Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly $7.$ Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?
$\textbf{(A) } \frac{7}{36} \qquad\textbf{(B) } \frac{5}{24} \qquad\textbf{(C) } \frac{2}{9} \qquad\textbf{(D) } \frac{17}{72} \qquad\textbf{(E) } \frac{1}{4}$
Consider the probability that rolling two dice gives a sum of $s$, where $s \leq 7$. There are $s - 1$ pairs that satisfy this, namely $(1, s - 1), (2, s - 2), \ldots, (s - 1, 1)$, out of $6^2 = 36$ possible pairs. The probability is $\frac{s - 1}{36}$.
Therefore, if one die has a value of $a$ and Jason rerolls the other two dice, then the probability of winning is $\frac{7 - a - 1}{36} = \frac{6 - a}{36}$.
In order to maximize the probability of winning, $a$ must be minimized. This means that if Jason rerolls two dice, he must choose the two dice with the maximum values.
Thus, we can let $a \leq b \leq c$ be the values of the three dice. Consider the case when $a + b  \frac{6 - a}{36}$, so rolling one die is always better than rolling two dice if $a + b  \frac5{72} \rightarrow a \leq 3$.
Thus, rerolling two dice is optimal if and only if $a \leq 3$ and $a + b \geq 7$. The possible triplets $(a, b, c)$ that satisfy these conditions, and the number of ways they can be permuted, are
$(3, 4, 4) \rightarrow 3$ ways.
$(3, 4, 5) \rightarrow 6$ ways.
$(3, 4, 6) \rightarrow 6$ ways.
$(3, 5, 5) \rightarrow 3$ ways.
$(3, 5, 6) \rightarrow 6$ ways.
$(3, 6, 6) \rightarrow 3$ ways.
$(2, 5, 5) \rightarrow 3$ ways.
$(2, 5, 6) \rightarrow 6$ ways.
$(2, 6, 6) \rightarrow 3$ ways.
$(1, 6, 6) \rightarrow 3$ ways.
There are $3 + 6 + 6 + 3 + 6 + 3 + 3 + 6 + 3 + 3 = 42$ ways in which rerolling two dice is optimal, out of $6^3 = 216$ possibilities, Therefore, the probability that Jason will reroll two dice is $\frac{42}{216} = \boxed{\textbf{(A) } \frac{7}{36}}$.