7. $\frac{27}{5005}$.

Let $P(n)$ be the probability that the backpack is empty after drawing cards according to the rules, starting with $n$ pairs of distinct cards.
Then $P(2)=1$.
Suppose the backpack contains $n(n \geqslant 2)$ pairs of distinct cards. The probability that two of the first three cards drawn form a pair is
$$
\frac{\mathrm{C}_{n}^{1} \mathrm{C}_{2 n-2}^{1}}{\mathrm{C}_{2 n}^{3}}=\frac{3}{2 n-1} \text {. }
$$

Thus, $P(n)=\frac{3}{2 n-1} P(n-1)(n \geqslant 3)$.
By repeatedly applying this recurrence relation, we get
$$
P(n)=\frac{3}{2 n-1} \cdot \frac{3}{2 n-3} \cdots \cdots \frac{3}{5} P(2) \text {. }
$$

Therefore, $P(7)=\frac{27}{5005}$.
7. The rules of a single-player card game are as follows: Place seven pairs of different cards into a backpack. The player randomly draws and returns cards each time, but when a pair is drawn, the paired cards are set aside. Each time, the player draws three cards (if there are fewer than three cards left, all remaining cards are drawn). If the three cards drawn are all different and do not form any pairs, the game ends; otherwise, drawing continues until the backpack is empty. What is the probability that the backpack is empty? $\qquad$