Probability of winning a game 
$A$ and $B$ play a game. 
  
  
*
  
*The probability of $A$ winning is $0.55$. 
  
*The probability of $B$ winning is $0.35$. 
  
*The probability of a tie is $0.10$. 
  
  
  The winner of the game is the person who first wins two rounds. What is the probability that $A$ wins?

The answer is $0.66$. I don't know how it's coming $0.66$. 
Please help.
EDIT :
The right combinations according to me are
{null,T,TT,TTT....}A{null,T,TT,TTT....}A
{null,T,TT,TTT....}A{null,T,TT,TTT....}B{null,T,TT,TTT....}A
{null,T,TT,TTT....}B{null,T,TT,TTT....}A{null,T,TT,TTT....}A

During the match, we swicth between three states: $t=$"tie", $a=$"$A$ leads", $b=$"$B$ leads". Let $p_t, p_a, p_b$ be the probabilizty that $A$ wins if the starting position is as indicated.
If we start in $t$, then we continue with $a$, $b$, or $t$ according to the given probabilities afor a single game, that is
$$p_t = .55p_a+.35p_b+.10 p_t $$
If we start in $a$, we have essentially the same, except that a won game decides the match. Thus
$$p_a = .55\cdot 1+.35p_t+.10 p_a $$
Finally, 
$$p_b = .55p_t+.35\cdot 0+.10 p_b $$
and this gives us three equations in three unknowns. The number we are looking for is $p_t$ (as we actually start in a tie situation).
