After discussing the possibility of finding any piece of information encoded somewhere in the digits of $\pi$, one of my year 12 students posed the following problem:

''Assuming the digits of $\pi$ form an infinitely long, essentially random sequence with no recurrence, is there definitely at least an initial repeat, in the sense that the first $n$ digits could repeat once before the rest continue 'randomly'?''

In other words, and more generally, is any irrational certainly in one of the forms $aabcdef...\,$, $ababcdef...\,$, $abcabcdef...\,$ and so on? (If you're worried about independence in subsequent digits then we can easily reformulate the problem to consider an infinite sequence of die rolls or coin flips or whatever.)

My first instinct was 'yes', perhaps because I'm pretty used to saying anything is possible given infinitely long. However, the anchoring at the start makes this a rather different and interesting problem. The probability of such a repeat seems to shrink much faster than the linear rate at which the digits accumulate. So, no then?

Well, it seems to be a bit more complicated than that...

After a few false starts, we came to the following formulation:

$P(2\text{-digit decimal is not of form }aa)=(1-\frac{1}{10})$

$P(4\text{-digit decimal is not of form }abab)=(1-\frac{1}{10^2})$

$P(6\text{-digit decimal is not of form }abcabc)=(1-\frac{1}{10^3})$

$P(2n\text{-digit decimal is not of form }a_1a_2...a_na_1a_2...a_n)=(1-\frac{1}{10^n})$

So, $P(2n\text{-digit decimal is not of the required form anywhere})=(1-\frac{1}{10})(1-\frac{1}{10^2})...(1-\frac{1}{10^n})=\Pi_{r=1}^{n}(1-\frac{1}{10^r})$.

For an irrational decimal then, the probability that it is not in the required form is $$\prod_{r=1}^{\infty}(1-\frac{1}{10^r})$$ I was stuck for a convenient way to calculate this, though, so I did what anyone else would do and stuck it in Wolfram Alpha. Following the link to the entertainingly named q-Pochhammer Symbol, I learned that this is actually an instance of a named function, Euler's phi, $\phi(x)=\prod_{r=1}^{\infty}(1-x^n)$. (Not to be confused with Euler's totient function, which is also a $\varphi$ but a curlier one.)

It turns out $\phi(0.1)=0.890010...$, so the probability that $\pi$ or any other irrational number is of the required form is apparently $0.109990...$. Just under a ninth of irrational numbers have this kind of initial repeat. What we don't know is whether $\pi$ is one of them, right?

Not so fast. $\prod_{r=1}^{100}(1-\frac{1}{10^r})\approx\phi(0.1)$ to a lot of digits and we know $\pi$ to a lot more than $100$ so $\pi$ almost certainly doesn't have such a repeat, but not because it's especially rare. After all, irrationals are uncountable and a ninth of uncountable is uncountably infinite, too.

(By similar reasoning, in any infinite sequence of coin tosses, the probability that the second $n$ results reproduce the first $n$ for at least one $n$ is $1-\phi(0.5)=0.71$ish.)