A general strategy is to compute the expected number of games $t(xy)$ played until a team is declared the winner, starting from every possible partial score of $x$ games won by a player vs $y$ games won by the other one. Then the expected total number of games is $t(00)$.
Some simple remarks: (i) $t(xy)=t(yx)$ by symmetry; (ii) $t(x4)=0$ for every $0\leqslant x\leqslant3$; (iii) looking at the result of the first game played yields a relation between $t(xy)$ and $t((x+1)y)$ and $t(x(y+1))$, for each $xy$.
Starting from the highest possible partial scores and going backwards, one gets successively, using remarks (i), (ii) and (iii),
$$t(33)=1,\quad t(23)=1+\tfrac12t(33)=\tfrac32,\quad t(22)=1+t(23)=\tfrac52,
$$
$$t(13)=1+\tfrac12t(23)=\tfrac74,\quad t(03)=1+\tfrac12t(13)=\tfrac{15}8,
$$
$$
t(12)=1+\tfrac12t(13)+\tfrac12t(22)=\tfrac{25}8,\quad t(02)=1+\tfrac12t(03)+\tfrac12t(12)=\tfrac72,
$$
$$
t(11)=1+t(12)=\tfrac{33}8,\quad t(01)=1+\tfrac12t(11)+\tfrac12t(02)=\tfrac{77}{16},
$$
and, finally, $t(00)=1+t(01)=\frac{93}{16}$.
Edit: To check this result, note that
$$
t(00)=\frac{2{3\choose 0}2^3\cdot4+2{4\choose 1}2^2\cdot5+2{5\choose 2}2^1\cdot6+{6\choose 3}2^0\cdot7}{2{3\choose 0}2^3+2{4\choose 1}2^2+2{5\choose 2}2^1+{6\choose 3}2^0}.
$$
You will have to sum up a number of mutually exclusive probabilities:
The last winning match has to be $B$, and $B$ must win any $2$ of the preceding ones,
hence $\binom22 + \binom32 + \binom42$ cases
$BBB,\;ABBB,\; BABB,\; BBAB,\; AABBB,\; ABABB,\; ......$
with probabilities $0.3*0.8*0.3 + 0.7*0.8*0.3*0.8 + .....$
Best Answer
Calculate the probability that the series ends after 3 games (AAA, BBB). Calculate the probability that it ends after 4 games (AABA, ABAA, BAAA, BBAB, BABB, ABBB). The remainder is the probability that it ends after 5 games. Look which of the 3 probabilities is the biggest.