Number of ternary strings of length n such that number of 0s is greater than or equal to number of occurrences of any other digit I understand how to count this for a binary string of a fixed length using combinations, so I think the way to go with this problem is to use an exponential generating function for each of the set {0, 1, 2} when counting the solutions. For example, if I want to count the number of ternary strings with an even number of 0s, we can use (1 + x^2/2! + x^4/4! + x^6 + ...) for 0's, (1 + x + x^2/2! + x^3/3! + ...) for 1's, and (1 + x + x^2/2! + x^3/3! + ...) for the number of 2's, and then we can combine exponential generating functions in the following way:
Exponential Generating Functions with odd num of 0's
Not sure how I would account for more 0's than any other term though. Thank you!

I don't think generating functions give the best approach.  You should exploit the symmetry of the situation instead.  The are $3^n$ ternary strings of length $n$.  In how many of them is $0$ a winner, that is, in how many of them are there at least as many $0$'s as $1$'s or $2$'s?  If we count all the winners, $\frac13$ of them will be $0$, $\frac13$ of them will be $1$, $\frac13$ of them will be $2$, by symmetry.
The only problem is that some strings have $2$ or $3$ winners.  Therefore, the problem reduces to counting the number of two-way ties and three-way ties.
For example, let $n=3$.  There are $27$ strings.  In $6$ of them there is a three-way tie.  There are no two-way ties.  In $21$ cases there is a single winner.  The total number of winners is $21+3\cdot6=39$, so $0$ comes in  first (including ties) $$\frac{39}3=13$$ times.
Can you do the general case?  (Note that three-way ties are possible only when $n$ is divisible by $3$.
