something different from "density function" in the sense of probability density. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 13 '15 at 19:08
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I can't tell you anything about Boltzmann's brains, but let me say something about probabilities. For simplicity, let's say that your countably infinite set is $S=\{1,2,3,\ldots\}$.

You need to assign a probability $p_n$ to each positive integer $n\geq 1$, such that $$p_1+p_2+p_3+\cdots = 1.$$

There are several ways of doing this, and the method you choose depends very much on your application. One way of doing it is to choose $p_n = (1/2)^n$. Then $p_1 = 1/2$, $p_2 = 1/4$, and so on, and it is true that $$1/2 + 1/4 + 1/8 + \cdots = 1.$$

For this choice of probabilities, the probability of getting an even number is

$$p_2+p_4+p_6+\cdots = 1/4 + 1/16+1/64+\cdots = 1/3.$$

But this is just one case of choosing the probabilities. One thing that is for certain is that you cannot choose all the $p_n$ to be the same number, but the way you otherwise choose them will depend on your application, and I'm sure you know more about which probabilities are appropriate for the Boltzmann's brains applications than I do.

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  • $\begingroup$ OK, so you can calculate probability on countable infinite sets if first, the sum of all the probabilities equals 1, and if your subset of probabilities converges to finite a value. So in the case of picking an even integer from the positive integers, this can't be done because it isn't possible to assign the same probability to each integer such that the total equals 1? So if it was equally likely that I am one of an infinite number of regular or Boltzmann brains , then we can't apply probability in this case (at least in terms of frequency)? $\endgroup$ – Michael Cerullo Dec 13 '15 at 17:42
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OK, I think I may have an answer based on the response from HowDoIMath, which I posted below their comment, but I want to expand on that possible answer.

My original question brings up a related statistical question about Boltzmann brains. First, let me give a little more background, Boltzmann brains are brains that spontaneously generate from random particles in the vacuum of space. According to some physicists, this is a problem for cosmological models that include infinite space. They worry that in this infinite space, an infinite number of Boltzmann brains would be generated and so (using anthropic reasoning) I (or anyone else) are more likely to be a random Boltzmann brain, and thus all my memories are false so we can't make any inductive arguments in science. When we unpack this statement a little, physicists acknowledge that a twin earth is also possible where an identical brain has evolved (although they assume this is less likely than a Boltzmann brain).

My problem with this whole line of reasoning is that there is no guarantee that even one Boltzmann or Twin earth brain will be generated, let alone infinite numbers of them. To make it easier, lets talk about flipping a coin. The probability is 1 that you will get a billion heads in a row if you flip the coin an infinite number of times, but by "1" here all we can say is "this is almost certain to happen" but we can never say for sure that it will happen. For example, it is possible that we could flip tails an infinite number of times. The probability of this is 0, but we need to interpret this 0 as "almost certainly won't happen" rather than that it won't happen. So with this line of thought the whole Boltzmann brain argument never gets off the ground because they may not even by one of them in an infinite universe. However, is my reasoning correct? Also, if I am reasoning correctly, it is equally likely (almost certain), that a billion heads will be flipped an countably infinite number of times.

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What do you think of the answer of Anubhav Balodhi here ? I have copied/pasted it in the following. He proposes a simple mathematical proof that leads to 0.5 probability to pick an even number among all natural numbers. His answer makes much more common sense than getting a 1/3 probability to pick an even number...

enter image description here

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  • $\begingroup$ What does this prove? All it shows is that as we increase $N$, the probability of an even integer gets closer to $1/2$ (it is exactly $1/2$ if $N$ is even, and close to $1/2$ if $N$ is odd). This says nothing about what happens if $N$ is actually infinite. $\endgroup$ – Bungo Apr 6 '18 at 15:56
  • $\begingroup$ Isn't it enough in order to prove that the probability of picking an even number among all natural numbers is $1/2$ ? $\endgroup$ – Luz Apr 6 '18 at 16:05
  • $\begingroup$ You haven't proven that. In fact, you haven't even defined a probability distribution on the set of all natural numbers; you can't talk about probabilities of subsets until you have done that. $\endgroup$ – Bungo Apr 6 '18 at 16:07
  • $\begingroup$ Don't you think that from his simple proof arises an easy probability distribution for all natural numbers that - this time - leads to a $1/2$ probability of picking an even number ? which is much closer to the common sense... (and therefore can have significant pedagogical value) $\endgroup$ – Luz Apr 6 '18 at 16:12
  • $\begingroup$ You haven't proven anything about the infinite case. In particular, you haven't explained how to define a probability distribution on $\mathbb N$. To do that, you have to be able to answer: given $n \in \mathbb N$, what is the probability of the singleton set $\{n\}$? And as the other answers show, there's no hope of assigning the same probability to each singleton, because these probabilities must sum to $1$. $\endgroup$ – Bungo Apr 6 '18 at 16:16

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