Imagine an experiment which outputs either $A$ or $B$ or $C$. $A$ happens with the probability $p_a$, $B$ happens with probability $p_b$ and $C$ happens with probability $p_c$. Here, $p_a > \frac{1}{2}$ and $p_b > p_c$ and $p_a + p_b + p_c = 1$. After running the experiment $n$ times, what is the probability of having $n_a$ times $A$ and $n_b$ times event $B$ such that $n_a<n_b$.
What I tried is the following:
$Pr[n_a < n_b] \leq \sum_{i=0}^{n_b-1}\binom{n}{i}p_{a}^i(1-p_{a})^{n-i}\leq \sum_{i = 0}^{n_b-1}\binom{n}{i}p_{a}^n$ since $p_a > p_c + p_b$. Then, I could not upper bound this with a nice formula which is what I need and also I am not sure if it is the best approach or correct.
In your solution you completely disregard the event $C $. The correct computation shall be: $$ \sum_{n_c=0}^{n-1}\sum_{n_a=0}^{\left\lceil\frac {n-n_c}2\right\rceil-1}\frac {n!}{n_a!n_c! (n-n_a-n_c)!}p_a^{n_a}p_b^{n-n_a-n_c}p_c^{n_c}. $$ I do not think the expression can be simplified in general case.