Let's Make a Tic-Tac-Deal!

The game of Tic-Tac-Deal $2.0$ has a $3$-by-$3$ square grid with the numbers $3$ through $11$, arranged as follows:

$3$	$4$	$5$
$6$	$7$	$8$
$9$	$10$	$11$

You start by rolling a standard pair of six-sided dice and add the two numbers rolled. You place an X on the board on the square that contains the sum. If the sum is a $2$ or $12$, your roll is wasted. If you have exactly three rolls of the dice, what are your chances of getting three Xs in a row (either horizontally, vertically, or diagonally)?

First, we see that $$p_n = \frac{\min\{n-1,13-n\}}{36}$$ is the probability of getting a vakue of $n$ on a roll of two standard dice. Secondly, we see that there are a total of 8 winning configurations: $\{3,4,5\}$, $\{3,6,9\}$, $\{3,7,11\}$, $\{4,7,10\}$, $\{5,7,9\}$, $\{5,8,11\}$, $\{6,7,8\}$, and $\{9,10,11\}.$

Since order doesn't matter here, the probability of getting a winning configuration $W=\{w_1, w_2, w_3\}$ is given by $$p(W)= 3! \prod_{i=1}^3 p_{w_i}.$$ So the overall probability of winning in three rolls is \begin{align*}P=\sum_W p(W)&= 3! \left( p_3p_4p_5 + p_3p_6p_9 + p_3p_7p_{11}\right. \\ &\quad\quad +p_4p_7p_{10} + p_5p_8p_{11} + p_5p_7p_9 \\ &\left.\quad\quad\quad + p_6p_7p_8 + p_9p_{10}p_{11} \right)\\ &= \frac{\left( 144 + 240 + 144 + 324 + 576 + 240 + 900 + 144 \right)}{46656} \\ &= \frac{113}{1944} \approx 5.812757\dots\%\end{align*}