Assume $X$ has a symmetric $\mathrm{Beta}(\alpha,\alpha)$ distribution, where $\alpha$ is large.
Can we give strong tail bounds (similar to Chernoff bounds for the binomial distribution), i.e., bounds of the form $$ \Pr\Bigl[\bigl|X-E[X]\bigr| \ge t \Bigr] \le \text{some small function of $t$ and $\alpha$?} $$
Of course, one can use Chebychev's inequality, but intuition says, stronger bounds should be obtainable. Also numerical experiments suggest that a bound of the form $e^{-C\cdot \alpha}$ holds.
Attempts: It is known that for $\alpha=m\in\mathbb N$, $X$ can be written as $$ X = \frac{E_1 + \cdots+ E_m}{E_1+\cdots+E_{2m}} $$ where $E_i$ are i.i.d. $\mathrm{Exp}(1)$ distributed; so $X$ can be written as a sum of variables, but they are not independent.
Since $t^{\alpha}(1-t)^{\alpha}$ is symmetric around $t=\frac{1}{2}$, the mean of a $\text{Beta}(\alpha,\alpha)$ random variable is always $\frac{1}{2}$, and accurate approximations for your probability just come from accurate approximations of the density function around $t=\frac{1}{2}$. Assuming $X\sim\text{Beta}(\alpha,\alpha)$, $$\begin{eqnarray*}\mathbb{P}\left[\left|X-\frac{1}{2}\right|\geq t\right]&=&\frac{2\,\Gamma(2\alpha)}{\Gamma(\alpha)^2}\int_{t}^{1/2}\left(\frac{1}{4}-u^2\right)^{\alpha-1}\,du\\&=&\frac{\Gamma(2\alpha)(1-4t^2)^{\alpha-1}}{4^{\alpha-1}\Gamma(\alpha)^2}\int_{2t}^{1}\left(\frac{1-u^2}{1-4t^2}\right)^{\alpha-1}\,du\end{eqnarray*} $$ and the last integrand function can be approximated by $\exp\left(-(\alpha-1)\frac{4t(u-2t)}{1-4t^2}\right)$, leading to a Chernoff-type bound: $$\begin{eqnarray*}\mathbb{P}\left[\left|X-\frac{1}{2}\right|\geq t\right]\leq\frac{\Gamma(2\alpha)(1-4t^2)^{\alpha}}{4^{\alpha}t(\alpha-1)\Gamma(\alpha)^2}.\end{eqnarray*}$$ Since $\alpha$ is assumed to be large, another effective upper bound comes from replacing $t(1-t)$ with $\frac{1}{4}\exp\left(-(2t-1)^2\right)$, then exploiting the well-known tail bounds of the normal distribution (thanks to Mike Spivey).