[Combinatorics]
Problem: A die, with the numbers $1,2,3,4,6$, and 8 on its six faces, is rolled. After this roll, if an odd number appears on the top face, all odd numbers on the die are doubled. If an even number appears on the top face, all the even numbers are halved. If the given die changes in this way, what is the probability that a 2 will appear on the second roll of the die?

Solution: There are only two possibilities on the first roll - it can either be even or odd.

Possibility 1 'The first roll is odd'

The probability of an odd outcome on the first roll is $\frac{1}{3}$.

After doubling all the numbers, the possible outcomes on the second roll would now be 2, 2, 6, $4,6,8$ with the probability of a 2 being $\frac{1}{3}$.

Thus the probability of a 2 on the second roll would be $\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}$.

Possibility 2 'The first is even'

The probability of an even outcome on the first roll is $\frac{2}{3}$.

After halving all the numbers, the possible outcomes on the second roll would be 1, 1, 3, 2, 3, 8 .

The probability of a 2 on the second die would now be $\frac{1}{6}$.

Thus the probability of a 2 on the second roll is $\frac{2}{3} \times \frac{1}{6}=\frac{1}{9}$.

The probability of a 2 appear on the top face is $\frac{1}{9}+\frac{1}{9}=\frac{2}{9}$.

Final answer: $\frac{2}{9}$