QMA error reduction ($ϵ\le (2\sqrt{2}/3{)}^r$)

Starting from completeness $2/3$ and soundness $1/3$, run an odd number $r$ of parallel copies and take majority. The QMA parallel repetition with entangled witnesses argument proves soundness for arbitrary repeated witnesses. Exponential Markov's inequality gives majority error at most $(2\sqrt{2}/3{)}^r$. Hence $O(\log (1/ϵ))$ copies suffice for error $ϵ$, with polynomial overhead when the target error is inverse-polynomial or exponentially small in input length.