Case 1: When A wins the next 3 games only :
B(lose)*A(win)*A(lose)*A(win)+
A(win)*B(lose)*A(lose)*A(win)+
A(win)*A(win)*B(lose)*A(lose)+
A(win)*A(win)*A(win)*B(lose)
A(win) = A win this round and was winnner in the previous round
A(lose) = A wins this round and was the loser
= $\frac{1}{3}*\frac{1}{3}*\frac{2}{3}*\frac{2}{3}+\frac{2}{3}*\frac{1}{3}*\frac{1}{3}*\frac{2}{3}+\frac{2}{3}*\frac{2}{3}*\frac{1}{3}*\frac{1}{3}+\frac{2}{3}*\frac{2}{3}*\frac{2}{3}*\frac{1}{3}=\frac{20}{81}$
Case 2: When A wins all the four games: A(win)*A(win)*A(win)*A(win)=$\frac{2}{3}*\frac{2}{3}*\frac{2}{3}*\frac{2}{3}=\frac{16}{81}$
required probability = $\frac{20}{81}+\frac{16}{81}=\frac{4}{9}$
Correc me if I'm wrong