smc

We have two cases: one where $A$ wins the first game and the other where $A$ loses. $\underline{\text{Case 1:}}$ In the $\frac{1}{2}$ chance that $A$ wins the first game, $A$ simply needs to win at least $1$ of the next $3$ games. We see that the probability of $A$ losing the next $3$ games is $(\frac{1}{2}{)}^3=\frac{1}{8}$, so, by complementary counting, the probability that $A$ wins at least $1$ of the next $3$ games is $1-\frac{1}{8}=\frac{7}{8}$. $\underline{\text{Case 2:}}$ In the $\frac{1}{2}$ chance that $A$ loses the first game, both teams need to win $2$ games, so $A$'s advantage completely disappears. Thus, the proability that $A$ wins the series from here is $\frac{1}{2}$. Combining the information from the two cases, we see that $A$'s probability of winning the series is $\frac{1}{2}\cdot \frac{7}{8}+\frac{1}{2}\cdot \frac{1}{2}=\frac{7}{16}+\frac{4}{16}=\frac{11}{16}$. Thus, our answer is $\boxed{11\text{ to }5}$.