If there is no limit on the number of sessions, it will be discussed in two situations:
First, regardless of the conditions of the draw (there is no agreed draw), that is, either negative or win. Then there are six situations in which Party A and Party B decide the outcome.
1, A wins continuously from the beginning: expressed as "A, A" (only the winner is marked, the same below);
2.B won a winning streak from the beginning: expressed as "B, B";
3.A loses one game in a row: it means "B, A, A";
4. Party B loses one game in a row first, which means "A, B, B";
5.A wins a game first and then wins: it means "A, B, A";
6. Party B wins one game before winning: expressed as "Party B, Party A and Party B".
The condition of "two-player competition, the one who wins the first two games wins" given in the title, coupled with the restriction of either losing or winning, can completely decide the outcome in three games at most. Everyone has the possibility of winning once in each game, and everyone has the possibility of winning two games three times in three games, so there are 2×3 winning situations in one game.
Second, if the agreement is tied, that is, if it is not negative, it may not win. Then they decide the outcome of the situation in addition to the above six, there may be the following situations:
Ping, Ping, Ping ... Ping, A, A (there may be n draws)
Ping, Ping ... Ping, Ping, A (there may be N draws)
Ping, ping, ping ... a, ping, a (there may be n draws)
Ping, Ping, Ping ... Ping, B, B (there may be n draws)
Ping, Ping ... Ping, Ping, B (there may be N draws)
Ping, ping, ping ... b, ping, b (there may be n draws)
Ping, Ping ... Ping, B, B (there may be n draws)
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So the question seems to have restrictive conditions. However, since the landlord has indicated that it is a third-grade olympiad, let's think as third-grade students. Either lose or win is the first condition to be set.