Does equal probability of rank imply two random variables symmetric around median? Let $X$ and $Y$ be two independent, continuous random variables with cumulative distribution functions $F_X$ and $F_Y$. Suppose that $X$ and $Y$ are not identically distributed, i.e. $F_X \ne F_Y$, and that
$$\mathbb{P}[X<Y] = \mathbb{P}[X>Y] = 1/2. \qquad (*)$$
It seems that a necessary condition for $(*)$ is that $X$ and $Y$ have the same medians, i.e. there exists some $m \in \mathbb{R}$ such that $F_X(m) = F_Y(m) = 1/2$. Must $X$ and $Y$ be symmetric around $m$?
 A: Here's a counter-example where both $X$ and $Y$ are supported only in $[0,1]$. The medians are different, and obviously neither of $X$ and $Y$ is symmetric with respect to their own median.
\begin{align}
F_X(t) &= t^2   \\
F_Y(t) &= 1 - (1 - t)^b   \quad \text{where} \quad b = \frac{-3 + \sqrt{17}}2 \approx 0.561553
\end{align}
That is, $X$ has a linear density $f_X(t) = 2t$ and $Y$ has a power law density flipped $f_Y(t) = (1-t)^b$ where the power $b$ shall by design render
$$\mathbb{P}[X > Y] = \int_{x = 0}^1 f_X(x) \cdot \Pr\{Y < x\} \,\mathrm{d}x = \frac12 $$
Or, as $\Pr\{Y < x\} = F_Y(x)$, more explicitly
\begin{align}
\mathbb{P}[X > Y] &= \int_{x = 0}^1 2x\left( 1 - (1-x)^b\right) \,\mathrm{d}x \\
&= 1-\frac2{b^2 + 3b + 2} \\
\end{align}
with $b$ being one of the two roots to $b^2 + 3b + 2 = 4$ (the root that is larger than $-1$; for $L_1$ integrability) such that $\mathbb{P}[X > Y] = 1-\frac24 = \frac12$ as desired.
Similarly, one can systematically generate examples of this kind: $f_X \propto t^a$ versus $f_Y \propto (1-t)^b$, with the two densities being power-laws "opposing each other".
For the record, here the medians are
\begin{align}
m_X &= \frac1{ \sqrt{2} } \approx 0.707107 \\
m_Y &= 1 - \frac1{ 2^{1/a} } \approx 0.708973
\end{align}
