Example 7 A backpack contains 12 cards labeled $1,1,2,2, \cdots, 6,6$. A person takes one card at a time from it, without putting it back; if he draws a card with the same label as one he has already drawn, he discards both cards; if he has 3 single cards in his hand or all the cards in the backpack have been taken, the operation ends. Find the probability that all the cards in the backpack are taken.
Let $P_{n}$ be the probability that there are $n$ pairs of cards and the conditions of the problem are met. We calculate the probability that the first 3 cards contain a pair: there are $n(2 n-2)$ ways to select cards that meet the condition, and there are $C_{2 n}^{3}$ ways to select in total, so $P=\frac{n(2 n-2)}{\bar{C}_{2 n}^{3}}=\frac{3}{2 n-1}$. Removing this pair of cards, it is equivalent to taking one card from $n-1$ pairs, so the probability that the bag is empty is $P_{n-1}$. Therefore, $P_{n}=\frac{3}{2 n-1} P_{n-1}$ and $P_{2}=1$. Thus, $P_{6}=\frac{3}{11} \times \frac{3}{9} \times \frac{3}{7} \times \frac{3}{5}=\frac{9}{385}$.
!Note Directly calculating the number of operations that meet the conditions is very difficult, here we cleverly use the recursive relationship of probabilities to solve the problem.