In some lecture notes that I have on discrete probability, after defining expectancy, it says "the expectancy doesn't depend on the random variable directly; it depends only on its distribution", where with "distribution" the function $ W:X(\Omega) \rightarrow \mathbb{R},\ W(x)=P(X=x), $ $X:\Omega \rightarrow \mathbb{R}$ being our random variable.
As an explanation for the above, the following line is given: $ \mathbb{E}X=\sum_{x\in X(\Omega) } x \cdot W(x)=\sum_{\omega \in \Omega} X(\omega) P(\omega). $
Now I understand why the above holds, but I don't understand why this line entitles one to say that expectancy depends only on the distribution of a random variable. If I would change my random variable $X$ to $X'$, so that $X(\omega')\neq X'(\omega')$, for some $\omega'\in \Omega$, than by the above line, of course $\mathbb{E}X \neq \mathbb{E}X'$ .
For a better understand: Could someone provide me with an example of two different r.v.'s having the same distribution ?