6. When rolling a die twice, the numbers obtained are $m, n$, and the vector $a=(m, n)$ is formed. Then the probability that the angle between $a$ and $b=(1,-1)$ is an angle in a right triangle is $\qquad$
6. $\frac{7}{12}$.

Since $m, n$ can both take values from $1 \sim 6$, the vector $a$ has
$6 \times 6=36$ ways to choose.
And $\cos \langle\boldsymbol{a}, \boldsymbol{b}\rangle=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}||\boldsymbol{b}|}=\frac{m-n}{\sqrt{2} \cdot \sqrt{m^{2}+n^{2}}}$.
Therefore, $\langle a, b\rangle$ being a right angle or an acute angle is equivalent to $m \geqslant n$.
Such $(m, n)$ pairs have
$$
1+2+\cdots+6=21 \text{ (pairs). }
$$

Thus, the required probability is $\frac{21}{36}=\frac{7}{12}$.