Let $W_t$ be a standard Brownian motion.
I know the arcsin-law, i.e. $P$[$W_t$ has no zeros on $(t_1,t_2)$]=$\frac{2}{\pi}arcsin\sqrt{\frac{t_1}{t_2}}$.
My goal is to compute $P[W_1<0,W_2>0]$.
Now a brief answer says,
$P[W_1<0,W_2>0]=P[W_2>0]-P[W_1>0,W_2>0]=\frac{1}{2}-\frac{1}{4}(1+\frac{2}{\pi}arcsin\sqrt\frac{1}{2})=\frac{1}{8}$
I just cannot figure out how to apply the arcsin-law on it.
Loosely speaking
$$P(W_1 > 0, W_2 > 0) = P(W_1 > 0, \text{no zeros}) + P(W_1 > 0, \text{ at least one zero}, W_2 > 0) \\ = 1/2 P(\text{no zeros}) + 1/4 P(\text{at least one zero}) \\ = 1/2 P(\text{no zeros}) + 1/4 (1- P(\text{no zeros}))$$
should do the trick, where "zeros" always refers to zeros on $(1,2)$.
The idea in the second equality is that once we reach the first zero it is equally likely that $W_2 > 0$ or $W_2 < 0$. So
$$ P(W_1 > 0, \text{ at least one zero}, W_2 > 0) = 1/2 P(W_1 > 0, \text{ at least one zero}).$$ The other identities hold similarly for symmetry reasons.