A draw is always possible in chess. So there are infinitely many possibilities. But let us assume, as presumably we are intended to, that a game always ends in a win by one of the players. We count the number of sequences of wins/losses that end with Alice winning the match, and multiply by $2$. For concreteness, it can be helpful to write A for "Alice wins" and B for "Bob wins."
There is only $1$ such $3$-game sequence, namely AAA. For reasons that will be clear later, I prefer to write $\binom{2}{2}$ instead of $1$.
Now let's count the $4$-game sequences that result in a win by Alice. So the $4$-th game must be won by Alice, the $4$-th and final letter of the sequence is an A. Of the first $3$ games, exactly $2$ were won by Alice. The location of these $2$ games can be chosen in $\binom{3}{2}$ ways.
Finally, let's count the $5$-game sequences that result in a win by Alice. She won the $5$-th game, so the $5$-th and final letter of the sequence was an A. Of the first $4$ letters, exactly $2$ were A's. The location of these $2$ A's can be chosen in $\binom{4}{2}$ ways.
Thus the total number of possibilities is $2\left(\binom{2}{2}+\binom{3}{2}+\binom{4}{2}\right).$
Remark: In the baseball World Series, and in the hockey Stanley Cup finals, the first team to win $4$ games wins the series. By reasoning very similar to the one above, if A and B are playing, there are $2\left(\binom{3}{3}+\binom{4}{3}+\binom{5}{3}+\binom{6}{3}\right)$ possible outcomes.