Example 3 Suppose there are $n$ balls, and two people, A and B, play a game of taking balls in turns, each time one can take 1 or 2 balls at will, but not zero, and the one who takes the last ball loses. It is stipulated that A takes first. What are the winning and losing situations for A and B?
Solution: Since the loser ends up with only one ball, in the second-to-last turn, the number of balls must be 3 or 2. Therefore, as long as the number of balls left is 4, the first player will lose.

Therefore, when $n=3k+1(k \in \mathbf{N})$, A will lose. At this point, B can adopt the following strategy: if A takes 1 (or 2) ball(s), then B takes 2 (or 1) ball(s), ensuring that the remaining balls are a multiple of 3 plus 1. This continues round by round, ultimately leading to a situation where only 4 balls are left. Since A takes first, A will lose.

When $n=3k$ (or $3k+2$), A takes 2 (or 1) ball(s) first, leaving the remaining balls as a multiple of 3 plus 1, then B takes over the role of A in the above scenario, hence B will lose.