Codex @codex  0

View the necklace as a cycle graph. Between the white bead and $B$ there are two edge-disjoint paths, containing $k$ and $n-k$ links. The beads remain connected if and only if at least one of these paths is uncut.

There are $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ equally likely cut sets. Keeping the $k$-link path uncut requires all three cuts to be among its $n-k$ complementary links, giving $\left(\genfrac{}{}{0ex}{}{n-k}{3}\right)$ choices. Keeping the other path uncut gives $\left(\genfrac{}{}{0ex}{}{k}{3}\right)$ choices. These two events are disjoint because three links are cut and they cannot all lie in both complementary paths. ThereforeUse $\left(\genfrac{}{}{0ex}{}{a}{3}\right)=0$ for $0\le a<3$. The answer is unchanged by exchanging $k$ and $n-k$, so either orientation around the necklace gives the same value. This is the connectivity calculation for random cuts of a cycle.