A probability question from sociology We know that $\frac{1}{2} \leq a \leq p \leq 1$. And, $n \geq 3$ is a positive odd number, and $t$ is an integer. $a$ satisfies the equation below.
\begin{equation} \small
\begin{aligned}
&\sum_{t=0}^{n-1}  \left(  {n-1 \choose t} [p a^{t} (1-a)^{n-t-1}+(1-p) (1-a)^{t} a^{n-{t}-1}] \cdot [\frac{a^{t+1} (1-a)^{n-t-1}}{a^{t+1} (1-a)^{n-t-1}+(1-a)^{t+1} a^{n-t-1}}-\frac{a^{t} (1-a)^{n-t}}{a^{t} (1-a)^{n-t}+(1-a)^{t} a^{n-t}}]   \right) \\[15pt]
&={n-1 \choose {\frac{n-1}{2}}} a^{\frac{n-1}{2}}(1-a)^{\frac{n-1}{2}}  (2p-1)
\end{aligned}
\end{equation}
We want to prove that $a$ converges to $\frac{1}{2}$.
Any references or remarks are appreciated!
 A: If I expand the left-hand-side of your equation around $a=1/2$ I find
$$
\sum_{t=0}^{n-1}    {n-1 \choose t} [p a^{t} (1-a)^{n-t-1}+(1-p) (1-a)^{t} a^{n-{t}-1}] $$
$$\times\left[\frac{a^{t+1} (1-a)^{n-t-1}}{a^{t+1} (1-a)^{n-t-1}+(1-a)^{t+1} a^{n-t-1}}-\frac{a^{t} (1-a)^{n-t}}{a^{t} (1-a)^{n-t}+(1-a)^{t} a^{n-t}}\right]   $$
$$=\sum_{t=0}^{n-1}\left(a-\frac{1}{2}\right) 2^{2-n} \binom{n-1}{t}+{\cal O}(a-1/2)^3=2a-1+{\cal O}(a-1/2)^3.$$
I similarly expand the right-hand-side,
$${n-1 \choose {\frac{n-1}{2}}} a^{\frac{n-1}{2}}(1-a)^{\frac{n-1}{2}}  (2p-1)=2^{1-n} (2 p-1) \binom{n-1}{\frac{n-1}{2}}-\left(a-\tfrac{1}{2}\right)^2 2^{2-n} (n-1) (2 p-1) \binom{n-1}{\frac{n-1}{2}}+{\cal O}(a-1/2)^4.$$
Equating left-hand-side and right-hand-side I solve for $a$,
$$a=\frac{1}{2}+\frac{2^n \left(\sqrt{2^{3-2 n} (n-1) (1-2 p)^2 q^2+1}-1\right)}{4(n-1) (2 p-1) q},\;\;q=\binom{n-1}{\frac{n-1}{2}}.$$
For $n\gg 1$ this solution tends to
$$a\rightarrow\frac{1}{2}+\frac{1}{2\sqrt{2n}}\frac{\sqrt{16 (p-1) p+\pi +4}-\sqrt{\pi }}{2 p-1}.$$
