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NB: This is a question and so there's not much context I can give.

It is claimed here that $\pi$ is almost surely normal; that is, the probability that $\pi$ is normal is one.

I haven't found a reference for this anywhere. If it is true, please would you provide one?

Context:

I know that whether $\pi$ is normal is unknown. It was in looking this up that I encountered the claim in question.

I'm just a group theory PhD student, so, besides the obvious search terms and places to look, I don't know what to look for or where, or whether what I have found is credible.

Please help :)

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    $\begingroup$ $\pi$ either is normal or it isn't; strictly speaking, probability doesn't work into it. The claim that $\pi$ is "almost surely" normal is probably based on the fact that almost all numbers are normal (in fact, almost all numbers are absolutely normal; that is, normal in all integer bases $\geq 2$). $\endgroup$
    – Brian Tung
    May 28, 2023 at 19:22
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    $\begingroup$ @Shaun A number is only a "probable prime" with respect to a particular test. It makes no sense to say that a number is a "probable prime" outside of that context unless it is provably prime. (In other words, we have Lucas pseudoprimes and Fermat pseudoprimes and pseudoprimes associated with any other test, but we don't have a set of just plain "pseudoprimes") $\endgroup$ May 28, 2023 at 19:58
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    $\begingroup$ The term "probable prime" simply means that the number passes a test which very vew composites pass , so we can safely assume that the number is prime without actually having a proof. We have nothing similar for the normality of $\pi$. If we would know , say , that every digit string of length $2$ appears in $\pi$ infinite many often , this would be a start. But in fact , we do not even know whether infinite many sevens appear. $\endgroup$
    – Peter
    May 31, 2023 at 14:47
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    $\begingroup$ Even worse , $\pi$ could have a "digit-tail" consisting only of zeros and ones. This would be very far from normal and still we cannot rule it out. $\endgroup$
    – Peter
    May 31, 2023 at 14:49
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    $\begingroup$ What the statement means : We do not know how exactly the digits in $\pi$ are distributed. And since this is unknown it makes sense to estimate the probability of the nomality of $\pi$. The normality is however an extremely strong property although a random chosen real number has probability $1$ to have it. So, I disagree with the statement that $\pi$ is "almost surely normal" in this sense. $\endgroup$
    – Peter
    May 31, 2023 at 15:06

3 Answers 3

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Any number randomly chosen from a continuous distribution is a.s. normal. If you took a variable uniformly distributed from 0 to 1, then each of its digits are uniformly distributed, so any sequence appears with the same probability in any position. Adding it up over all such positions and applying the Law of Large Numbers gives that any sequence in base $b$ of length $n$ appears with density $1/b^n$ a.s. There are a countable number of such sequences, so it’s a.s. true for all sequences, so the number is a.s. normal. Since any real number is normal if its fractional part is, any random real number is a.s. normal.

$\pi$ like pretty much every non-artifically constructed irrational number isn’t obviously non-normal, so it’s probably normal from the above heuristic, but this isn’t really a mathematical fact. It’s hard (impossible?) to rigorously assign probabilities to mathematical facts without entirely proving or disproving them.

It’s possible to apply heuristics to estimate probabilities but these are mostly just being hand wavey with mathematics and aren’t really rigorous. Some examples are that the googolth digit of $\pi$ is 0 with probability $1/10$ or that the probability $p$ is prime is $1/\log(p)$. Neither of these are really true, but we can use hand-wavey metrics like this for practical purposes to build intuition about what is likely true or false. I have seen some suggestions for using models of computationally bounded agents to assign probabilities to theorems in a way that can’t be arbitraged by similar agents but nothing that’s practically applicable.

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  • $\begingroup$ The only evidence of the normality of $\pi$ are the awfully many digits that have been computed. They exhibit exactly the normal property. But there is no reason to believe that this is still true for the first $10^{100}$ or $10^{1000}$ digits. $\endgroup$
    – Peter
    May 31, 2023 at 14:55
  • $\begingroup$ Err, yes? I don’t think I said anything implying differently. $\endgroup$
    – Eric
    May 31, 2023 at 16:35
  • $\begingroup$ Though I personally believe it’s true that $\pi$ is normal dispute the lack of rigorous proof based on the a.s. proof and numerical evidence. $\endgroup$
    – Eric
    May 31, 2023 at 16:41
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I have never seen any statement formalizing this, but have read quite a lot of times the slogan "$\pi$ is almost surely normal". I’m almost sure (joke intended) that whenever this sentence is written, it is just a fancy way to say that with respect to normality, $\pi$ doesn’t look very different from the other reals, which are themselves almost all normal. So it would be shocking to learn that $\pi$ is not normal. Of course, it may well be that some day in the future, a proof that $\pi$ isn't normal will be found, and that this proof, after a lot of work, will become simple enough so that people will think that it is obvious that $\pi$ isn't normal. But today, we are forced to say that $\pi$ isn't "obviously nonnormal"!

More precisely, for a property $P$ on a set $X$, and some element $x$ of $X$, some people find it nice to say "$x$ almost surely has property $P$" when what they really want to say is "$X$ carries a natural probability measure $\mu$, and $P$ is $\mu$-almost surely true; moreover, I don't see any good reason to think that $x$ is very different from the elements in the support of $\mu$".

This happens for $X := [0,4]$, $P := \mbox{"being normal"}$, $x := \pi$, as I just said. This also happens for $X := \mathcal{P}(\mathbb{N}^*)$, $P := \mbox{"contains infinitely many twins"}$, $x := \mathcal{P} = \{p \in \mathbb{N}^* \ \vert \ p \mbox{ is prime}\}$ and $\mu$ is $\otimes_{n \in \mathbb{N}^*} Bernoulli\left(\frac{1}{\ln n}\right)$ (this measure $\mu$ is called Cramér's model). In this case, I have read sentences like « the twin prime number conjecture is almost surely true ». This just meant that, given this specific model of random subsets of the integers that are constrained to behave, in a density fashion, like the prime numbers, one can show that almost all such sets have infinitely many twins. See for example the section about the heuristic here.

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    $\begingroup$ $\pi$ is a very special number , and in particular not a random number since it is computable. I would rather believe that $\pi$ does not even contain every finite digit string. So, the argument "almost all reals are normal" for the normality of $\pi$ is not at all on par with the evidence of the truth of the twin prime conjecture. $\endgroup$
    – Peter
    May 31, 2023 at 14:42
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    $\begingroup$ I never said $\pi$ was random... And I did not talk at all about evidence of the truth of the twin prime conjecture... I think you mistunderstood my answer, and I edited to make it clearer. $\endgroup$
    – Plop
    May 31, 2023 at 15:05
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This hasn't been stated as clearly as I would like, so to be very clear about this: this usage of the term "almost surely" is technically incorrect, and somewhere between informal language and a joke.

In mathematics "$X$ is almost surely $P$" means that $X$ is a random variable and that the probability that the output of $X$ has property $P$ is equal to $1$. However, $\pi$ isn't a random variable so it doesn't make formal mathematical sense to talk about the probability that it has any property. It either has that property or it doesn't.

$\pi$ is conjectured to be normal, because nobody has a compelling reason to believe that its digits behave like anything other than a random sequence of digits, and a random sequence of digits is almost surely normal (in the technical sense; the probability that a random number chosen uniformly at random in any closed interval, say $[0, 1]$, is normal is $1$). So someone might say that $\pi$ is "almost surely" normal as somewhere between a reference to this technical fact, and a jokey way to say it is "very likely" that the conjecture is true, or just as a use of informal language without noticing that it conflicts with more formal language here.

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