$X \sim Bin(n,1/2), n-X \sim Bin(n,1/2)$ Alice flips a fair coin $n$ times and Bob flips another fair coin $n+ 1$ times, resulting
in independent $X \sim Bin(n,1/2)$ and $Y \sim Bin(n+1,1/2)$. Is it true that $P(X < Y) = P(n-X < n+1-Y)$
I said the answer is no because if $X,Y$ count the number of tails, then $n-x,n+1-Y$ count the number of heads.
So, $P(X < Y) \neq P(n-X < n+1-Y)$ because if $n =8$ and $X=5,Y=8$, then $P(5<8) \neq P(3 < 1)$
But, as seen here (Simple Probability Distribution Question) this is wrong.
 A: Since $X$, $Y$ and $n$ are all integers, you have the event equivalence:
$$\begin{align}
X < Y 
&\quad \iff \quad n+X < n+Y \\[6pt]
&\quad \iff \quad n-Y < n - X \\[6pt]
&\quad \iff \quad n+1-Y \leqslant  n - X \\[6pt]
&\quad \iff \quad \neg (n-X < n+1-Y), \\[6pt]
\end{align}$$
so you have:
$$\mathbb{P}(X < Y) = 1- \mathbb{P}(n-X < n-Y+1).$$
If you combine this result with your conjectured equation you get:
$$\mathbb{P}(n-X < n-Y+1) = \mathbb{P}(X < Y) = 1- \mathbb{P}(n-X < n-Y+1),$$
which would imply that:
$$\mathbb{P}(X < Y) = \mathbb{P}(n-X < n-Y+1) = \frac{1}{2}.$$
So, it remains to check if this equation holds.
A: 
So, $P(X < Y) \neq P(n-X < n+1-Y)$ because if $n =8$ and $X=5,Y=8$, then $P(5<8) \neq P(3 < 1)$

To the point, you're mixing up equality of probabilities of events with equality of the events themselves.  That is, while $A=B$ always implies $P(A) = P(B)$, we may have $A\neq B$ while still having $P(A)=P(B)$.  This isn't a contradiction, so showing $A\neq B$ does not show us $P(A) \neq P(B)$.
In your case, you have two events:

*

*$X < Y$

*$n-X < n+1-Y$
These are not the same events (in fact, they're complementary).  That is, setting $n,X,Y$ may make one event occur while the other doesn't occur, but importantly, this doesn't impact whether these two events are equally likely.  While we can consider a particular value of the parameter $n$, once we choose particular values of $X$ or $Y$, we're no longer talking about probabilities; we're talking about events.
