Let $S$ be the unit sphere in $\mathbb{R}^3$ and $\Theta=\{(x,y,z), x\geq 0, z^2+y^2 \leq \frac{1}{10}e^{-1/x^2}\}$. Try to show that $\exists \delta>0,\forall x\in (-1,0)$,$$ P^x(W_t \text{ hit the unit sphere} \text{ before hitting } \Theta)\geq \delta,$$ where $W_t$ is a $3d$ standard brownian motion and under $P^x$ it starts at $x$.
I have thought about some 2 dimension projection and hitting time of annulus but they seem not to work. I can not get a uniform $\delta$ from the above two approaches. Any ideas?
First, note that if the Brownian motion is started from $(x_0,y_0,z_0)$, then the probability to hit the cylinder $\{y^2+z^2= \delta^2\}$ before the cylinder $\{y^2+z^2=1\}$ equals$$\frac{\log(y_0^2+z_0^2)}{\log \delta^2}.$$ Let $$ \Theta_{\delta;r}:=\{y^2+z^2\leq \delta^2;x\geq r\}. $$ Stopping the Brownian motion at the first time it hits the sphere of radius $r$ around the starting point $(x_0,0,0)$, $-1<x_0<0$, and using the above computation, we see that the probability to hit $\Theta_{\delta,r}$ before $S$ is bounded above by$$C\cdot\frac{\log r}{\log\delta},$$with some absoulute constant $C$. Now, the cusp $\Theta$ is contained in the union of the semi-cylinders $$\Theta_{\frac{1}{10}\exp(-{2^{2n-2}});2^{-n}},$$for $n\geq 1$. The union bound on the corresponding events gives a convergent series, thus the probability to hit the tip of the cusp is bounded away from $1$.
Remark. This paper here seems to address some aspects of your problem.