I wanted to know if the digits of $\pi$ are truly uniform, so I attempted a proof of it as a homework question. Can you guys let me know if it is right?
—————-
Let $\Pi$ represent the ordered set of $\pi$'s digits such that \begin{align*} \Pi &= \{\{0, 0, \cdots, 0\}, \{1, 1, \cdots, 1\}, \cdots, \{9, 9, \cdots, 9\}\} \\ &= \{S_0, S_1, \cdots, S_9 \}, \end{align*} where $S_l$ represents the set of digits whose values are all equal to $l$. Next, let $n_{l}$ represent the cardinality of subset $S_{l}$ such that $$n_l = \sum_{i=1}^{|\Pi| = \infty}{\Delta_{li}},$$ where $$\Delta_{li} = \left\{ \begin{array}{rcl} 1 & \mbox{if} \ \mathcal{D}_{i} = l \\ 0 & \mbox{otherwise}\end{array}\right.$$ for the $i^{th}$ digit $\mathcal{D}_{i}$. If $ 0 < p_{l} \leq 1$ represents the fixed probability that any digit $\mathcal{D}$ is equal to $l$, then $$\Delta_{i} \sim \mathrm{Bernouli}\left(p = p_{l}\right),$$ and consequently $$n_{l} \sim \mathrm{Binomial}\left(n = \infty, p = p_{l}\right),$$ which expresses $n_{l}$ as a function of independent and identically distributed events $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{i}, \cdots, \Delta_{\infty}.$ Since \begin{align*} \sum_{i = 1}^{\infty}{p_{l}} &= \infty \times p_{l} \\ &= \infty, \end{align*} it follows from the second Borel-Cantelli lemma that $$\Pr\left(n_{l} = \infty \right) =1$$ as long as $p_{l} > 0$. And since $$ \bigcup_{l=1}^{9}S_{l} = \Pi$$ and $$\Pr\left(n_{0} = n_{1} = \cdots = n_{9}\right) = 1,$$ the probability that $D_{i} = l$ is equal for all $l$. ————————-
I apologize in advance if the proof looks sloppy. I didn’t know it was an open question, I just thought it was from homework. Any thoughts would be highly appreciated!
The premise that $0, \ldots, 9$ even have a definite probability of appearing among the digits of $\pi$ is an open problem, let alone that such probabilities are related to Bernoulli distributions. So there is no proof here, just some heuristic reasoning.
That you don't make an actual connection to the number $\pi$ is another gap, but even without that we can't say that your argument applies to any specific real number. Borel proved in $1909$ that for almost all numbers in $[0,1]$, each base $b$ digit appears with frequency $1/b$, for every base $b \geq 2$. That is, the set of such numbers in $[0,1]$ has measure $1$. Note that part of his proof involves showing the digit frequencies even exist at all; their existence is not an assumption. That a set in $[0,1]$ has measure $1$ does not mean we can easily give examples of numbers in that set: to this day, no specific number that anyone actually cares about has been proved to be normal.