6. A uniform cubic die is rolled twice, and the numbers that appear in succession are $a$ and $b$, respectively. The probability that the equation $x^{2}+a x+b=0$ has real roots is $\qquad$ (answer in simplest fraction form).
6. $\frac{19}{36}$.

From the problem, since $a, b \in \{1,2, \cdots, 6\}$, the total number of basic events is $6^{2}=36$.

The necessary and sufficient condition for the equation $x^{2}+a x+b=0$ to have real roots is $\Delta=a^{2}-4 b \geqslant 0$.

For the values of $a^{2}, 4 b$ and the sign of $a^{2}-4 b$ as shown in Table 1. Among them, $\Delta=a^{2}-4 b$ has 19 non-negative values. Therefore, the required probability is $P=\frac{19}{36}$.