From a tweet: if you write the digits of e = 2.7182... as a concatenation of primes, you get the sequence 2, 7, 182818284...429 (649 digits), 5, 3, 11, ... The third element is huge! Assuming e is a normal number, is the expected length of these numbers infinite?
More generally let $X_1,X_2,X_3,\dots$ be a random sequence of digits in base b. On average, what is the smallest positive integer such that $X_{1:N}$ (the concatenation of the first N digits) is prime?
It is not too hard to directly calculate the first few values in base 10 via counting primes: $$ P(N=1) = P(X_1 \text{ prime}) = \frac{4}{10}$$
$$ \begin{aligned} P(N=2) = P(X_{1:2} \in \{11, 13, 17, 19, 23, 29, 41, 43, 47, 61, 67, 83, 89, 97\})= \frac{14}{100}\end{aligned}$$ $$P(N=3) =\frac{60}{1000}$$
Et cetera. For large $n$, we can write the probability mass as
$$ \begin{aligned} P(N=n) &= P(X_1\text{ not prime}) \\ &\times P(X_{1:2}\text{ not prime} \mid X_{1}\text{ not prime}) \\ & \vdots \\ &\times P(X_{1:(n-1)}\text{ not prime} \mid X_{1:(n-2)}\text{ not prime}) \\ &\times P(X_{1:n}\text{ prime} \mid X_{1:(n-1)}\text{ not prime}) \end{aligned}$$
If we just ignore the conditional assuming the primality of a number and adding a random digit are unrelated and use the prime counting function then $$ P(N=n) \approx \frac{\pi(b^n)}{b^n}\prod_{k=1}^{n-1}\left(1-\frac{\pi(b^k)}{b^k}\right)$$ From the prime number theorem: $$ P(N=n) \approx \frac{1}{n\log b}\prod_{k=1}^{n-1}\left(1-\frac{1}{k \log b}\right)= \frac{1}{n(\log b)^n}\prod_{k=1}^{n-1} \left(\log b - \frac{1}{k}\right) $$ So finally, the expected value is approximately $$ E[N] \approx \sum_{n=1}^{\infty} \frac{1}{(\log b)^n}\prod_{k=1}^{n-1} \left(\log b - \frac{1}{k}\right) $$ which indeed appears to diverge:
Heuristically, the probability of a $d$-digit base-$b$ number being prime is approximately $c/d$ where $c = 1/\ln(b)$, so if you randomly add digits one at a time, the probability that no prime has occurred after $d$ digits have been added is about $$\prod_{j=1}^d (1 - c/j) \sim \frac{1}{\Gamma(1-c) d^c}$$ This does go to $0$ as $d \to \infty$, so you should eventually get a prime, but (unless the base $b = 2$) the expected value of the number of digits required will be infinite because $\sum_d 1/d^c$ diverges.