I am aware that $P(A\cap B)$ is less than $P(B)$. I have already seen Probability Question: Would A always have a greater chance of $A\cap B$?.
My issue is regarding the conjunction fallacy and order of events. Below is a common example of the fallacy, where logically $2$ should be more likely than $1$ as $P(A\cap B)$ is less than $P(B)$:
1. The Soviet Union will attack Poland and then the USA will cut off diplomatic relations with it.
2. The USA will cut off diplomatic relations with the Soviet Union.
Now assume A is the attack event and B is the event cutting off diplomatic relations.
The problem here is that looking at $P(A\cap B)$ in this way doesn't seem to take order of events into account. Won't the posterior probability $P(B|A)$ (in situation $1$) increase dramatically as compared to the prior $P(B)$ (situation $2$)?
Now if the probability of an attack, $P(A)$ is high as well (let's assume it is for mathematical purposes) this should be enough to make $P(A)\times P(B|A)$ more likely than $P(B)$ alone.
But this argument justifies the fallacy. So where is the logical error, or is this example invalid?