7.20 A single-player card game has the following rules: Place 6 pairs of different cards into a backpack. The player draws cards randomly from the backpack and returns them, but when a pair is drawn, it is set aside. If the player always draws three cards at a 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 the bag becomes empty when starting to draw cards according to the rule in the problem, given that there are $n$ pairs of distinct cards in the bag.
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
$$P(n)=\frac{3}{2 n-1} \cdot \frac{3}{2 n-3} \cdots \cdots \frac{3}{5} \cdot P(2) .$$

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, $p+q=9+385=394$.