In the Wikipedia for Hoeffding's Inequality, it says that $\alpha \leq 2e^{-2\varepsilon^2n}$ implies that $n \geq \frac{\log(2/\alpha)}{2\varepsilon^2}$. When I work through the intermediate steps, I get an inequality flip from dividing by a negative:
\begin{equation} \begin{aligned} &\frac{\alpha}{2} \leq e^{-2\varepsilon^2n} \\ \implies &\log(\frac{2}{\alpha}) \geq 2\varepsilon^2n \\ \implies &\frac{\log(2/\alpha)}{2\varepsilon^2} \geq n. \end{aligned} \end{equation} What am I missing?
The phrasing in Wikipedia is a bit misleading. Instead of writing $$ \alpha = P(\bar X \notin [p-\epsilon, p+\epsilon]) $$ what we are actually looking for is a sufficient condition for $$ P(\bar X \notin [p-\epsilon, p+\epsilon]) \leq \alpha .$$ Since the probability on the left is smaller than $2e^{-2\epsilon^2 n}$, it is enough to ensure that $$ 2e^{-2\epsilon^2 n} \leq \alpha . $$ From there you can obtain Wikipedia’s inequality.