7. Xiao Zhang, Xiao Li, Xiao Hua, and Xiao Ming are playing a game of taking turns to roll a standard die. If one person rolls the smallest number, and no one ties with him, he is declared the winner; if more than one person rolls the smallest number, those who did not roll the smallest number are eliminated first, and then the remaining people play another round, continuing this process until someone wins. It is known that Xiao Zhang, who rolls the die first, rolls a 3. Then the probability of him winning is
7. $\frac{175}{864}$.

Consider the four possible scenarios in the first round.
(1) Xiao Zhang wins. The probability of this is
$$
P_{1}=\left(\frac{3}{6}\right)^{3}=\frac{1}{8} \text {. }
$$
(2) Xiao Zhang ties with one other person. The probability of this is
$$
\mathrm{C}_{3}^{1} \times \frac{1}{6} \times\left(\frac{3}{6}\right)^{2}=\frac{1}{8},
$$

Thus, the probability of this scenario occurring and Xiao Zhang ultimately winning is
$$
P_{2}=\frac{1}{8} \times \frac{1}{2}=\frac{1}{16}
$$
(Note that the probability of the game continuing indefinitely is 0, the same applies below).
(3) Xiao Zhang ties with two other people. The probability of this is
$$
C_{3}^{2} \times\left(\frac{1}{6}\right)^{2} \times \frac{3}{6}=\frac{1}{24},
$$

Thus, the probability of this scenario occurring and Xiao Zhang ultimately winning is
$$
P_{3}=\frac{1}{24} \times \frac{1}{3}=\frac{1}{72} \text {. }
$$
(4) Everyone ties. The probability of this is
$$
\left(\frac{1}{6}\right)^{3}=\frac{1}{216} \text {, }
$$

Thus, the probability of this scenario occurring and Xiao Zhang ultimately winning is
$$
P_{4}=\frac{1}{216} \times \frac{1}{4}=\frac{1}{864} .
$$

In summary, the probability of Xiao Zhang winning the game is
$$
\begin{array}{l}
P=P_{1}+P_{2}+P_{3}+P_{4} \\
=\frac{1}{8}+\frac{1}{16}+\frac{1}{72}+\frac{1}{864}=\frac{175}{864} .
\end{array}
$$