Placing $4$ numbers on (-1,1) following some conditions! I want to place 4 numbers, call them $A,B,C$ and $D$ on (-1,1) subject to the following conditions:
(1) $A>B>0>C>D$ i.e. $A$ and $B$ are positive (with $A>B$) while $C$ and $D$ are negative (with $C>D$).
(2) $A-B >C-D$
(3) $ A*C > B*D $
On the space of all placements of numbers satisfying condition (1), how many cases (measure), do also satisfy (2) and (3)?
I failed so far to conceptualize the problem better, if you have intuition about the discrete case, please let me know! Your input will be highly appreciated. Thanks.
 A: Put $A+B=1+x$, $A-B=y$. Then the point $(X,Y)$ is distributed according to area measure in the triangle $\{(x,y)\ |\ 0\leq y\leq 1-|x|\}$; see the following figure.

Similarly, let $(-D)+(-C)=1+u$, $(-D)-(-C)=v$. Then the point $(U,V)$ is equidistributed in the triangle $\{(u,v)\ |\ 0\leq v\leq 1-|u|\}$. 
The restriction $(2)$ amounts to $V\leq Y$. The probability that it is fulfilled is obviously ${1\over2}$, on account of symmetry. In the following we condition on  $(Y,V)$, where $0\leq V\leq Y\leq1$. For a given $y\in[0,1]$ the probability that $y\leq Y\leq y+dy$ is given by $2(1-y)\>dy\ $ (= the area of the dark green strip in the above figure). Therefore the a-priori probability density of $Y$ is $f_Y(y)=2(1-y)$, and that of $v$ is $f_V(v)=2(1-v)$. It follows that the  joint probability density for $(Y,V)$ is given by
$$f(y,v)=8(1-y)(1-v)\qquad(0\leq v\leq y\leq1)$$
and $=0$ elsewhere.

For given admissible $Y$ and $V$ the point $(X,U)$ encoding the remaining two coordinates is equidistributed in the rectangle $R_{y,v}:=[-1+y,1-y]\times[-1+v,1-v]$ of the $(x,u)$ plane, see the above figure. The condition $(3)$ amounts to
$$y(1+u)\leq v(1+x)\ ,$$
and defines a triangle $G_{y,v}\subset R_{y,v}$ of "good" points in $R_{y,v}\>$. Computation shows that
$${{\rm area}(G_{y,v})\over {\rm area}(R_{y,v})}={v(1-y)\over2(1-v)y}=:p(y,v)\ .$$
The overall probability $P$ that a given $(1)$-admissible quadruple $A$, $B$, $C$, $D$ is good is then given by
$$P={1\over2}\int_0^1\int_0^y p(y,v) f(y,v)\ dv\ dy={1\over12}\ .$$
Update: For corroboration I did a simulation with $12\cdot 10^6$ trials satisfying condition $(1)$. I got $1\,000\,718$ good quadruples.
A: Maybe something in the line of: $$\frac{\int_0^1\int_0^x\int_{-1}^0\int_{-1}^z[x-y>z-u][xz>yu]dudzdydx}{\int_0^1\int_0^x\int_{-1}^0\int_{-1}^zdudzdydx}$$  where $[x-y>z-u]$ and $[xz>yu]$ denote the characteristic functions on $(-1,1)^4$ related to the conditions.
