If $X$ is a random variable from binomial distribution $Bin(n,p)$, then $$P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$$ where $p$ is the probability of one success. The expected value of random variable $X$ is $EX = n\cdot p$.
I wonder, how "close" is the expected value, to the value of $X$ which appears with the biggest probability. I checked for some random values, and (it seems that it is not very close) but I am looking for more theoretical or general explanation, how I can express how close are this to values.
I believe you mean the mode of the distribution, which is indeed not equal to the mean. You can find the equation here.