6. (1985 American High School Mathematics Examination) Suppose non-zero digits are selected in such a way that the probability of selecting the digit $d$ is $\lg (d+1)-\lg d$. What is the set of digits from which the probability of selecting the digit 2 is $\frac{1}{2}$ of the probability of selecting a digit from the following sets?
A. $\{2,3\}$
B. $\{3,4\}$
C. $\{4,5,6,7,8\}$
D. $\{5,6,7,8,9\}$
E. $\{4,5,6,7,8,9\}$
6. C. Reason: The probability that the number 2 is selected according to the problem is
$$
\lg (2+1)-\lg 2=\lg \frac{3}{2},
$$

Notice that the probability of the numbers in set C being selected is
$$
\begin{array}{l}
\lg \frac{5}{4}+\lg \frac{6}{5}+\lg \frac{7}{6}+\lg \frac{8}{7}+\lg \frac{9}{8} \\
=\lg \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6} \cdot \frac{8}{7} \cdot \frac{9}{8}=\lg \frac{9}{4}=\lg \left(\frac{3}{2}\right)^{2}=2 \lg \frac{3}{2},
\end{array}
$$

Thus, the probability of the numbers in set C being selected is twice the probability of 2 being selected.