Get dice from matrix of win chances. Is it always possible? As an example, consider a set of nontransitive dice 
$D_1: 2, 2, 4, 4, 9, 9$
$D_2: 1, 1, 6, 6, 8, 8$        
$D_3: 3, 3, 5, 5, 7, 7$
On the long run D1 wins vs D2, D2 wins vs D3 and D3 wins vs D1. The win chances are
$P=\left(\begin{matrix}\frac{1}{2}&\frac{5}{9}&\frac{4}{9}\\\frac{4}{9}&\frac{1}{2}&\frac{5}{9}\\\frac{5}{9}&\frac{4}{9}&\frac{1}{2}\end{matrix}\right)$
$P(i,j)$: chance that $D_i$ rolls higher than $D_j$
More general, let $X$ be a nxn-matrix with:
$X(i,j) = 1 - X(j,i)$
e.g. with $n=3$
$X=\left(\begin{matrix}\frac{1}{2}&a&b\\1-a&\frac{1}{2}&c\\1-b&1-c&\frac{1}{2}\end{matrix}\right)$
What I am interested in is:
Is it always possible to create a set of $m$-sided dices labelled with integers ($0\leq x\leq q$), i.e. a $n\times m$-matrix D with 
$0 \leq D(i,j) \leq q < \infty$, $D(i,j)$ is the number on side $j$ of dice $i$
such that $X(i,j)$ is the chance that dice $i$ wins vs dice $j$ (i.e. the chance that $D(i,:) > D(j,:)$)?
In some cases it is rather easy to construct the dice, but in other cases it is not. $X(i,j)$ should be rational. However, apart from that I don't really know how to find conditions on $X(i,j)$ more systematically. 

There are some limits. See Usiskin, 1964 which considers the slightly relaxed problem of finding the distribution of $n$ independent variables to maximize:
$$min\{P(X_1>X_2),P(X_2>X_3),...,P(X_n>X_1)\}$$
The implied limit for $4$ dice is $2/3$ (see table 1) which is achieved by Efron's Dice.
