Let the numbers shown in Jason's first roll be $a \le b \le c$, respectively. Then,

• If $a+b+c=7$, then Jason will not re-roll.
• If Jason re-roll three dices, the probability of having a sum of $7$ equals to the number of positive integer solution of $x+y+z=7$ divided by $6^3$ which is $\frac{5}{12}$
• If Jason re-roll two dices and leaving one showing $x$ where $1 \le x \le 6$, then he must have the sum of the two re-rolled dice be $(7-x)$ in order to win. This event occurs at a probability of $\frac{6-x}{36}$. (This conclusion can be verified by calculating the results for $x=1$, $2$, $\cdots$ etc.)
• If follows that if $a+b < 7$, it is always wise to re-roll just one dice which gives a winning probability of $\frac{1}{6}$ which is larger than re-rolling two or three dices.

Therefore, we only need to consider the case where $a+b \ge 7$. This is the only scenario when Jason may choose to re-roll two dices. In this case, because his winning probability is $\frac{6-x}{36}$ where $x$ is the remaining number, it is always wise to leave the smallest number $a$ alone. However, as $\frac{6-4}{36} < \frac{5}{12}$, if the smallest number is at least $4$, then it is wise for Jason to re-roll all three dices. This means the final result equals to the probability when $a + b \ge 7$ and the smallest number $a$ is no more than $3$.

• When $a=1$, we must have $b=c=6$. There are totally $3$ qualified ways.
• When $a=2$, we have $5\le b\le c$. There are totally $3+6+3=12$ qualified ways.
• When $a=3$, we have $4\le b\le c$. There are totally $3+6+6+3+6+3=27$ qualified ways.

These number are calculated by listing all the possible combinations such as $(1, 6, 6)$. Then if all the three numbers are different, there will be totally $3!=6$ different ways for each combination. If two numbers are the same, then there are only $3$ ways (let the unique number to choose from one of the three position positions.)

Hence, the final answer is $$\frac{3+12+27}{6^3}=\boxed{\frac{7}{36}}$$