Whats the probability for person A to win the dice game? The rules: Two people rolls a single dice. If the dice rolls 1,2,3 or 4, person A gets a point. For the rolls 5 and 6, person B gets a point. One person needs a 2 point lead to win the game.
This is a question taken from my math book. The answer says the probability is $\frac{4}{5}$ for person A to win the game. Which I dont understand.
My thought process: Lets look at all the four possible outcomes with the two first roles. These would be AA, BB, AB or BA. AA means person A gets a point two times a row. Under is the probability for all these scenarios:
$P(AA)=(\frac{2}{3})^2=\frac{4}{9}$
$P(BB)=(\frac{1}{3})^2=\frac{1}{9}$
$P(AB)=\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$
$P(BA)=\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$
If AB or BA happens they have an equal amount of points again, no matter how far they are into the game. The probability of this would then be $2\cdot\frac{2}{9}=\frac{4}{9}$. Since they have an equal amount of points, you can look at that as the game has restarted.
Meaning person A has to get two points a row to win no matter what. Would that not mean the probability is $\frac{4}{9}$ for person A to win? Can someone tell me where my logic is flawed and what the correct logic would be?

As you showed, person $A$ can win the first round of two rolls with probability $4/9$.  However, as Fishbane pointed out in the comments, person $A$ can also win if both person $A$ and person $B$ fail to win in the first round and person $A$ obtains two points in a subsequent round before person $B$ does.  Hence, the probability that person $A$ wins should be higher than $4/9$ since person $A$ does not have to win the game in the first round of two rolls.
Method 1:  We add the probabilities that person $A$ wins in the $k$th round of two rolls.
Since the probability neither $A$ nor $B$ wins a round is $4/9$, the probability neither $A$ nor $B$ wins in any of the first $k - 1$ rounds of two rolls is $(4/9)^{k - 1}$.  The probability that person $A$ then wins the $k$th round of two rolls is $4/9$.  Hence, the probability that person $A$ wins in the $k$th round is $(4/9)^k$.  Therefore, the probability that person $A$ wins is
\begin{align*}
\sum_{k = 1}^{\infty} \left(\frac{4}{9}\right)^k & = \frac{4}{9}\sum_{k = 1}^{\infty} \left(\frac{4}{9}\right)^{k - 1}\\
& = \frac{4}{9} \cdot \frac{1}{1 - \frac{4}{9}}\\
& = \frac{4}{9} \cdot \frac{1}{\frac{5}{9}}\\
& = \frac{4}{9} \cdot \frac{9}{5}\\
& = \frac{4}{5}
\end{align*}
Method 2:  Let $p$ be the probability that $A$ wins the game.  You showed that the probability $A$ wins in the first round is $4/9$.  You also showed that the probability that neither $A$ nor $B$ wins the first round is $4/9$, at which point the game restarts, so $A$ again has probability $p$ of winning.  Hence,
$$p = \frac{4}{9} + \frac{4}{9}p$$
Solving for $p$ yields $p = 4/5$, as before.
