A $p$-Biased Centered Random Walk never leaves [-1,1] wp. $\ge e^{-np(1-p)}$. Let $X, X_1, \dots, X_n$ be iid. $\sim \text{Bernoulli}(p)$.
I can show that with probability
$\ge e^{-n\min\{ \frac{p}{1-p} , \frac{1-p}{p} \}}$
the centered random walk
$S_j = \sum_{i=0}^{j}(X_i-p)$
stays in the interval $[-1,1]$ at all times.
That is, for all $j\in\{0,\dots,n\}$ we have $-1\le S_j\le 1$.
From experiments, it seems that this can be improved to $$e^{-np(1-p)}, $$
and perhaps something even better as $n\to\infty$.
In the picture below I have plotted $\log\Pr[\text{survival}] / n$ for various $n$ and $p$.

I wonder if this is a well-known result? My proof also isn't that elegant, so I would love to see other ways to prove it.
It is relatively easy to get the bound $\exp(-n H(p))$ as well, where $H(p)=p\log\frac1p + (1-p)\log\frac1{1-p}$ is the entropy function.
I also tried using a coupling to Brownian motion, but that caused a constant factor loss in the exponent, which seems unavoidable.
The cases where $p$ is a simple fraction are intuitively easier.
The case $p=1/2$ is quite simple, we get $\Pr[\text{survival}]=(3/4)^{\lfloor n/2\rfloor - 1}$.
Perhaps one needs to use a combinatoric approach, considering other $p\in\mathbb{Q}$ like that.
It seems pretty hairy to carry out though...
 A: One heuristic approach: If you assume $p$ is irrational, the interval $|S - p i| ≤ 1$ always contain exactly two values $S$. Let $a_n$ and $b_n$ be the probability of being at the low/high position at time $n$, then
$$
(a, b)_n =\begin{cases}
a_{n-1}(1-p), & a_{n-1} p + b_{n-1}(1-p) & \text{if } \lfloor np\rfloor = \lfloor (n-1)p\rfloor
\\
a_{n-1}p + b_{n-1}(1-p), & b_{n-1}p & \text{if } \lfloor np\rfloor = \lfloor (n-1)p\rfloor+1
\end{cases}$$
with $a_0=1$, $b_0=0$.
I don't know how to solve that recurrence exactly, but we may notice that $\lfloor np\rfloor = \lfloor (n-1)p\rfloor+1$ happens at a fraction $p$ of the time-steps. Thus we can approximate the recurrence as
$$
\begin{align}
(a, b)_n =
(1-p)[&a_{n-1}(1-p), & a_{n-1} p + b_{n-1}(1-p)]
\\+p[&a_{n-1}p + b_{n-1}(1-p), & b_{n-1}p].
\end{align}
$$
This we can solve, and we get $a_n+b_n = (1-p(1-p))^n$, which is just a bit smaller than $\exp(-np(1-p))$.
I wonder if you can prove the "approximate" recurrence solution is always below the "real" one.
