25. Person A throws a standard cubic die three times, then he examines the dice and selects a subset of them (which could be empty or all three dice) to re-roll. After the re-roll, he wins if and only if the sum of the numbers on the top faces of the three dice is exactly 7. To increase his chances of winning, the probability that A chooses to re-roll two dice is ( ).
(A) $\frac{7}{36}$
(B) $\frac{5}{24}$
(C) $\frac{2}{9}$
(D) $\frac{17}{72}$
(E) $\frac{1}{4}$
25. A.

Consider the probability of getting a total score $s(s \leqslant 7)$ when rolling two dice.
At this time, the two dice are
$$
(1, s-1),(2, s-2), \cdots,(s-1,1) \text {, }
$$

There are $s-1$ pairs that satisfy this condition, and there are 36 possible outcomes for the two dice, so the probability of getting a total score $s(s \leqslant 7)$ is $\frac{s-1}{36}$.

Thus, if one of the dice has a value of $a$, and player A recalculates the values of the other two dice, the probability of winning is
$$
\frac{7-a-1}{36}=\frac{6-a}{36} \text {. }
$$

To maximize the probability of winning, $a$ must be as small as possible. This indicates that if player A rerolls two dice, he must choose the two dice with the highest values.

Let the values of the three dice $A, B, C$ be $a, b, c$ respectively.
Consider $a+b\frac{6-a}{36}$.
Thus, to make $a+b7$ the situation.
Rerolling one die will not win, because the sum of the three dice is always greater than 7. If rerolling two dice, the probability of winning is still $\frac{6-a}{36}$. To find a winning strategy, recalculate the probability of winning for all three dice, which can be done by setting each die to 1 and finding the ways to distribute the remaining 4, which is $\mathrm{C}_{6}^{2}=15$ ways, with a winning probability of
$$
\frac{15}{6^{3}}=\frac{5}{72} \text {. }
$$

Thus, to make rerolling two dice more advantageous than rerolling three dice,
$$
\text { Therefore, } \frac{6-a}{36}>\frac{5}{72} \Rightarrow a \leqslant 3 \text {. }
$$

Therefore, when and only when $a \leqslant 3$ and $a+b \geqslant 7$, rerolling two dice is the optimal strategy. The possible triples $(a, b, c)$ that satisfy these conditions, and the number of ways they can be arranged are
$\{3,4,4\}$ with 3 cases,
$\{3,4,5\}$ with 6 cases,
$\{3,4,6\}$ with 6 cases,
$\{3,5,5\}$ with 3 cases,
$\{3,5,6\}$ with 6 cases,
$\{2,6,6\}$ with 3 cases,
$\{1,6,6\}$ with 3 cases.
Thus, the number of optimal methods for rerolling two dice is
$$
3+6+6+3+6+3+3+6+3+3=42 \text {. }
$$

Therefore, the required probability is $\frac{42}{6^{3}}=\frac{7}{36}$.