Let $(X_n)_{n\geq 1}$ be an i.i.d. sequence of variables with unequal Bernoulli distribution : $P(X=0)=p, P(X=1)=1-p$ with $p\gt \frac{1}{2}$ (so getting a $0$ is more likely than getting a $1$).
Let $S_n=\sum_{k=0}^{n} X_k$, and let $E_n$ be the event $S_{2n+1} \leq n$ (thus $E_n$ says that there are strictly more occurrences of the likeliest outcome).
I have two questions : it seems "very clear" that $P(E_n) \to 1$ when $n\to\infty$, but how to show it ? Also, can we evaluate how fast $P(E_n)$ converges to $1$ ?
My thoughts : it seems one cannot express the question in terms of tail events and apply the usual machinery of the laws of large numbers. Also, the complement of $E_n$ is defined by $S_{2n+1} \gt (2n+1)x_n$ where $x_n=\frac{n}{2n+1}$. If $(x_n)$ was a constant sequence, we could apply Cramer's theorem.
The variable $S_{2n+1}$ has mean $(2n+1)p > n$ and is a sum of i.i.d. variables taking values in $[0,1]$, so by a Chernoff bound, $$ \Pr[S_n \le n] \le \exp\left(-\frac{((2n+1)p - n)^2}{2(2n+1)p}\right) $$ and now some algebra answers all your questions. The expression on the right-hand side looks complicated, but it is approximately $e^{-\lambda n}$ for $\lambda = \frac{(2p-1)^2}{4p}$.