Maximum occupation is not a hitting probability (${\sup }_{t>0}\mathbb{P}(Y_t>\ell )\ne \mathbb{P}({\sup }_{t\ge 0}{Y}_t\ge \ell )$)

For $Y_t=-vt+\sqrt{2D}{B}_t$ with $v,D,\ell >0$ and standard Brownian motion, the one-time occupation probability is $\frac{1}{2}\mathrm{erfc}[(\ell +vt)/\sqrt{4Dt}]$. It is maximized at $t=\ell /v$, giving $\frac{1}{2}\mathrm{erfc}(\sqrt{v\ell /D})$. The actual hitting probability is $e^{-v\ell /D}$. To derive it, solve $D{h}^{\prime \prime }-v{h}^{\prime }=0$ on $(-A,\ell )$ with $h(-A)=0$, $h(\ell )=1$, obtaining $h(0)=(1-e^{-vA/D})/(e^{v\ell /D}-e^{-vA/D})$, and let $A\to \infty$. A path that reached the level and returned contributes to the latter event but not to a given one-time tail.