431 There is a card game called "Twelve Months" that is only played during the Spring Festival. The rules are as follows:

First, take a brand new deck of cards, remove the big joker, the small joker, and the 4 Ks, leaving 48 cards, which are then shuffled.

Second, lay out the shuffled cards face down in 12 columns, with 4 cards in each column.

Third, start by flipping the first card in the first column. If you flip over an $N (N=1,2, \cdots, 12$, where $\mathrm{J}$ and $\mathrm{Q}$ represent $11$ and $12$ respectively, without distinguishing suits), place this face-up card at the end of the $N$th column.

Fourth, continue to flip the first face-down card in the $N$th column, and operate according to the method in the third step.

Fifth, continue this process until it can no longer proceed. If all 12 columns are eventually flipped over, it signifies that everything will go well in the coming 12 months; otherwise, if some columns remain unflipped when the game ends, it indicates that there will be difficult events in the corresponding months.
Find the probability of the event "all flipped over" occurring.
Solve the generalization, construct an $m \times n$ number table, where $a_{ij}$ represents the number in the $i$-th row and $j$-th column. Extract some or all of these numbers according to a predetermined rule to form a sequence $\left\{b_{n}\right\}$. The rule is as follows:
(i) First, extract $a_{11}$, and let $b_{1}=a_{11}$; if $b_{1}=j$, then extract the number with the smallest row index from the $j$-th column, and denote it as $b_{2}$.
Continue in this manner.
(ii) When $b_{k}=j$, extract the number with the smallest row index from the $j$-th column and denote it as $b_{k+1}$; continue until no more numbers can be extracted.

Answer in two steps: (1) Prove that the last number in the sequence $\left\{b_{n}\right\}$ must be 1;
(2) Find the probability that the sequence $\left\{b_{n}\right\}$ has exactly $m n$ terms.
(1) The proof is obvious, noting that except for the first column, only if the previous extraction was $j$, can a number be extracted from the $j$-th column this time. In other words, as long as $j$ (not 1) is extracted this time, there must still be a number in the $j$-th column for the next extraction.
(2) In fact, the number of all scenarios where 1 appears exactly at the $m n$-th extraction is $\frac{(m n-1)!}{(m!)^{n-1}(m-1)!}$, while the total number of basic events is $\frac{(m n)!}{(m!)^{n}}$. The ratio of the two gives $\frac{1}{n}$.

Therefore, in the game, the probability of the event "all numbers are flipped out" is $\frac{1}{12}$.
(Zhang Jianqiang, Shenzhen Middle School, Guangdong Province, 518001)