I am considering the following problem:
Given the following equation: \begin{equation*} c = \sum_{k=n}^{2n-1} \binom{2n-1}{k} p(c)^k (1-p(c))^{2n-k-1} \end{equation*}
Which is the probability that total number of success is larger than half of total events ($2n$).
I would like to know when $n$ goes to infinity, what's asymptotic behavior of $p(c)$.
What we know is that for any finite $n$, $p(0)=0,p(1)=1$. My guess is that $p(c)$ converges to 1 if $c>0.5$ and converges to 0 if $c<0.5$. But I'm not sure how to prove it.
Thanks in advance!
Let $S_n$ be a binomial($2n,p$) random variable, then $$\mathbb{P}(S_n>n)=\mathbb{P}({S_n/n}>1).\tag 1 $$ By the weak law of large numbers, $S_n/n\to 2p$ in probability, so the probability in (1) converges to one if $2p>1$, and it converges to zero if $2p<1$.