How can I find the probability $$\lim_{n \to \infty} P\left(\sum_{i=1}^n X_i>0\right) = \lim_{n \to \infty} P\left(\frac{1}{n}\sum_{i=1}^n X_i>0\right)$$ if $X_i$ are iid? If the expectation of $X_i$ is less than $0$, then by the law of large numbers, the limit probability would be $0$. Conversely, if the expectation is greater than $0$, the limit probability would be $1$. What I'm having trouble with is if the expectation is exactly $0$.
If the distribution is symmetric about $x=0$, then the probability is $\frac{1}{2}$ by symmetry. I think it may actually always be $\frac{1}{2}$ since the average of $n$ iid variables would approximately be the normal distribution $N\left(0, \frac{\sigma^2}{n}\right)$ by the central limit theorem where $\sigma^2$ is the variance of $X_i$.
Is it true that the limit probability would be $\frac{1}{2}$ no matter what the exact distribution is (assuming the mean is $0$)? To me, this seems somewhat counterintuitive, but it is the only result I get.