One-sided normal approximation for a rejection proposal budget ($n\simeq [\{z\sqrt{M-1}+\sqrt{z^2(M-1)+4MN}\}/2{]}^2$)

For rejection sampling with acceptance probability $1/M$, $M>1$, the accepted count in $n$ trials has binomial distribution, mean $n/M$ and variance $n(M-1)/M^2$. To obtain at least $N$ acceptances with approximate probability $1-\delta$, use the standard normal distribution quantile $z={\Phi }^{-1}(1-\delta )$. The central limit theorem without continuity correction gives $n-MN\ge z\sqrt{n(M-1)}$. Solving this quadratic inequality in $\sqrt{n}$ gives the displayed threshold, rounded upward. It exceeds the average stopping cost $MN$ because it allows for fluctuations in the accepted count. For $M=1$, the count is deterministic and $n=N$ exactly. Exact coverage uses the binomial distribution tail rather than the approximation.