If only one tuple in $S$ has $b\_id=102$ out of a billion total, then between $\sigma_{b\_id=102}(R \bowtie S)$ and $R \bowtie \sigma_{b\_id=102}(S)$, the truth is

Both forms are logically equivalent (commutativity of σ with ⋈ when σ touches only one relation). The cost differs hugely: filter-then-join joins a billion rows down to one row first, then does a tiny join. Join-then-filter joins a billion against the other side before filtering. Same answer, very different cost.