Reaching one before time one

Idea. Use the symmetry of Brownian motion after it first hits the level. Paths that hit $1$ and end below it are matched with paths that end above it.

Step 1: split. Let $\tau$ be the first time $W$ hits $1$. Then $$\begin{array}{rl}\mathbb{P}(\tau \le 1)& =\mathbb{P}(\tau \le 1,W_1\ge 1)\\ & +\mathbb{P}(\tau \le 1,W_1<1).\end{array}$$

Step 2: reflect. After $\tau$, the process $W_{\tau +s}-1$ is a Brownian motion independent of the past (strong Markov property), so it is as likely to end above $0$ as below. The two terms are equal. And $W_1\ge 1$ already implies $\tau \le 1$, so the first term is $\mathbb{P}(W_1\ge 1)$.

Step 3. $$\begin{array}{rl}\mathbb{P}(\tau \le 1)& =2\mathbb{P}(W_1\ge 1)=2(1-\Phi (1))\\ & \approx 2\times 0.15866=\mathrm{0.3173.}\end{array}$$

In general. $\mathbb{P}({\max }_{t\le T}{W}_t\ge a)=2\mathbb{P}(W_T\ge a)$ for $a>0$: the running maximum has the same law as $|W_T|$.