What is the probability of getting the exact number of expected digits ($0-9$) in $10^6$ digits of $\pi$? I noticed that at $1$ million digits of $\pi$, none of the digits has the "perfect" expected $100{,}000$ occurrences.  My question is what is the probability (if the digits are truly random) of at least one of the digits having the "perfect" # of occurrences (in this case $100{,}000$)?  To be more accurate as one reader pointed out, what is the probability of $1$ million randomly generated digits from $0$ to $9$ having this property?
 A: The problem is equivalent to throwing $N=10^6$ balls into $m=10$ urns (equiprobably) and asking for the probability that at least one urn $i$ has $X_i=N/m$ balls.
This can be approximated (Poissonization) as $m$ iid Poisson variables $Y_i$ with mean $\lambda=E(X_i)=N/m$
The probability that one given urn gets $Y_i=\lambda$ balls is
$$p = \frac{\lambda^\lambda}{\lambda!} e^{-\lambda} \approx \frac{\lambda^\lambda}{(\lambda/e)^\lambda \sqrt{2 \pi \lambda}} e^{-\lambda}= \sqrt{ \frac{m}{2 \pi N}} \approx 0.00126$$
The probabilty that some ball gets $Y_i=\lambda$ is $ 1-(1-p)^m \approx $ (which can be approximated by $m \, p$ -  if you want to).
Then, the desired probability is $0.012544\cdots$
Both approximations (the Poissonization and the Stirling formula) can be refined. Anyway, it's seen that the probability decreases as $1/\sqrt{N}$. Notice, BTW, that this gives the probability of "success" for fixed $N=10^6$ , not for all "tries" $n\le N$ - which would be a more difficult problem.
A: Consider a specific digit ($7$, say). The probability of $k$ $7$'s in $n$ random digits is (see here)
$$\binom{n}{k}\left(\frac{1}{10}\right)^k\left(\frac{9}{10}\right)^{n-k}$$
With $n=1,000,000$ and $k=100,000$, Wolfram Alpha evaluates this as 0.001329806480909191...
This agrees with user187373's approximation $\frac{1}{300\sqrt{2\pi}} \approx 0.001329807601338108...$ to six significant figures, which supports user187373's final probability of $\frac{1}{30\sqrt{2\pi}}$ (about $1$ in $75$).
