In balanced allocations, the goal is to place $m$ balls into $n$ bins, so as to minimize the gap (difference of max to average load). The One-Choice process places each ball to a bin sampled independently and uniformly at random. The Two-Choice process places balls in the least loaded of two sampled bins. Finally, the $(1+β)$-process mixes these processes, meaning each ball is allocated using Two-Choice with probability $β\in(0,1)$, and using One-Choice otherwise. Despite Two-Choice being optimal in the sequential setting, it has been observed in practice that it does not perform well in a parallel environment, where load information may be outdated. Following [BCEFN12], we study such a parallel setting where balls are allocated in batches of size $b$, and balls within the same batch are allocated with the same strategy and based on the same load information. For small batch sizes $b\in[n,n\log n]$, it was shown in [LS22a] that Two-Choice achieves an asymptotically optimal gap among all processes with a constant number of samples. In this work, we focus on larger batch sizes $b\in[n\log n,n^3]$. It was proved in [LS22c] that Two-Choice leads to a gap of $Θ(b/n)$. As our main result, we prove that the gap reduces to $O(\sqrt{(b/n)\cdot\log n})$, if one runs the $(1+β)$-process with an appropriately chosen $β$ (in fact this result holds for a larger class of processes). This not only proves the phenomenon that Two-Choice is not the best (leading to the formation of "towers" over previously light bins), but also that mixing two processes (One-Choice and Two-Choice) leads to a process which achieves a gap that is asymptotically smaller than both. We also derive a matching lower bound of $Ω(\sqrt{(b/n)\cdot\log n})$ for any allocation process, which demonstrates that the above $(1+β)$-process is asymptotically optimal.