What is $\sup_X \{ \inf_Y \{P(X > Y)\}\}$? v2.0 Suppose $X$ and $Y$ are two independent non-negative random variables, such that $E[X] = E[Y] = 1$. What is $\sup_X \{ \inf_Y \{P(X > Y)\}\}$?
I managed to derive the following facts:
$\frac{1}{4} \leq \sup_X \{ \inf_Y \{P(X > Y)\}\} \leq \frac{1}{2}$.
One can get the upper bound simply by examining the case, when $X$ and $Y$ are i.i.d.
To get the lower bound we have to look at $X$, that takes values $0$ and $2$ with equal probability. In this case, for any $Y$ satisfying our condition, we have
$$P(X > Y) = P(X = 2)(1 - P(Y \geq 2)) = \frac{1}{2}(1 - P(Y \geq 2)) \geq \frac{1}{4}$$
by Markov inequality.
However, I do not know, whether any of those bounds is actually tight.
Here is a similar question for random variables without non-negativity condition (and with answer $0$).

Let $X$ be uniform in $[0,2]$. Then every non-negative variable $Y$ independent of $X$ satisfies $$P(X \le Y\, |\, Y) \le Y/2 \,,\; \; \; (*)$$
with equality iff $Y \le 2$. Next, recall the assumption $E(Y)=1$. Taking expectation in (*) using the tower property [1] gives
$$P(X \le Y) \le E(Y/2)= 1/2 \,,$$
with equality iff $Y \le 2 $ a.s.
Thus for this choice of $X$ and for every $Y \ge 0$  independent of $X$ that satisfies $E(Y)=1$, we have
$$P(X>Y)\ge 1/2 \,,$$
so
$$\sup_X \inf_Y P(X>Y)=1/2 \,,$$
where $X,Y$ are assumed to be independent, non-negative of mean 1.
[1] https://en.wikipedia.org/wiki/Law_of_total_expectation
