= Solution
Count the ordered roll pairs producing each value. The <probability mass functions> are
$$
\mathbb P(X=i)=\frac{6-|i-7|}{36}\quad(i=2,\ldots,12),\qquad\mathbb P(Y=j)=\frac{6-|j|}{36}\quad(j=-5,\ldots,5),
$$
and are zero outside these ranges. Both counts have a unique peak:
$$
\boxed{i=7,\quad j=0;\qquad\mathbb P(X=7)=\mathbb P(Y=0)=\frac16.}
$$
However, $Y=0$ forces equal rolls, whose sum is even. Hence $\mathbb P(X=7,Y=0)=0$, whereas the product of the two marginal <probabilities> is $1/36>0$. This disproves <independence of random variables>:
$$
\boxed{X\text{ and }Y\text{ are not independent}.}
$$
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/mathematics-2013-dice-sum-difference.png]
{title=The 36 possible sum and difference pairs for two independent fair dice; the two marginal modes cannot occur together}
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