If $A,B$ are random variables of different distributions, and $\mathbb{E}(A)=\mathbb{E}(B)$, is it necessary for $P(A>B)=P(A<B)$

Question

As the title asks, If $A,B$ are random variables of different distributions, and $\mathbb{E}(A)=\mathbb{E}(B)$, is it necessary for $P(A>B)=P(A<B)$? I tried letting $A\sim Uniform()$ and $B\sim Bin(n,p)$. Made sure the expectation is the same, then calculated $Pr(A>B), Pr(A<B)$ using python for some values of $n,p$ at it was true, but I am wondering if it holds in general.

Answer 1

Expanding on MichaelBurr's hint...

Let $A$ be a random variable with support $\{0,1\}$ with expectation $0.5$

It follows that $Pr(A = k) =\begin{cases} 0.5 & \text{if }k=0\\0.5&\text{if }k=1\end{cases}$

Let $B$ be a random variable with support $\{0,1000\}$ with expectation $0.5$ and be independent of $A$.

It follows that $Pr(B = k) = \begin{cases}\frac{1999}{2000}&\text{if }k=0 \\\frac{1}{2000}&\text{if }k=1000\end{cases}$

You get that $B>A$ if and only if $B = 1000$. You get then that $Pr(B>A)=\frac{1}{2000}$

You get that $A>B$ if and only if $A=1$ and $B=0$. You get then that $Pr(B<A)=Pr(B=0)Pr(A=1)=\frac{1999}{2000}\cdot 0.5 = \frac{1999}{4000}$ which is not equal to $\frac{1}{2000}$, thus disproving your claim.