Can BPP be bounded around any constant other than 1/2?
If $c &lt; 1/2$ then for any problem there is an algorithm that answers correctly with probability at least $c+1/n$, say. For small $n$, the algorithm just outputs the hardwired correct answer. When $n$ is large enough so that $c + 1/n \leq 1/2$, the algorithm just tosses a coin.

What went wrong? For BPP amplification to work, we need a gap between the promise for a Yes instance and the promise for a No instance. In your definition of "weak BPP" (not a standard term), you are given an algorithm such that:


On a Yes instance, accepts with probability at least $s(n) + 1/p(n)$.
On a No instance, accepts with probability at most $s(n)$.


In contrast, in your suggested definition, we have the following promise:


On a Yes instance, accepts with probability at least $c + 1/p(n)$.
On a No instance, accepts with probability at most $1 - c - 1/p(n)$.


If $c &lt; 1/2$ then for large enough $n$, the two acceptance intervals overlap (at acceptance probability $1/2$), and so the definition is meaningless. When $c = 1/2$, there is an inverse polynomial gap, which can be amplified to a constant gap, to match the standard definition of BPP.