  By now, the Maker-Breaker connectivity game on a complete graph $K_n$ or on a
random graph $G\sim G_{n,p}$ is well studied. Recently, London and Pluh\'ar
suggested a variant in which Maker always needs to choose her edges in such a
way that her graph stays connected. By their results it follows that for this
connected version of the game, the threshold bias on $K_n$ and the threshold
probability on $G\sim G_{n,p}$ for winning the game drastically differ from the
corresponding values for the usual Maker-Breaker version, assuming Maker's bias
to be $1$. However, they observed that the threshold biases of both versions
played on $K_n$ are still of the same order if instead Maker is allowed to
claim two edges in every round. Naturally, this made London and Pluh\'ar ask
whether a similar phenomenon can be observed when a $(2:2)$ game is played on
$G_{n,p}$. We prove that this is not the case, and determine the threshold
probability for winning this game to be of size $n^{-2/3+o(1)}$.
