7.20 A kind of single-player card game, its rules are as follows: Place 6 pairs of different cards into a backpack, the player randomly draws and returns cards each time, but when a pair is drawn, it is placed aside. If the player always draws three cards each time, and if the three cards drawn are all different (i.e., no pairs), the game ends; otherwise, drawing continues until the backpack is empty. Let the probability that the backpack is empty be $\frac{p}{q}$, where $p$ and $q$ are coprime positive integers, find $p+q$.
[Solution] Let there be $n$ pairs of distinct cards in the bag, where $n \geqslant 2$. Then the probability that 2 out of the first 3 cards drawn form a pair is
the number of ways to draw 3 cards with 2 forming a pair
the number of ways to draw 3 cards
$$
\begin{array}{l}
=\frac{C_{n}^{1} \cdot C_{2 n-2}^{1}}{C_{2 n}^{3}} \\
=\frac{3}{2 n-1}
\end{array}
$$

Let $P(n)$ be the probability that, when there are $n$ pairs of distinct cards in the bag, the bag can be emptied by following the rules described in the problem.
Clearly,
$$
\left\{\begin{array}{l}
P(2)=1 \\
P(n)=\frac{3}{2 n-1} \cdot P(n-1), n \geqslant 3 .
\end{array}\right.
$$

By repeatedly using this recursive formula, we get
$$
P(n)=\frac{3}{2 n-1} \cdot \frac{3}{2 n-3} \cdots \cdot \frac{3}{5} \cdot P(2) \text {. }
$$

Let $n=6$, then
$$
\begin{aligned}
P(6) & =\frac{3}{11} \cdot \frac{3}{9} \cdot \frac{3}{7} \cdot \frac{3}{5} \cdot P(2) \\
& =\frac{9}{385} .
\end{aligned}
$$

Thus, $\frac{p}{q}=\frac{9}{385}$.
Therefore, $p+q=9+385=394$.