Probability that a Brownian motion with drift hits +1 before hitting -1 before time 1, and similar events [Searched for this question and found something similar, but not identical. I hope this is not a duplicate. I don't know if this is a very easy question, I took my stochastic processes course too much time ago.]
Let $(B_t)_{0 \leq t \leq 1}$ be a standard Brownian motion and consider the process (for $0 \leq t \leq 1$, $\mu \in \mathbb{R}$, $\sigma > 0 \in \mathbb{R}$)
$$
X_t = \mu t + \sigma B_t
$$
Consider the two stopping times 
$$\tau_1 = \inf \left\{ 0 \leq t \leq 1: X_t =1 \right\}$$ 
and 
$$\tau_{-1} = \inf \left\{ 0 \leq t \leq 1: X_t = -1 \right\} $$
How can I compute the following quantities?
$$
\mathbb{P}(\tau_1 \leq \tau_{-1} \leq 1)
$$
$$
\mathbb{P}(\tau_{-1} \leq \tau_1 \leq 1)
$$
$$
\mathbb{P}( \left\{ \tau_1 > 1 \right\} \cap \left\{\tau_{-1} > 1\right\})
$$
 A: Using renewal theory (please ask in the comments if you want some details), you can show that if $u(x,t)$ is the probability for $\sqrt{2}B_t$ to hit $1$ before $0$ given a time budget of $t$ starting at $x$, then $u$ satisfies the IBVP:
$$u_t= u_{xx} \\
u(x,0)=\begin{cases} 0 & x \in [0,1) \\
1 & x=1 \end{cases} \\
u(0,t)=0 \\
u(1,t)=1.$$
The idea of this argument is to consider a Markov process $(t,\sqrt{2}B_t)$ and wait for it to hit the boundary of $[0,1]^2$ starting at some point $(0,x)$. Notably, the initial condition in this problem is not continuous, so the sense of any solution to it must be somewhat weak.
The solution to this IBVP is
$$u(x,t)=x+\sum_{k=1}^\infty \frac{(-1)^k}{\pi k} e^{-k^2 \pi^2 t} \sin \left ( k \pi x \right ).$$
You are welcome to translate this solution to $[-1,1]$. I don't know any simpler expression for this. Notably however, this solution actually happens to capture precisely the type of discontinuity in the initial data that we wanted in this case.
The answer to (a modification of) your question is $u(x,1)$. Notably, with these numbers, by $t=1$ only the first correction term is really significant anymore; $\frac{e^{-4 \pi^2}}{2 \pi}$ is about $10^{-18}$, and thereafter the series is converging faster than a geometric series with common ratio $e^{-4 \pi^2}$. So a good approximation of the solution here is just $x-\frac{e^{-\pi^2}}{\pi} \sin(\pi x)$. If you tweak the numbers you might need more terms; for example, with an interval of length $2$ and a standard Brownian motion, the eigenvalues are $-k^2 \pi^2/8$, so you'll want at least two correction terms, perhaps three.
You can imitate this argument to handle the general case with constant drift and diffusion; you will have $u_t=\mu u_x + \frac{1}{2} \sigma^2 u_{xx}$ with the same boundary conditions.
