$G(3,2)=\frac{1}{2}$.
For any set $F$ composed of 3-element sets, let's assume there's a sequence of choices $a_1,b_1,\dots a_k,b_k$ where B gains the upper hand and has two elements chosen out of more than half the sets in $F$, despite A playing optimally. If B can force such a sequence, then A can force the sequence $b_1,?,b_2,?\dots,b_k$ for itself, leaving it with more than half the sets in $F$, despite B playing optimally. Therefore, B can never gain the upper hand at any move in $G(3,2)$, so it can have at most half the sets with two of its chosen elements in them.