Law of total probability and absolute value paradox I am trying to reconcile a difference between two methods of finding $\mathbb{P}(|X|\geq y)$ for some RV $X$ and some $y > 0$.
Method 1
Consider the following line of reasoning:
\begin{align*}
&|X| \geq y\\
\iff& (X > 0\text{ and }X \geq y)\text{ or } (X < 0\text{ and } -X\geq y)\\
\iff& (X \geq y)\text{ or } (-X\geq y) &\text{since $y>0$}\\
\iff& (X \geq y) \text{ or } (X \leq -y)
\end{align*}
Note that on the last line, the two events are disjoint. This leads us to believe
$$
\mathbb{P}(|X| \geq y) = \mathbb{P}(X\geq y) + \mathbb{P}(X\leq -y)
$$
Method 2
We have that $\{\{X > 0\}, \{X \leq 0\}\}$ is an event space. Then by the law of total probability, we have
\begin{align*}
\mathbb{P}(|X|\geq y) &= \mathbb{P}(|X|\geq y\:\vert\:X>0)\,\mathbb{P}(X>0) + \mathbb{P}(|X|\geq y\:\vert\:X\leq 0)\,\mathbb{P}(X\leq 0)\\
&= \mathbb{P}(X\geq y)\,\mathbb{P}(X > 0)+\mathbb{P}(-X\geq y)\,\mathbb{P}(X\leq 0)\\\\
\mathbb{P}(|X|\geq y)&= \mathbb{P}(X\geq y)\,\mathbb{P}(X>0)+\mathbb{P}(X\leq-y)\,\mathbb{P}(X\leq 0)
\end{align*}
It seems that these two methods should always give the same answer, however they only coincide when further assumptions about the probabilities are made.
What is the resolution here?
 A: You do not always have 
$\mathbb{P}(|X|\ge y | X > 0) = \mathbb{P}(X\ge y)$.
Instead, you have 
$\mathbb{P}(|X|\ge y | X > 0)\mathbb{P}(X > 0) = \mathbb{P}(X \ge y)$.
When you replace that in your formula (and similarly for the negative), you get exactly the same result as Method 1.
A: Since in general
$$ \Bbb P(|X|\ge y\mid X>0)\ne \Bbb P(|X|\ge y\land X>0),$$
your method 2 is incorrect
A: $P(|X|\geq y | X>0) \ne P(X\geq y)$ 
Consider the problem where X=2 or X=-2 with probability $\frac 12$ and $y=1$:
$P(|X|\geq y | X>0)  $ means : what is the probability of $|X|>1$ given that $X>0$. Answer : 1.
$P(X\geq y)  $ means : what is the probability of $|X|>1$. Answer : $\frac 12$.
In $P(A|B)$, conditioning by B tells you the event B has happened. It is "deterministic". It truncates your universe. This event can not be consider at the same level as A which is still random and you want to access the probability that A happens in your smaller universe defined by B. You can not intersect them and simplify the probability. What I try to explain in Layman words is the Bayes formula.
Indeed, $P(|X|\geq y \cap X>0) = P(X\geq y)$  is true and I think it corresponds to the reasoning you did. 
