<s>[INST] What's $\Pr(X=0|X=Y)$? Let $X$ and $Y$ be independent random variables that take values in $\{0,1\}$.
My question is:  how to solve our $\Pr(X=0 | X=Y)$?  We only know that $\Pr(Y=0) = \Pr(Y=1) = 0.5$.

My attempt:
$$\begin{split}
\Pr(X=0|X=Y) &= \frac{
\Pr(X=0)
}{
\begin{split}
 &\Pr(X=0,Y=0)\\
+&\Pr(X=1,Y=1)
\end{split}
}\\
&= \frac{
\Pr(X=0)
}{
\begin{split}
 &\Pr(X=0)\Pr(Y=0)\\
+&\Pr(X=1)\Pr(Y=1)
\end{split}
}\\
&= \frac{
\Pr(X=0)
}{
\begin{split}
 &\Pr(X=0)\Pr(Y=0)\\
+&(1-\Pr(X=0))(1-\Pr(Y=0))
\end{split}
}
\end{split}$$
Then, if I try $\Pr(Y=0)=.5$, I get:
$$\begin{split}
\Pr(X=0|X=Y) &= 
\frac{
\Pr(X=0)
}{\Pr(X=0)\times .5 + (1-\Pr(X=0)) \times .5}\\
 &= 
\frac{
\Pr(X=0)
}{\Pr(X=0)\times .5 + (1 \times .5-\Pr(X=0) \times .5)}\\
 &= 
\frac{
\Pr(X=0)
}{\Pr(X=0)\times .5 + .5-\Pr(X=0) \times .5}\\
 &= 
\frac{
\Pr(X=0)
}{.5}\\
 &= 2\Pr(X=0)\\
\end{split}$$
Then, if $\Pr(X=0)=.6$, the answer is $2\times.6 = 1.2 > 1$.  Madness!

My attempt (NEW):
$$\begin{split}
\Pr(X=0|X=Y) 
&= \frac{\Pr(X=0,X=Y)}{\Pr(X=Y)} \\
&= \frac{\Pr(X=0)\Pr(X=Y|X=0)}{\Pr(X=Y)} \\
&= \frac{\Pr(X=0)\Pr(X=Y|X=0)}{\Pr(X=0)\Pr(Y=0)+\Pr(X=1)\Pr(Y=1)} \\
&= \frac{\Pr(X=0)\Pr(Y=0)}{\Pr(X=0)\Pr(Y=0)+\Pr(X=1)\Pr(Y=1)} \\
\end{split}$$
Plugging $\Pr(Y=0) = \Pr(Y=1) = .5$:
$$\begin{split}
\Pr(X=0|X=Y) 
&= \frac{.5\Pr(X=0)}{.5\Pr(X=0)+.5\Pr(X=1)} \\
&= \frac{.5\Pr(X=0)}{.5(\Pr(X=0)+\Pr(X=1))} \\
&= \frac{.5\Pr(X=0)}{.5(1)} \\
&= \frac{.5\Pr(X=0)}{.5} \\
&= \Pr(X=0) \\
\end{split}$$
 [/INST] 
Let $X$ and $Y$ be independent random variables that take values in
$\{0,1\}.$
We know that $\Pr(Y=0) = \Pr(Y=1) = 0.5\tag2.$
What is $\Pr(X=0 | X=Y) \;?$

Your new attempt is all good.
Note that armed only with the information in the first line but without statement $(2),$ $X$ and $Y$ attaining the same value, i.e., $\{X=Y\},$ and $\{X=0\}$ are not necessarily independent events. (After all, the probability that $Y$ equals $x$ generally does depend on whether $x$ equals $0.)$ Verify this by setting $P(X=0)P(X=Y)=P(X=0=Y).$
These two events are independent precisely when $\{X=0\}$ is impossible or certain or $Y$ is uniformly distributed. Here, statement $(2)$ informs us of the latter.
</s>