Let $\omega$ be a complex cube root of unity with $\omega \neq 1.$ A fair die is thrown three times. If $r_{1},\ r_{2}$ and $r_{3}$ are the numbers obtained on the die, then the probability that $\omega ^{r_{1}}+\omega ^{r_{2}}+\omega ^{r_{3}}=0$ is

For omega raised to the power of r to equal 1, omega squared, or negative omega, r modulo 3 must be 1, 2, or 0 respectively. Therefore, the equation omega^r1 + omega^r2 + omega^r3 = 0 requires exactly one outcome from each of these three residue classes modulo 3. When a fair die is rolled three times, the total number of possible outcomes is 6^3 = 216. Each residue class modulo 3 contains exactly two numbers from 1 to 6, so there are 2C1 times 2C1 times 2C1 = 8 favorable outcomes, which can be permuted in 3! = 6 ways. The probability is 8 times 6 divided by 216, which equals 2/9.

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