Is $\mathbb{P}( X > Y) = \mathbb{P}( X+k > Y+k )$ true, where X and Y are random variables?

Question

My intuition tells me that $\mathbb{P}(X+k > Y+k) = \mathbb{P}(X > Y)$ should be true, since there (should?) be a bijection between every single result between these two probability distributions. But clearly there's a misunderstanding somewhere here, but I'm having trouble pinning it down.

A counterexample would be: Suppose $X \sim \operatorname{Bin}(n, 0.5)$ and $Y \sim \operatorname{Bin}(n+1, 0.5)$. $\mathbb{P}(X < Y) = \mathbb{P}(n-X < n+1-Y)$, since the probability distribution of $\operatorname{Bin}(n, 0.5)$ and $n - X$ should be exactly the same.

Thanks in advance for your help!

Answer 1

HINT: $\mathbb{P}(X>Y)=\mathbb{P}(\lbrace\omega\in \Omega:X(\omega)>Y(\omega)\rbrace)$, but $X(\omega),Y(\omega)$ are real numbers, then $X(\omega)>Y(\omega)$ if and only if $X(\omega)+k>Y(\omega)+k$.