Discuss the two cases separately.

1。 Let Beibei play first 1.

Jingjing put 3, so Beibei can't put 3, because Jingjing put 1 again, and Beibei lost, so Beibei chose to put 1 in the second round, but after choosing to put 1, how much will Jingjing put? Jingjing won't choose to put 1, but will only choose to put 3. Because Jingjing won, Beibei lost.

Therefore, it is impossible for Beibei to win 1 in the first round. In other words, if both sides are smart enough, Beibei will not put 1 in the first round.

2 Beibei puts 3 first.

If Jingjing also puts 3, Beibei will win if she puts 3 again, so Jingjing won't put 3, only 1.

If Beibei plays 3 in the second round, Jingjing will definitely play 1, so Jingjing won, so Beibei can only play 1 in the second round.

Beibei put 1 in the second round. In fact, there is no doubt that she will lose, because Jingjing only chose to put 3, so Beibei lost.

To sum up, we come to the conclusion that under the condition that both sides are completely rational, that is, smart enough not to make selective mistakes, the person released later will win. It is not difficult to analyze the plans that the released people should adopt in the face of their opponents' different strategies.

Generally speaking, the first player puts 1 first, and then the second player wins by putting 3 in the first round.

If the first player puts 3 first, then the second player only needs to choose to put 1 in the first round to win.

In fact, this is a typical dynamic game, and my method is forward induction, which is a relatively simple solution.

As for backward induction, it will be relatively difficult.