6. Given that $P, Q, R, S$ are four points inside the tetrahedron $A-BCD$, and $Q, R, S, P$ are the midpoints of $PA, QB, RC, SD$ respectively. If $V_{P-ABC}$ represents the volume of the tetrahedron $P-ABC$, and similarly for the others, then $V_{P-ABC}: V_{P-BCD}: V_{P-DAB}: V_{P-DAC}=$ $\qquad$ .
6. $8: 1: 4: 2$

Points $A, P$ to the distance of plane $BCD$ are $h_{A}, h_{P}$, and similarly for the others. Therefore, we have
$$
\begin{array}{l}
h_{Q}=\frac{1}{2}\left(h_{A}+h_{P}\right) \\
h_{R}=\frac{1}{2} h_{Q} \\
=\frac{1}{4}\left(h_{A}+h_{P}\right) \\
h_{S}=\frac{1}{2} h_{R}=\frac{1}{8}\left(h_{A}+h_{P}\right), \\
h_{P}=\frac{1}{2} h_{S}=\frac{1}{16}\left(h_{A}+h_{P}\right) .
\end{array}
$$

Thus, $h_{P}=\frac{1}{15} h_{A}$.
Hence $V_{P-BCD}: V_{A-BCD}=1: 15$.
Similarly, $V_{P-CAD}: V_{A-BCD}=2: 15$.
$$
\begin{array}{l}
V_{P-DAB}: V_{A-BCD}=4: 15, \quad V_{P-ABC}: V_{A-BCD}=8: 15 \\
V_{P-ABC}: V_{P-BCD}: V_{P-DAB}: V_{P-DAC}=8: 1: 4: 2
\end{array}
$$