Is it true that P(X|Y)=P(Z|Y) implies P(X) = P(Z)? Something feels intuitively off about this but I can't find a way to algebraically manipulate it to be false. 
Thanks :)
 A: Using conditional probability,
$$P(X\mid Y) = \frac{P(X \cap Y)}{P(Y)}$$
So, if $P(X\mid Y) = P(Z\mid Y)$,  then
$$ P(X\cap Y) = P(Z\cap Y)$$
This leads to $P(X) = P(Z),$ only if $Y$ is independent of both $X$ and $Z$. Because, this leads to
$$P(X)\cdot P(Y) = P(Z) \cdot P(Y)$$
$$P(X) = P(Z)$$
I hope this make sense to you. 
A: This is not true in general. Contradiction is one way to solve this. Here is an example:
Let's toss a coin that we don't know if it is fair ($P($Head$)=p=\frac{1}{2}$) or biased with $p$ something different than $\frac{1}{2}$, e.g. $p=\frac{2}{3}$. And, call the events $X$,$Y$,$Z$ as follows: 
$X$: Toss comes up Head, $Y:$ Coin is Fair, i.e. $p=\frac{1}{2}$, $Z$: Toss comes up Tails. Also, let $P(Y)=\frac{1}{2}$ to ease things.
We know that $P(Z|Y)=P(X|Y)=\frac{1}{2}$. But, using total probability theorem, 
$P(Z)=P(Z|Y)P(Y)+P(Z|Y')P(Y')=\frac{1}{2}\frac{1}{2}+\frac{1}{3}\frac{1}{2}=\frac{5}{12}$
Similarly, 
$P(X)=\frac{1}{2}\frac{1}{2}+\frac{2}{3}\frac{1}{2}=\frac{7}{12}$. So, they're not equal.
