Problem 598

When a dot is removed from a face with an even number of dots, that face then has an odd number of dots.
When a dot is removed from a face with an odd number of dots, that face then has an even number of dots.
Initially, there are 3 faces with an even number of dots and 3 faces with an odd number of dots.
If a dot is removed from a face with an even number of dots, there are then 4 faces with an odd number of dots and 2 faces with an even number of dots. This means that the probability of rolling an odd number after a dot is removed is $\frac{4}{6}$ in this case.
If a dot is removed from a face with an odd number of dots, there are then 2 faces with an odd number of dots and 4 faces with an even number of dots. This means that the probability of rolling an odd number after a dot is removed is $\frac{2}{6}$ in this case.
Since there are $2+3+4+5+6+7=27$ dots on the faces, then the probability that a dot is removed from the face with 2 dots is $\frac{2}{27}$, from the face with 3 dots is $\frac{3}{27}$, and so on.
Thus, the probability that a dot is removed from the face with 2 dots and then an odd number is rolled is the product of the probabilities, which is $\frac{2}{27} \cdot \frac{2}{3}$, since there are now 4 odd faces and 2 even faces.
Similarly, the probability that a dot is removed from the face with 3 dots and then an odd number is rolled is $\frac{3}{27} \cdot \frac{1}{3}$.
Continuing in this way, the probability of rolling an odd number after a dot is removed is $\frac{2}{27} \cdot \frac{2}{3} + \frac{3}{27} \cdot \frac{1}{3} + \frac{4}{27} \cdot \frac{2}{3} + \frac{5}{27} \cdot \frac{1}{3} + \frac{6}{27} \cdot \frac{2}{3} + \frac{7}{27} \cdot \frac{1}{3}$.
This equals $\frac{2}{3}\cdot(\frac{2}{27} + \frac{4}{27} + \frac{6}{27}) + \frac{1}{3}(\frac{3}{27} + \frac{5}{27} + \frac{7}{27}) = \frac{2}{3} \cdot \frac{12}{27} + \frac{1}{3} \cdot \frac{15}{27} = \frac{8}{27} + \frac{5}{27} = \frac{13}{27}$.