Let $N$ be a natural number. We uniformly choose a random number between $1$ and $N$ (both inclusive) and subtract it from $N$. If we repeat this on the new number obtained, what is the expected number of trials needed to reach $0$?
The crux is that the limits of random distribution change depending on the previous number. How can we handle this?
Edit: Random number in range [1, N] i.e. you have N choices for the random number, each one having the same probability of 1/N (uniform distribution).
Then after subtracting, the resultant number obtained becomes new N. So, N is not constant. N is changing according to the previous random number.
A more practical version: you have N chocolates and you eat some random number (as described above) each day. After how many days will the chocolates get over?
As a follow-up to the discrete case, what if the uniform distribution is continuous?
Let $z(N)$ be your expected number of subtractions. Then $z(1) = 1$ and $$z(n+1) = 1 + {1 \over n+1} \left(0 + z(1) + \dots + z(n)\right)$$ From here, for $n > 1$: \begin{align}z(n+1) &= 1 + {1 \over n+1}\left(z(1) + \dots + z(n)\right) \\ (n+1)z(n+1) &= z(1) + \dots + z(n-1) + z(n) + n+1 \\ (n+1)z(n+1) &= nz(n) + z(n) + 1 \\ z(n+1) &= z(n) + {1 \over n+1} \\ z(n) &= 1 + {1 \over 2} + \dots + {1 \over n} = H_n\end{align} where $H_n$ is $n$-th harmonic number.