Notice that the order in which the people leave the line is irrelevant. Give each right-facing person a weight of 1, and each left-facing person a weight of -1. We claim the answer for some arrangement of these $2n$ people is -2 times the minimum prefix sum. For instance: $$\text{LRRLRLLLRRRRL} \rightarrow(-2)(-2) \rightarrow 4$$ $$\text{RRRLLRLLLRRL} \rightarrow(-2)(-1) \rightarrow 2$$ Proof. The final configuration is always of the form $$\underbrace{\mathrm{LL} \ldots \mathrm{LL}}_{k} \underbrace{\mathrm{RR} \ldots \mathrm{RR}}_{k}$$ and the minimum prefix sum is invariant. As the final configuration has minimum prefix sum is $k$, we are done. So, we want to find the expected value of the minimum prefix sum across all such strings of 1s and -1s. To find this, we will instead compute the equivalent value $$\sum_{k=1}^{\infty} \operatorname{Pr}[\text{maximum prefix sum is } \geq k]$$ Consider the $k$th term of this sum, and the corresponding walk from $(0,0)$ to $(2n, 0)$ with L corresponding to a step of $(1,-1)$ and R corresponding to a step of $(1,1)$. Consider the point $P$ at $y=k$ with minimal $x$-coordinate, and reflect the remainder of the walk across $y=k$. This gives a path that ends at $(2n, 2k)$. Noting that this is a bijection between walks from $(0,0)$ to $(2n, 2k)$ and walks that reach $y=k$, we have $$\begin{aligned} \sum_{k=1}^{\infty} \operatorname{Pr}[\text{maximum prefix sum is } \geq k] & =\sum_{k=1}^{\infty} \frac{\binom{2n}{n-k}}{\binom{2n}{n}} \\ & =\frac{1}{2}\left[\left(\sum_{k=-\infty}^{\infty} \frac{\binom{2n}{n-k}}{\binom{2n}{n}}\right)-1\right] \\ & =\frac{1}{2}\left(\frac{2^{2n}}{\binom{2n}{n}}-1\right) \end{aligned}$$ Adjusting for the factor of 2 we saved at the beginning, our final answer for $n=50$ is $\frac{2^{100}}{\binom{100}{50}}-1$.