TITLE: Probability of zero in a random matrix
QUESTION [31 upvotes]: Let $M(n,k)$ be the set of $n\times n$ matrices of nonnegative integers such that every row and every column sums to $k$.  Let $P(n,k)$ be the fraction of such matrices which have no zero entries, equivalently the probability that a random matrix from the uniform distribution on $M(n,k)$ has no zero entries.
One thing to note is that $$P(n,k)=\frac{|M(n,k-n)|}{|M(n,k)|}$$
(think about subtracting 1 from every entry).  Also note that $P(n,k)$ is the fraction of integer points in the $k$-dilated Birkhoff polytope that lie in the interior.
It seems "obvious" that $P(n,k)$ is a non-decreasing function of $k$. For large enough $k$ it is strictly increasing by Ehrhart theory, but I'd like to see a proof for all $k$.  So the problem is: 
Prove that $P(n,k)$ is a non-decreasing function of $k$ for fixed $n$.
COMMENT: By a result of Stanley (see David Speyer's answer for refs) there are non-negative integers $\{h_i\}$ such that
$$|M(n,k)| = \sum_{i=0}^d h_i \binom{k+i}{d}$$
where $d=(n-1)^2$. I'm wondering if this is enough. Does every polynomial of that form have the desired properties? [Gjergji showed not.]
COMMENT2: The reciprocity theorem for Ehrhart series provides a formula for the number of points in the interior in terms of the number in the whole (closed) polytope. Making use of the above, we find that if the $H_n(k)$ is the polynomial equal to $|M(n,k)|$ for positive integers $k$, then the number of interior points (already identified as $H_n(k-n)$) equals $(-1)^{n+1}H_n(-k)$.  So what we have to prove is that
$$(-1)^{n+1}\frac{H_n(-k)}{H_n(k)}$$
is non-decreasing for integer $k\ge n$. Experimentally, it is not increasing for real $k$ until $k$ is larger.

REPLY [6 votes]: Here are a few more ways to rephrase the problem. Let $A_k$ denote a $kn\times kn$ matrix, composed of $k\times k$ blocks, where the $ij$ block is filled with copies of a standard exponential random variable $\gamma_{ij}$. The $\gamma_{ij}$'s are taken to be independent. A. Barvinok showed that $$M(n,k)=\frac{\mathbb{E}(\operatorname{per}(A_k))}{(k!)^{2n}}$$
and used this to prove that $M(n,k)$ is almost log-concave in $k$. More specifically, he showed in "Brunn-Minkowski inequalities for contingency tables and integer flows",
Adv. in Math., 211 (2007), 105-122, that the following inequality holds
$$\alpha(k)M(n,k)^2\geq M(n,k-1)M(n,k+1)$$
with $\alpha(k)=O(k^n)$. He conjectures that this inequality holds for $\alpha=1$, but as far as I know, this remains open. I'm not sure what's the best upper bound for $$\frac{M(n,k)M(n,n+k+1)}{M(n,k+1)M(n,k+n)}$$ 
that one gets using this analytic approach.

Another way of stating the problem is through RSK, and turn it into an inequality in terms of Kostka numbers. If $\tau(k)$ stands for the partition $(k,k,\dots,k)$, with $k$ parts. Then log-concavity is equivalent to
$$\left(\sum_{\lambda} K_{\lambda \tau(k)}\right)^2\geq \left(\sum_{\lambda} K_{\lambda \tau(k-1)}\right)\left(\sum_{\lambda} K_{\tau(k+1)}\right)$$
and your inequality can be written similarly. One might hope that there is a proof making use of known inequalities between Kostka numbers or known Schur positivity results.
Yet another equivalent formulation comes from Pietro's answer. We need to prove that the coefficient of $(x_1y_1\cdots x_ny_n)^k(z_1t_1\cdots z_nt_n)^{k+1}$ is non-negative in 
$$\frac{\left(z_1t_1\cdots z_nt_n-x_1y_1\cdots x_ny_n\right)}{\prod_{i,j=1}^n(1-x_iy_j)(1-t_iz_j)}.$$

In a different direction, we have $$M(n,k)=\sum_{i=0}^d h_i\binom{k+i}{d}$$ 
where $d=(n-1)^2$. In "Ehrhart polynomials, simplicial polytopes, magic squares and a conjecture of Stanley", C.A. Athanasiadis proved that $(h_0,h_1-h_0,\dots,h_{\lfloor d/2\rfloor}-h_{\lfloor d/2\rfloor-1})$ is a g-vector, so it satisfies the inequalities of Mcmullen's g-theorem. In particular $h$ is a symmetric unimodal sequence. I doubt that this is enough to conclude log-concavity of $M(n,k)$, or your property for that matter, but perhaps it gives some insight on how hard the problem is.