6. From the set $\{1,2, \cdots, 2014\}$, randomly and without replacement, take three numbers $a_{1} 、 a_{2} 、 a_{3}$, and then from the remaining 2011 numbers, similarly randomly and without replacement, take three numbers $b_{1} 、 b_{2} 、 b_{3}$. The probability that a brick with dimensions $a_{1} \times a_{2} \times a_{3}$ can fit into a box with dimensions $b_{1} \times b_{2} \times b_{3}$ is $\qquad$
6. $\frac{1}{4}$.

Assume $a_{1}<a_{2}<a_{3}, b_{1}<b_{2}<b_{3}$, the brick can be placed in the box if and only if $a_{1}<b_{1}, a_{2}<b_{2}, a_{3}<b_{3}$.

Let $c_{1}<c_{2}<\cdots<c_{6}$ be six numbers selected from $\{1,2, \cdots, 2014\}$, and then select three out of these six, with $\mathrm{C}_{6}^{3}=$ 20 ways. These three will be $a_{1}, a_{2}, a_{3}$, and the remaining three will be $b_{1}, b_{2}, b_{3}$. The suitable $a_{1}$ can only be $c_{1}$.
If $a_{2}$ is $c_{2}$, then $a_{3}$ can be $c_{3}$ or $c_{4}$ or $c_{5} ;$ if $a_{2}$ is $c_{3}$, then $a_{3}$ can be $c_{4}$ or $c_{5}$.
Therefore, there are 5 suitable ways, and the probability is
$$
p=\frac{5}{20}=\frac{1}{4} \text {. }
$$