$X \sim Bin(n,1/2), n-X \sim Bin(n,1/2)$

Solution 1:

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.

Solution 2:

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.