10. For the string of three letters $\hat{\mathrm{i}}$ "aaa" and "bbb", they are transmitted through a circuit, with each string being sent one letter at a time. Due to issues with the line, each of the 6 letters has a $\frac{1}{3}$ chance of being transmitted incorrectly (an $a$ is received as a $b$, or a $b$ is received as an $a$), and the correctness of each letter's reception is independent of the others. Let $S_{\mathrm{a}}$ be the message received when "aaa" is transmitted, and $S_{\mathrm{b}}$ be the message received when "bbb" is transmitted. Let $P$ be the probability that $S_{\mathrm{a}}$ is lexicographically before $S_{\mathrm{b}}$. When $P$ is written as a reduced fraction, what is the numerator?
$x_{1}=a, y_{1}=b, i=1,2,3$.

Let $S_{1} \propto S_{2}$ represent that $S_{1}$ is lexicographically before $S_{2}$, and $P\left(S_{1} \propto S_{2}\right)$ represent the probability that $S_{1}$ is lexicographically before $S_{2}$. Due to the independence of the correct transmission of letters, we have
$$
\begin{array}{c}
P\left(S, \propto S_{\mathrm{b}}\right)=P\left(x_{1} x_{2} x_{3} \propto y_{1} y_{2} y_{\mathrm{s}}\right) \\
=P\left(x_{1} \propto y_{1}\right)+P\left(x_{1}=y_{1}, x_{2} \propto y_{2}\right) \\
+P\left(x_{1}=y_{1}, x_{2}=y_{2}, x_{3} \propto y_{3}\right) \\
=P\left(x_{1} \propto y_{1}\right)+P\left(x_{1}=y_{1}\right) P\left(x_{2} \propto y_{2}\right) \\
+P\left(x_{1}=y_{1}\right) P\left(x_{2} \propto y_{2}\right) P\left(x_{3} \propto y_{3}\right) .
\end{array}
$$

Since the necessary and sufficient condition for $x_{1} \propto y_{1}$ is $x_{1}=a, y_{1}=b$,
we have $P\left(x_{1} \propto y_{1}\right)=\frac{2}{3} \cdot \frac{2}{3}=\frac{4}{9}$.
Similarly, $P\left(x_{2} \propto y_{2}\right)=P\left(x_{3} \propto y_{3}\right)=\frac{4}{9}$.
Also, the necessary and sufficient condition for $x_{1}=y_{1}$ is that one transmission is correct and the other is incorrect.
Thus, $P\left(x_{1}=y_{1}\right)$
$$
\begin{array}{l}
=P\left(x_{1}=y_{1}=a\right)+P\left(x_{1}=y_{1}=b\right) \\
=\frac{2}{3} \cdot \frac{1}{3}+\frac{1}{3} \cdot \frac{2}{3}=\frac{4}{9}
\end{array}
$$

Similarly, $P\left(x_{2}=y_{2}\right)=\frac{4}{9}$.
Therefore, $P\left(S_{\mathrm{a}} \propto S_{\mathrm{b}}\right)$
$$
=\frac{4}{9}+\left(\frac{4}{9}\right)^{2}+\left(\frac{4}{9}\right)^{3}=\frac{532}{729} \text {. }
$$

Hence, the required number is 532.