16. A die is a uniform cube, with the numbers $1,2,3,4,5,6$ engraved on its 6 faces. Now, on the table, there are 3 dice made of wood, bone, and plastic, respectively. Repeat the following operation until there are no dice left on the table: roll all the dice on the table, then remove those showing an odd number. Find the probability that the number of times the above operation is performed is more than three.
16. First consider the probability of completing the operation in at most three attempts, where:
(1) The probability of getting 3 odd numbers on the first attempt is $\left(\frac{1}{2}\right)^{3}$ (the operation is completed in just one attempt);
(2) The probability of getting 2 odd numbers on the first attempt is (the operation is completed in two or three attempts)
$$
\mathrm{C}_{3}^{1}\left(\frac{1}{2}\right)^{2} \times \frac{1}{2}\left[\frac{1}{2}+\left(\frac{1}{2}\right)^{2}\right] \text {; }
$$
(3) The probability of getting 1 odd number on the first attempt is (the operation is completed in two or three attempts)
$$
\begin{array}{l}
\mathrm{C}_{3}^{1} \times \frac{1}{2}\left(\frac{1}{2}\right)^{2}\left[\left(\frac{1}{2}\right)^{2}+\mathrm{C}_{2}^{1}\left(\frac{1}{2}\right)^{2} \times \frac{1}{2}+\right. \\
\left.\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{2}\right] ;
\end{array}
$$
(4) The probability of getting 3 even numbers on the first attempt is (the operation is completed in two or three attempts)
$$
\begin{array}{l}
\left(\frac{1}{2}\right)^{3}\left[\left(\frac{1}{2}\right)^{3}+\mathrm{C}_{3}^{1}\left(\frac{1}{2}\right)^{2} \frac{1}{2} \times \frac{1}{2}+\right. \\
\left.\mathrm{C}_{3}^{1} \times \frac{1}{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{3}\right]
\end{array}
$$

In summary, the probability of completing the operation in at most three attempts is
$$
\frac{2^{6}}{2^{9}}+\frac{9 \times 2^{4}}{2^{9}}+\frac{27 \times 2^{2}}{2^{9}}+\frac{27}{2^{9}}=\frac{343}{512} \text {. }
$$

Therefore, the probability of completing the operation in more than three attempts is
$$
1-\frac{343}{512}=\frac{169}{512} \text {. }
$$