6. A die is rolled four times consecutively, and from the second roll onwards, the number of points that appear each time is not less than the number of points that appeared in the previous roll. The probability of this happening is $\qquad$ .
6. $\frac{7}{72}$.

Let $a_{1}, a_{2}, a_{3}, a_{4}$ be the numbers obtained from four dice rolls, respectively. Then $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ has $6^{4}$ different possibilities.

If from the second roll onwards, each number is not less than the previous one, then $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4}$.

If $a_{1}, a_{2}, a_{3}, a_{4}$ take exactly $1, 2, 3, 4$ different values, then $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ has $\mathrm{C}_{6}^{1}, 3 \mathrm{C}_{6}^{2}, 3 \mathrm{C}_{6}^{3}, \mathrm{C}_{6}^{4}$ different possibilities.
Therefore, the number of cases satisfying $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4}$ is $\mathrm{C}_{6}^{1}+3 \mathrm{C}_{6}^{2}+3 \mathrm{C}_{6}^{3}+\mathrm{C}_{6}^{4}=126$ (cases).
Thus, the required probability is $\frac{126}{6^{4}}=\frac{7}{72}$.