Let $S_n$ be a symmetric, simple random walk starting at $0$. Find $$ P(30 \leq \max(S_k)_{k \leq 100} < 40) $$
So basically I want to find the probability that a simple random walk of length 100 at least once touches the line of 30 and does not cross or touch the line of 40.
Could anyone give me a hand or solve this? I wish I could show my attempt but there isn't much to show.
(Edited. Sorry for the calculation mistakes committed earlier.)
You can apply the reflection principle to find out the probability $$\mathbb{P}\left(\max_{1\le i\le 100} S_i < M,\ S_{100} = 2k\right),$$ where $2k < M.$ Consider it's complement, i.e. the event where it touches the line $y=M.$ Take the part till the first time it touches that line, and reflect it in that line, so that $(0,0)$ goes to $(0, 2M).$ Now you can easily see that $$\mathbb{P}\left(\max_{1\le i\le 100} S_i \geq M,\ S_{100} = 2k\right) = 2^{-100}\binom{100}{50+k-M}.$$ Here we have $2k<M$ and we need $k\geq M -50$ for the binomial coefficient to be positive (let's assume $M>0,$ then we would have $k\geq M-50>-50.$)
Now take $M=2\ell.$ Hence $$\mathbb{P}\left(\max_{1\le i\le 100} S_i < 2\ell\right)=\sum_{-50\le k<M/2}\mathbb{P}(S_{100}=2k) - \sum_{M-50\leq k < M/2} - \mathbb{P}\left(\max_{1\le i\le 100} S_i \geq 2\ell,\ S_{100}=2k\right)$$ which can be evaluated as $$2^{-100}\left[\sum_{k=-50}^{\ell-1} \binom{100}{50+k} -\sum_{k=M-50}^{\ell-1} \binom{100}{50+k-M}\right].$$ Using this, you can evaluate $\mathbb{P}(30\leq \max_{1\le k\le 100} S_k < 40).$