, 2021

Solution:

Consider this as a Markov chain on $\mathbb{Z} / 2021 \mathbb{Z}$. This Markov chain is aperiodic (since $0$ can go to $0$) and any number can be reached from any other number (by adding $1$), so it has a unique stationary distribution $\pi$, which is uniform (since the uniform distribution is stationary).

It is a well-known theorem on Markov chains that the expected return time from a state $i$ back to $i$ is equal to the inverse of the probability $\pi_{i}$ of $i$ in the stationary distribution. (One way to see this is to take a length $n \rightarrow \infty$ random walk on this chain, and note that $i$ occurs roughly $\pi_{i}$ of the time.) Since the probability of $0$ is $\frac{1}{2021}$, the expected return time from $0$ to $0$ is $2021$.

After the first step (from $0$), we are at $1$ with probability $1/2$ and $0$ with probability $1/2$, so the number of turns it takes to get from $1$ to $0$ on expectation is $2 \cdot 2021 - 2 = 4040$.