Probability distribution of runs given k bits set in an n-bit circular buffer
Note first that there is no need to randomly select a location in the buffer – we can fix one as we please. From that location, the probability that a run will be at least $t$ long is the probability that the $k$-set includes the first $t$ locations, which is
$$
\Pr[R \geq t] = \frac{\binom{n-t}{k-t}}{\binom{n}{k}}.
$$
Similarly, the probability that a run will be exactly $t$ long is the probability that the $k$-set includes the first $t$ locations but not the one after it, which is
$$
\Pr[R = t] = \frac{\binom{n-t-1}{k-t}}{\binom{n}{k}}.
$$



The expectation of $R$ is
$$
\mathbb{E}[R] = \sum_{t \geq 1} \Pr[R \geq t] = \frac{\sum_{t \geq 1} \binom{n-t}{n-k}}{\binom{n}{k}} = \frac{\binom{n}{n-k+1}}{\binom{n}{k}} = \frac{k}{n-k+1},
$$
using the formula $\sum_{x=r}^m \binom{x}{r} = \binom{m+1}{r+1}$.