Cut a number to a random integer between 0 and that number. Keep going until that number is 0. How many cuts do we need? Start with an integer like n = 100 and set it equal to a uniformaly random integer between [0,n] inclusive. Keep cutting it this way until n = 0. What's the expected value of the number of cuts needed?
For me, intuition gives an expected value of $\log_2 n ≈ 6.64$, but empirical simulation in Python:
import random
 
cuts = 0
expectedValue = 0
trials = 100000
 
for i in range(trials):
  startingValue = 100
  while startingValue > 0:
    startingValue = random.randint(0, startingValue)
    cuts += 1
 
expectedValue = cuts / trials
print(expectedValue)

results in $≈6.18$.
Does there exist an explicit solution for n = 100 or for any integer n?
 A: This problem is similar to the Frog problem, see the description on cross validated "The Frog Problem (puzzle in YouTube video)" and this variant here.
The short form of the recurrence relationship can alternatively also be seen more directly by
$$e_n = \underbrace{\frac{1}{n+1} (e_n+1)}_{\text{remain in same place}} +  \underbrace{\frac{n}{n+1} (e_{n-1})}_{\text{the other options}}$$
There is $\frac{1}{n+1}$ probability that you stay in place (get the same integer), and $\frac{n}{n+1}$ probability that you advance to the same possibilities as if you would have been in position $n-1$.
This gives $$e_n = e_{n-1} + \frac{1}{n}$$ and along with $e_1 =2$ you get $$e_n = 1 + \sum_{k=1}^n \frac{1}{k}$$
A: If $\ e_n\ $ is the expected number of cuts to reach $\ \{0\}\ $ from $\ \bigcup_\limits{i=0}^n\{i\}\ $, then $\ e_n\ $ satisfies the recursion
\begin{align}
e_0&=0\\
e_n&=1+\sum_{i=0}^n\frac{e_i}{n+1}\ .
\end{align}
It's not difficult${}^\dagger$ to show by induction that the solution of this recursion is given by
$$
e_n=1+\sum_{i=1}^n\frac{1}{i}
$$
for $\ n\ge1\ $. As is well-known, $\ \lim_\limits{n\rightarrow\infty}\left(\sum_\limits{i=1}^n\frac{1}{i}-\ln n\right)=\gamma\ $, where $\ \gamma\approx0.57722\ $ is the Euler-Mascheroni constant, so therefore $\ \lim_\limits{n\rightarrow\infty}\left(e_n-\ln n\right)=\gamma+1\ $, and
$$
e_n\approx1+\gamma+\ln n
$$
for sufficiently large $\ n $. For $\ n=100\ $ this gives
\begin{align}
e_{100}&\approx1.57722+\ln100\\
&\approx6.1824
\end{align}
in good agreement with the result of your python simulation.
${}^\dagger$ Especially if you use the observation made by TheBestMagician in a comment below.
Addendum
According to Wolfram alpha  the exact value of $\ e_{100}\ $ rounded to $20$ significant figures is $$\color{green}{6.1873775176396}\color{red}{202608},$$which agrees with the value obtained by Stinking Bishop to its $13^\text{th}$ decimal place.
