Candy time

There are $7^9$ possible sequences of candies $(c_1, \dots, c_7)$ we can draw.

One type "success" sequence has one color three times; there are 7 choices for that special color. For the remaining six colors, there are 9 places in the sequence for the first color, then 8 places left for the second color, etc. All in all, this makes for $N_1 = 7\times 9\times 8 \times 7 \times 6 \times 5 \times 4$ sequences of this type.

The other type of "success" sequence has two colors two times each; they can appear in the orders AABB, ABAB, and ABBA, giving a total of $7 \times 6 \times 3$ orderings. For the remaining five colors, there are 9 places in the sequence for the first color, then 8 places left for the second color, etc. All in all, this makes for $N_2 = 7\times 6 \times 3 \times 9\times 8 \times 7 \times 6 \times 5$ sequences of this type.

In total, the number of "success" sequences is $N_1 + N_2 = 7 \times 9 \times 8 \times 7 \times 6 \times 5 \times (4 + 6\times 3),$ and the odds therefore are $\frac{N_1+N_2}{7^9} = \frac{9\times 8 \times 6\times 5 \times 22}{7^7} \approx 0.0577.$ Thus the answer is slightly more than $\boxed{5\%}$ .