So we have two biased coins, one comes out head w.p. $1/2+\epsilon$ and the other w.p. $1/2-\epsilon$. How many times should we flip these two coins to be able to tell them apart w.p. at least $\delta$? 
Using the Chernoff bound we know that $\frac{1}{\epsilon^2}\log(1/\delta)$ is enough. And I know that this is also a lower bound but I cannot prove it. By doing coupling I can only show that the lower bound is $\frac{1}{\epsilon}\log(1/\delta)$. Does anybody know the proof?