I know that the following sum $\sum_{i=0}^{n} C_{n}^{i} \cdot (1-p)^i \cdot p^{n-i}$ is equal to 1. It is also not hard (but tedious) for me to calculate
$$\sum_{i=0}^{n/2} C_{n}^{i} \cdot (1-p)^i \cdot p^{n-i}$$
where $n$ is the odd number and $p > 0.5$. Just to make sure I wrote the formula correct, here is an example of what I want to calculate for n=5 and p=0.7:
$$C_{5}^{0} \cdot 0.7^5 \cdot 0.3^0 + C_{5}^{1} \cdot 0.7^4 \cdot 0.3^1 + C_{5}^{2} \cdot 0.7^3 \cdot 0.3^2$$
I am almost sure that there is an easy way to estimate this sum (for a big number of $n$), but I can't find it.