Recursive random draw Let $R(n)$ be a random draw of integers between $0$ and $n − 1$ (inclusive). I
repeatedly apply $R$, starting at $10^{100}$. What’s the expected number of repeated applications until I get zero?

Just to fill in the algebra in the solution of BruceZ why $E(n) = E(n-1) + \frac{1}{n}, \:\:n > 1:$
Firstly, we re-arrange $E(n) = 1 + \frac{1}{n}\sum_{k=1}^{n-1}E(k)$ to $\sum_{k=1}^{n-1}E(k) = n \cdot [E(n)-1].$ (*)
We also have: 
$E(n+1) = 1 + \frac{1}{n+1}\sum_{k=1}^{n}E(k) =  1 + \frac{1}{n+1} \sum_{k=1}^{n-1}E(k) + \frac{1}{n+1} E(n)$
After that we plug in (*) to get
$E(n+1) = 1 + \frac{1}{n+1}\sum_{k=1}^{n}E(k) =  1 + \frac{n}{n+1} [E(n)-1] + \frac{1}{n+1} E(n)$
and simplify:
$E(n+1) = 1 + \frac{(n+1) \cdot E(n) - n}{n+1} = 1 + E(n) - \frac{1}{n+1}$ 
to finally get 
$E(n+1) = E(n) + \frac{1}{n+1}$
BTW, $E(n=10^{100}) = 100 \cdot \log_{10}{e} + \gamma = 230.2585 + 0.5772 =  230.8357$
