Without loss of generality, let's assume the duplicated number and color of these cards are $1$ and red, respectively. In order for the qualified event to happen, one of these cards must be a red $1$ because Sharon should be able to eliminate the duplicate number and color by removing a single card.

If one card is red-$1$, then the other red color can only be assigned to $2$ to $7$ which has $6$ choices. Afterwards, the remaining $6$ colors can be assigned arbitrarily. Hence, the number of qualified cases equals $6\times 6!$.

The number of all possible cases can be calculated by adding the number of qualified cases (i.e. the above) and the number of unqualified cases. The later equals $\binom{6}{2} \times 6!$. This is because if no card is red-$1$, then the two reds can match of two of the other six numbers and then the other $6$ colors have no restriction.

Therefore, the final answer is $$\frac{6\times 6!}{\binom{6}{2}\times 6! + 6\times 6!}=\boxed{\frac{4}{9}}$$