Does pi's decimal expansion certainly repeat once (of course not infinitely many times)??

Question

My 8-year-old asked me this and I could see arguments for either answer.

Is pi guaranteed to have an expansion of the form

3.NNM

where N is a finite sequence of digits? For example, 3.14151415696745... (N=1415) or 3.1415926141592696033... (N=1415926). Of course these are not the true value of N.

Argument for:

At any position P in the expansion, The chance of the next P digits being equal to the first P is nonzero. Since there are an infinite number of positions P, eventually we will succeed.

Argument against:

The chance of succeeding at P=1 is 1 in 10. (We fail, as pi does not begin with 3.11). The chance of succeeding at P=2 is 1 in 100. (Fail again: pi does not begin with 3.1414). Etc. If I play a game where I attempt to succeed at P=1, and then at P=2, and repeat infinitely until I succeed, the fact that the chance at each step becomes 10 times less causes the total probability to converge on a number less than 100%.

Which argument is correct, if either? If it's the latter, what is the overall probability that we converge upon?

Answer 1

I think that this question deserves some sort of conclusion. Pretty much everything that makes sense has been covered in the comments, this is just a bit more sorted write-up.

First of all, assume that digits of $\pi$ can be treated as random digits. Which is unproven and there could still be a hidden regularity in the sequence of the digits.

The OP's first argument doesn't really work. The sum of infinitely many numbers (in this case small probabilities) does not need to be infinite. It doesn't even need to be $1$. Whatever misconception is behind this argument, it is certainly reminiscent of Zeno's paradoxes, which makes us think that Achilles would never catch up with the tortoise.

As for the second argument, let's say we already know $k$ digits of $\pi$ after the decimal point (or a few more), and we know that $\pi$ is not representable in the form 3.NNM where $N$ is up to $k$ digits long. For example, apparently we know $10^{14}$ digits of $\pi$ and I will assume that repetition is not observed anywhere in those digits. (Or at least in $k=10^{14}-1$ digits: as per the above article, the $10^{14}$th digit is $0$ so repetition is not happening at that point either.)

The probability that $\pi$ will be of the form 3.NNM is bounded from above by the following sum:

• The probability that the subsequent $k+1$ digits will match the first $k+1$ digits,
• The probability that the subsequent $k+2$ digits will match the first $k+2$ digits,
• The probability that the subsequent $k+3$ digits will match the first $k+3$ digits,
• etc.

I am not saying the probability is equal to the sum above because the above events are not mutually exclusive, but it is certainly less than or equal to the sum above.

Now, the probabilities above are, respectively: $10^{-(k+1)}, 10^{-(k+2)}, 10^{-(k+3)}, \ldots$ - and the sum of those is:

$$\sum_{i=k+1}^\infty 10^{-i}=10^{-k-1}\frac{1}{1-\frac{1}{10}}=\frac{1}{9}10^{-k}$$

(summation of a geometric series). This is an incredibly small number if $k$ is large (in the trillions): it is really:

$$0.\underbrace{000\cdots0}_{k\text{ zeros}}111\cdots$$

and the chance that the repetition happens is, as I said, even smaller than that.

So, the conclusion would be: assuming that digits of $\pi$ can be treated as random digits, and knowing that the repetition does not happen within the first $k$ digits, where $k\approx 10^{14}$, the chances that the repetition will happen later are vanishingly small. (Not zero, but incredibly small indeed.) Note that this sadly does not prove that the repetition does not happen, so with all the discussion above, all we can tell your $8$yr old is that, most likely, there will be no repetition of the form 3.NNM in the digits of $\pi$.

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A funny thing is that, for the number $e\approx 2.718281828459\ldots$ you can see sort-of repetition, not on the first decimal but on the second decimal. However, if we allowed repetitions after an arbitrary decimal, then the whole problem turns on its head and you are bound to find repetitions of any size, somewhere in $\pi$, or in $e$, as long as those are normal numbers. (Which has been conjectured but not proven, by the way!) Say, if you want the same repetition of $1828$ found in $e$ to be found in $\pi$, you would just search for the sequence "$18281828$" in the digits of $\pi$, and you should find one such sequence every $10^8$ digits or so. (If $\pi$ is indeed a normal number.)

Answer 2

Let $X$ be the target Sequence with $N$ Digits.
Let $A$ , $B$ , $C$ be arbitrary Sequences of non-Zero length.

Due to normality of transcendental number $\pi$ & taking the Digits to be some random Sequence ( assumption ! ) , these have a "guarantee" , in terms of Probability :
$3.AXBXC$ ( Due to having Infinite Cases varying over $A$ & $B$ )
$3.XAXB$ ( Due to having Infinite Cases varying over $A$ )

Whether that assumption is true or not , this has no "guarantee" with $\pi$ :
$3.XXA$ (Due to no choice when $X$ is selected)
When some $N$ Digit $X$ will not work , we might take the next Digit to make new $N+1$ Digit $X$ : Probability will then reduce by $10^2$.
Overall Probability is $P=100^{-1}+100^{-2}+100^{-3}+... = 1/99$

We might have that "guarantee" when we consider $\pi$ & $e$ & all transcendental numbers together , where we have the stronger "guarantee" that it always Deterministically Exists without Probability.
Eg these will work :
$0.123456123456 + \pi \times 10^{-13}$
$0.123456789123456789 + e \times 10^{-19}$

There are numbers where it Deterministically will not Exist Eg :
$0.1020020002000020000020000002....$ where the $0$ Sequences will have lengths $1,2,3,4,5,6....$ terminating with $2$ & the Initial $1$ never reoccurs.