$P(X > Y) = ?$ when $X,Y \stackrel{iid}{\sim} \text{Bin} \left(n, \frac{1}{2} \right)$

Question

If $X,Y \stackrel{iid}{\sim} \text{Bin} \left(n, \frac{1}{2} \right)$, how do I calculate $P(X^2 > Y^2)$ or $P(X^2 = Y^2)$?

Attempt:

$\begin{aligned} P(X^2 > Y^2) &= P((X-Y)(X+Y) > 0) = P((X-Y) > 0) \cdot P((X+Y) > 0) + P((X-Y) < 0) \cdot P((X-Y) < 0) \\ &= [ \cdots ] \end{aligned}$

How do I calculate $P(X > Y)$? I think if I know this, I'll be able to do the rest of it.

[This is not a homework problem. I was doing a longer question and this part is bothering me for sometime.]

Answer 1

The random variable $n-Y$ has the same law as $Y$, and is independent of $X$ (like $Y$). Hence $(X,n-Y)$ has the same distribution as $(X,Y)$. Moreover, $X+Y$ is Binomial with parameters $2n$ and $1/2$. As a result $$P[X=Y] = P[X=n-Y] = P[X+Y=n] = \frac{{2n \choose n}}{2^{2n}}.$$ Moreover, $(X,Y)$ has the same distribution as $(Y,X)$, so $$2P[X>Y] = P[X>Y] + P[X<Y] = 1-P[X=Y] = 1-\frac{{2n \choose n}}{2^{2n}}.$$

Answer 2

Another approach without complementary probability is simply summing all cases: iterating through all possibilities of $X$ and summing the probability that $Y$ is any number less than each value of $X$. In other words, by independence of $X$ and $Y$, \begin{align*} P(X > Y) &= \sum_{i = 0}^n P(X = i) \cdot P(Y < i) \\ &= \sum_{i = 0}^n P(X = i) \sum_{j = 0}^{i - 1} P(Y = j) \\ &= \sum_{i = 0}^n \binom{n}{i} \frac{1}{2^n} \sum_{j = 0}^{i - 1} \binom{n}{j} \frac{1}{2^n} \\ &= \frac{1}{2^{2n}} \sum_{i = 0}^n \binom{n}{i} \sum_{j = 0}^{i - 1} \binom{n}{j} .\end{align*} By binomial identities, the double sum is $2^{2n - 1} - \binom{2n - 1}{n}$, giving us $$ P(X > Y) = \frac{1}{2} - \frac{1}{2^{2n}} \binom{2n - 1}{n} = \frac{1}{2} - \frac{1}{2^{2n + 1}} \binom{2n}{n}. $$ As Christophe Leuridan points out in the comments, this is also $P(X^2 > Y^2)$ because $X$ and $Y$ are non-negative.