Given a polynomial $p(x\_1,\ldots, x\_k)$ in $k$ variables with maximum degree $n$, and $x\_1,\ldots, x\_k \in [0,1]$. Suppose $\max\_{x \in [0,1]^k} p(x) = 1$, can we get an upper bound on the probability $\mathbb{P}(|p(x)| < \epsilon)$ where we treat the product space $[0,1]^k$ as a natural probability space? Using techniques from [this](http://www.its.caltech.edu/~kaloshin/research/dioph-SO3.pdf) paper: 
one can show that the above probability is small provided $\epsilon$ is of tower exponential order $e^{-e^n}$. But I need $\epsilon$ to be of poly-exponential order, i.e., $e^{-(nk)^c}$. Looking at the polynomial $x\_1^n x\_2^n \ldots x\_k^n$, one deduce by central limit theorem that $\mathbb{P}(|p(x)| < \epsilon)$ is small provided $\epsilon = \mathcal{O}(e^{-kn})$. But I can't prove that this polynomial has the the highest probability of staying near zero. 


If needed, one can also bound $\epsilon$ in terms of the total degree, i.e., the sum of degrees of all the monomial terms in $p$. 



bounding the probability that a polynomial is near 0
I believe that the result you want follows from the results in [this very cool paper.](http://arxiv.org/pdf/math/0108212v2) (see, in particular page 9). There are probably improvements since then...


