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?
 A: 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, $E(n=10^{100}) = 100 \cdot \log_{10}{e} + \gamma = 230.2585 + 0.5772 =  230.8357$
A: @PT272
great you added this part, but I think there is a little typo. Unfortunately I do not have the permissions to comment or even edit so I put it as a new answer.
After "and simplify" I it should be
$E(n+1) = 1+E(n)-\frac{\color{red}{n}}{n+1}$
A: Let $E(n)$ denote the expected number. Then
$$E(1) = 1$$
$$E(n) = 1 + \frac{1}{n}\sum_{k=1}^{n-1}E(k),\:\: n > 1$$
since we always require at least 1, and with probability $1/n$ we land on one of the other $n-1$ numbers $k = 1$ to $n-1$, at which point we will require an additional $E(k)$.  Inspection of how $E(n+1)$ is produced form $E(n)$ reveals
$$E(n) = E(n-1) + \frac{1}{n}, \:\:n > 1$$
so
$$E(n) = \sum_{k=1}^{n}\frac{1}{k}=H_n, \:\: n > 0$$
or the $n^{th}$ harmonic number.  So
$$E(10^{100}) = H_{10^{100}}$$
This will be approximately $\ln(10^{100})+\gamma$ where $\gamma$ is the  Euler-Mascheroni constant, or $100/\log_{10}{e} + \gamma \approx 230.8.$
