Note after the first game we have three used and six unused balls.
There are just two cases to have exactly one unused ball left.
3N; 3N; 2N & 1U Three new, three new, then one used with 2 new.
3N; 2N & 1U; 3N Three new, one used with 2 new then three new.
It is easy to find the probability of each of the two.