Probability that a particular football team wins a two-game series against another team Question

Football teams $T_1$ and $T_2$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T_1$ winning, drawing and losing a game against $T_2$ are $\frac{1}{2},\frac{1}{6},\frac{1}{3}$ respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two games. Find $$P(X>Y)$$

My Approach
I took simple approach that $X>Y$
$T_1~\text{wins and } T_2~\text{loses}=\frac{1}{2} \times\frac{1}{3} $
$T_1~\text{wins and } T_2~\text{draws}\frac{1}{2} \times\frac{1}{6}$
$T_1~\text{draws and } T_2~\text{loose}\frac{1}{6} \times\frac{1}{3}$
Required probability$$\frac{1}{6} +\frac{1}{12}+\frac{1}{18}=\frac{11}{36}$$
I know it is wrong.  But I am unable to move on.
Please help!

Lets think about the full distribution for $X$ and $Y$. What you have done is combined the distributions of $T_1$ and $T_2$. We want the combination of $X$ and $Y$. When $X=6$, $Y=0$. When $X=4$, $Y=1$. When $X=1$, $Y=4$. When $X=3$, $Y=3$. When $X=0$, $Y=6$. If $W$ is win, $D$ is draw and $L$ is loss, we can: $WW, LL, WL, LW, WD, DW, DD$
We want the combinations where $X \gt Y$ strictly. That is $X=6$, $Y=0$ AND $X=4$, $Y=1$. This is represented by $WW$ or $WD$ or $DW$. Sum this:
$2$ Wins: $\frac 12 \times \frac 12$
$1$ Win $1$ Draw: $\frac 12 \times \frac 16$
So
$$P(X>Y)=P(X=6,Y=0)+P(X=4,Y=1)$$ 
$$=P(WW)+2 \times P(WD)$$
$$=\frac 12 \times \frac 12 \space + \space 2 \times\frac 12 \times \frac 16= \frac 14+ \frac 1{6}= \frac 5{12}$$
Edit ### - Looks like my first guess was right.
So 5/12 I hope is your answer.
