<s>[INST] Probability that a $(n, \frac12)$-binomial random variable is even Let $X$ be a $(n, \frac12)$-Binomial RV. Show that $X$ is even with probability $\frac12$.
 [/INST] A more general question might be "What is the probability that a $(n,p)$-Binomial random variable $X$ is even?"  Generalizing on my hint on the main question,
$$
\begin{align*}
\left((1-p) + p\right)^n &= \sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}\\
&= \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}p^{2k}(1-p)^{n-2k} + 
\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k+1}p^{2k+1}(1-p)^{n-(2k+1)}\\
&= P\{X~\text{even}\} + P\{X~\text{odd}\}
\end{align*}
$$
and 
$$
\begin{align*}
\left((1-p)-p\right)^n &= \sum_{k=0}^n \binom{n}{k}(-p)^k(1-p)^{n-k}\\
&= \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}p^{2k}(1-p)^{n-2k} - 
\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k+1}p^{2k+1}(1-p)^{n-(2k+1)}\\
&= P\{X~\text{even}\} - P\{X~\text{odd}\}
\end{align*}
$$
giving that
$$P\{X~\text{even}\} = \frac{1}{2}(1 + (1-2p)^n) 
= \frac{1}{2} + \frac{1}{2}(1-2p)^n$$
which is $\frac{1}{2}$ if $p = \frac{1}{2}$ and converges to $\frac{1}{2}$
as $n \to\infty$ except when $p = 1$, in which case the probability alternates
between $0$ and $1$ depending on the parity of $n$ (as is obvious without
all the above calculation).
</s>