Mean-preserving spread, conditional probability

Question

Suppose real variable $Y$ is a mean-preserving spread of $X$, e.g. $Y = X + \epsilon$, where $\epsilon$ has expected value zero and is independent of $X$. $X$ takes on value $r$ with positive probability. Is it then true that

$$ P(X \geq r) = P(Y \geq r) ?$$

Answer 1

Suppose $X$ is uniformly chosen from $\{1,2,3,4,5,6\}$ (you roll a die) and $\epsilon$ is uniformly chosen from $\{-100,100\}$. Then $\Pr[X \ge 6] = \frac16$ but $\Pr[X+\epsilon \ge 6] = \frac12$.

Answer 2

Not it is not true.

Take this for an example, where $P(X > r) \not= P(Y > r)$ for any case where $P(X=r)>0$

r   P(X=r) P(Y=r) P(X>r) P(Y>r)
0    0      0.1    1      0.9  
1    0.5    0.5    0.6    0.4
2    0.5    0.2    0.4    0.2
3    0      0.2    0      0