Let X be the result of a roll with a 6-sided dice. In this case the PMF of X is uniform and has the following form
If the individual bars were weights on a ruler, the expected value would be the center of mass, i.e. the position on the ruler from which the same weight lies on the left and right. In this case, the expected value for X is calculated as follows
$E(X)=\sum_{x}x*p_X(x) = 1/6*(1+2+3+4+5+6)=3.5$
Similarly, for a bivariate PMF $p_{X,Y}(x,y) = P(X=x, Y=y)$ there would have to be a point on the plane loaded with the probability weights and spanned by X and Y, on which one could balance it. How can I calculate this point?
I believe you can generalize the mean by computing each component separately, considering the balance point to be the mean vector, $\vec{\mu}$. For example:
$$ E[X] = \Sigma_y \Sigma_x x p_{X,Y}(x,y) \\ E[Y] = \Sigma_y \Sigma_x y p_{X,Y}(x,y) $$ The balance point will then be $\vec{\mu}= (E[X],E[Y])$.
I hope this helps.