This is a continuation of Number of drifted Brownian motions that never hit zero under allocation

For each $n\ge 1$, consider $X^i_t=1+\beta t + W^i_t$ for $i=1,\ldots n$ and $t\ge 0$, where $\beta>0$ and $(W^i_t)_{t\ge 0}$ are independent Brownian motions. $\phi\equiv \big((\phi^1_t)_{t\ge 0},\ldots, (\phi^n_t)_{t\ge 0}\big)$ is called an allocation strategy if every $(\phi^i_t)_{t\ge 0}$ is progressively measurable w.r.t. the Brownian filtration $\big(\mathcal F_t:=\sigma(W^1_s,\ldots, W^n_s, s\le t)\big)_{t\ge 0}$,

$$\phi^i_t\ge 0 \quad\mbox{ and }\quad \sum_{i=1}^n\phi^i_t\le 1,\quad \forall t\ge 0.$$

Denote

$$X^{\phi,i}_t:=X^i_t+\int_0^t \phi^i_sds \quad \mbox{and} \quad \tau^{\phi}_i:=\inf\{t\ge 0: X^{\phi,i}_t\le 0\}.$$

Let $S^{\phi}_n:=\sum_{1\le i\le n}{\bf 1}_{\{\tau^{\phi}_i=\infty\}}$ be the number of $X^{\phi,i}$ that never hits zero. Clearly,

$$\frac{\mathbb E[S^{\bf 0}]}{n}~=~\mathbb P[X^1_t>0, \forall t\ge 0]~=~1-e^{-\beta},$$

where $\bf 0$ stands for the strategy with $\phi^i\equiv 0$ for $i=1,\ldots, n$. Can we can show

$$\lim_{n\to\infty}\frac{\mathbb E[S^{\phi}]}{n}~~=~~1-e^{-\beta}$$

for all the strategies $\phi$? Any answers, comments or references are highly appreciated!