Two Birds Side by Side

By symmetry, the set of two positions occupied by the blue birds is uniformly distributed over the $\binom{8}{2}=28$ unordered pairs of positions among $1,\dots,8$. The blue birds are adjacent exactly when this pair is $\{i,i+1\}$ for some $i\in\{1,\dots,7\}$, which gives $7$ favourable pairs. Hence

Alternative. Condition on the position of the first blue bird. If it is at an end (probability $\tfrac{2}{8}$), exactly one of the $7$ remaining spots is adjacent to it. Otherwise (probability $\tfrac{6}{8}$), two of the $7$ are. The probability is therefore $\tfrac{2}{8}\cdot\tfrac17+\tfrac68\cdot\tfrac27=\tfrac{14}{56}=\tfrac14$.

Generalisation. With $n$ spots, the probability is $(n-1)/\binom{n}{2}=2/n$.