Example 6 There are 1994 matches on the table, two children, A and B, take turns to take 1, 2 or 3 matches each time, the one who can take the last match wins. Now A takes first, which child will win? How should he play to win this game?
Solution: When it's A's turn, if the number of matches is $4k+1$, $4k+2$, or $4k+3$, then A can win. The specific strategy is: for $4k+1$, $4k+2$, or $4k+3$ matches (favorable state), A takes $1$, $2$, or $3$ matches respectively, leaving $4k$ matches (unfavorable state) for B to take. When it's A's turn, it is always a favorable state (the number of matches is not a multiple of 4); and after A takes, when it's B's turn, it is always an unfavorable state (the number of matches is a multiple of 4), forcing B to disrupt this state and leave a favorable state for A. Since $1994=4 \times 498+2$, A will win.