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*}
No. In general, for $a,b,c,d\in \Bbb N$ with $\frac{a}{b} < \frac{c}{d}$, $$\frac{a}{b} <\frac{a+c}{b+d}<\frac{c}{d}.$$ Proof: The left inequality is equivalent to $$a(b+d) < b(a+c) \iff ad < bc \iff \frac{a}{b}<\frac{c}{d}.$$ Likewise for the right inequality.