In one version of this question, it is required to compute expected number of rolls to see each side of a dice. In that case, the answer is
$$\frac{6}{6}+\frac{6}{5}+\frac{6}{4}+\frac{6}{3}+\frac{6}{2}+\frac{6}{1}=14.7.$$
I wonder whether the requirement to see each side twice is equivalent to to seeing each side of 12-sided dice at least once, i.e.,
$$\frac{12}{12}+\frac{12}{11}+\frac{12}{10}+\frac{12}{9}+\frac{12}{8}+\frac{12}{7}+\frac{12}{6}+\frac{12}{5}+\frac{12}{4}+\frac{12}{3}+\frac{12}{2}+\frac{12}{1}\approx37.24.$$