Distribution of $X=\binom m Y \left(\frac 1 2\right)^Y \left(\frac 1 2\right)^{m-Y}$ where $Y$ is a Binomial variable? If $X=\dbinom m Y \left(\dfrac 1 2\right)^Y \left(\dfrac 1 2\right)^{m-Y}$, and $Y$ has a Binomial distribution with parameters $m$ and $p$, What can we say about the distribution of $X$?
 A: If $Y \sim \text{Binomial}(m,p)$, then its pmf is $f_Y(y;p) = {m \choose y} p^y(1-p)^{m-y}$. You are transforming $Y$ by applying this pmf to it, and you're using the value $p=1/2$. So
$$
X = f_Y(Y;.5),
$$
which means it has support or takes on values in the range of all possible probabilities for $Y$'s pmf when using $p=.5$. You could easily find this range for any specific $m$ by evaluating the pmf of $Y$ on the values $0,1,\ldots,m$.
To find the probabilities of each one of these support values $x$, notice that 
$$
P(X=x) = P(f(Y;.5) = x) = P(Y = f^{-1}(x;.5)) = {m \choose f^{-1}(x;.5)}p^{f^{-1}(x;.5)}(1-p)^{f^{-1}(x;.5)}
$$ 
where $f^{-1}$ is the pre-image (the inverse might not always exist).
Edit: Thanks to @DilipSarwate and @jbownman for pointing the following out:
Since we're using the parameter $.5$ to transform $Y$ into $X$ in this case, the transformation
$$
f(y;.5) = {m \choose y} .5^y.5^{m-y} = {m \choose y} .5^m = {m \choose m-y} .5^m = f(m-y;.5)
$$
is symmetric for $0 \le y \le m$. 
This means
\begin{align*}
P(X=x) &= P(f(Y;.5) = x) \\
&= P(Y = k) + P(Y = m-k)
\end{align*}
as long as $k < m-k$ is such that $f(k;.5) = x$. This only doesn't happen when $m$ is even and $k=m/2$. In that case, there is only one $k$ such that $f(k;.5) = x$, which means 
\begin{align*}
P(X=x) &= P(f(Y;.5) = x) \\
&= P(Y = k).
\end{align*}
This is the case that @jbownman mentioned: the case where the inverse exists.
