<s>[INST] Are there $a,b,c,d \in \mathbb N$ such that $\frac{a + b}{a + b + c + d} < \frac{a}{a + c} < \frac{b}{b + d}$? Consider the following $2 \times 2$ contingency table:
\begin{array}{c|cc|c}
 & C & \overline C & \Sigma \\ \hline
V & 4000 & 3500 & 7500 \\
\overline V & 2000 & 500 & 2500 \\ \hline
\Sigma & 6000 & 4000 & 10000
\end{array}
In data mining, we informally say that the association rule $C \to V$ is misleading. We could justify this by observing that:
$$
\Pr[V \mid C] = \frac{4000}{6000} = 0.\overline{6} < 0.75 = \frac{7500}{10000} = \Pr[V]
$$
Alternatively, we could justify this by observing that:
$$
\Pr[V \mid C] = \frac{4000}{6000} = 0.\overline{6} < 0.875 = \frac{3500}{4000} = \Pr[V \mid \overline C]
$$
I'm trying to figure out which justification is more compelling. When I change the numbers in the contingency table, both inequalities always seem to be simultaneously satisfied. I can't seem to make one inequality true and the other false.

More generally, consider the contingency table:
\begin{array}{c|cc|c}
 & C & \overline C & \Sigma \\ \hline
V & a & b & a+b \\
\overline V & c & d & c + d\\ \hline
\Sigma & a + c & b + d & a + b + c + d
\end{array}

Question: Do there exist $a,b,c,d \in \mathbb N$ such that either one of the following two inequalities are satisfied:
  \begin{align*}
\frac{a + b}{a + b + c + d} &< \frac{a}{a + c} < \frac{b}{b + d} \tag{1} \\
\frac{b}{b + d} &< \frac{a}{a + c} < \frac{a + b}{a + b + c + d} \tag{2}
\end{align*}

 [/INST] None of the two double-inequalities can hold for any $a,b,c,d\in\Bbb R^+$.
$$\frac{a + b}{a + b + c + d} < \frac{a}{a + c} \iff (a+b)(a+c)<a(a+b+c+d)$$
$$\iff \frac{a}{b}>\frac{c}{d}$$
The same way you get:
$$ \frac{a}{a+c} < \frac{b}{b + d} \iff \frac{a}{b}<\frac{c}{d}$$
The two inequalities contradict each other.
Use the same method for the other double-inequality.
</s>