Two players $A,B$ throw two dice....

Question

Two players $A,B$ throw two dice.
A throw first, and they throw it in turns (i.e. $A,B,A,B,A...$).

If $A$ gets sum of $10$ at the dice he wins, if $B$ gets $9$ - he wins.
What is the probability that $A$ thrown the dice last?

I used this EQ: $$P(E|F)=P(E|FG)\cdot P(G|F)+P(E|FG^C)\cdot P(G^C|F)$$ $E$ = A got sum of $10$.
$F$ = The sum of the dice was $9$ or $10$.
$G$ = $B$ got $9$.

My result was $\frac{2}{5}$. I'm right?
And I need to find another way (i.e. not by this equation).
Can you give ideas how?

Thank you!!

Answer 1

Let $p$ be the chance that $A$ wins. Then $B$ wins $1-p$. $A$ can win by throwing $10$, or by throwing something else, having $B$ not throw a $9$, and then winning from start. As the chance of a $10$ is $\frac 3{36}$ and a $9$ is $\frac 4{36}$, we have $p=\frac 3{36}+(1-\frac 3{36})(1-\frac 4{36})p$