Is the intuition correct?

Question

Let $X$ be a random variable such that $E|X|<\infty$ and $$P\left(X\ge\frac{1}{2}+x\right)=P\left(X\le\frac{1}{2}-x\right)\ \forall x\in \mathbb{R}.$$ Then

1. $E(X)=\frac{1}{2}$ and Median($X$)$=\frac{1}{2}$
2. $E(X)=\frac{1}{2}$ and Median($X$)$>\frac{1}{2}$
3. $E(X)<\frac{1}{2}$ and Median($X$)$=\frac{1}{2}$
4. $E(X)<\frac{1}{2}$ and Median($X$)$>\frac{1}{2}$

I think option $1$ is correct as it looks (still in doubt) from the given condition that the median is $\frac{1}{2}$ and about the expectation the equation can be rewritten as $$P\left(X-\frac{1}{2}\ge x\right)=P\left(X-\frac{1}{2}\le -x \right)\\ P(Y\ge x)=P(Y\le -x)$$ this implies symmetry hence $E(Y)=0\implies E(X)=\frac{1}{2}$$