Let $A$, $B$ and $C$ be disjoint events and let $X$ be an event. I wanted to show that
$$\text{if}\quad P(X|A)>P(X|B)>P(X|C)\quad\text{then}\quad P(X|A\cup B)>P(X|A\cup C)\,,$$
but I found a counter-example myself ($P(X|A)=\frac{10}{14}$, $P(X|B)=\frac{14}{25}$, $P(X|C)=\frac{2}{5}$).
However, I find the incorrectness of the statement very counter-intuitive. If I was to choose two out of the three ($A$, $B$ and $C$) in order to maximize the probability of getting an $X$ (when multiple attempts are allowed if the unchosen letter occurs), intuitively, I would choose $A$ and $B$.
Can anybody explain why the statement is not correct or explain under what circumstances it is? Thank you!
Let's say $\mathbb P (A) = 0.6, \mathbb P(B) = 0.3, \mathbb P(C) = 0.1$ and A,B,C are disjoint. Meanwhile define $\mathbb P(X\cap A) = 0.55$, $\mathbb P(X\cap B) = 0.2$, $\mathbb P(X\cap C) = 0.05$. Hence clearly, $\mathbb P(X|A) = \frac {11}{12} > \mathbb P(X|B) = \frac 23 > \mathbb P(X|C) = \frac 12$. However, $\mathbb P (X|A\cup B) = \frac 56 < \mathbb P(X|A\cup C) = \frac 67$.
I guess the intuition would be, $A,B, C$ may itself have different probabilities, therefore when $\mathbb P(X|A\cup C) $ could be arbitrarily close to $\mathbb P(X|A) $ as long as $A$ has much larger probability comparing with $C$. Similarly, $\mathbb P(X|A\cup B) $ could be arbitrarily close to $\mathbb P(X|B) $, which may give you the 'counter-intuitive' result.
This might also be similar to Simpson's Paradox https://en.wikipedia.org/wiki/Simpson%27s_paradox.