Drawing balls of an urn, probability of one colour run out (with replacement)

Since the problem statement does not say how many times you can remove and replace balls, presumably the question is about the likelihood that you can keep on doing this forever without running out of any color.

If the numbers of balls of each color are $(n - 2, n, n + 2)$ at the start, then it will always be true that there is at least one color with $n$ balls or fewer. (Actually there will always be a color with $n-1$ balls or fewer, but that's a stronger statement than we need so I won't take the trouble to prove it.)

If there are $n$ or fewer balls of one color, there is at least an $\frac{2^nn!}{(3n)^n}$ probability that the next $n$ moves will reduce the number of balls of that color to zero. Hence there is at most a $1 - \frac{2^nn!}{(3n)^n}$ probability that all three colors are still in the urn after $n$ moves.

Now consider the probability you still have all three colors after $2n$ moves, or after $3n$ moves.

With these considerations you should be able to put an upper bound on the expected number of moves until one color has run out. More to the point, you should be able to put a lower bound on the probability that you run out of one color in some finite number of moves.