An unbiased chance of a quiet minute

Idea. The total count is sufficient for $\lambda$. Conditioning the crude estimator on it averages away the noise, and given the total, the calls are spread over the minutes like balls thrown at random into 5 boxes.

Step 1: given the total. Given $S=s$, the counts of the 5 minutes are multinomial: each of the $s$ calls falls in minute 1 with probability $\frac{1}{5}$, independently.

Step 2: condition the indicator. $$\mathbb{E}\left[1\{X_1=0\}\mid S=s\right]={\left(1-\frac{1}{5}\right)}^s={\left(\frac{4}{5}\right)}^s.$$ By the Rao-Blackwell and Lehmann-Scheffé theorems, ${\left(\frac{4}{5}\right)}^S$ is the unbiased estimator of $e^{-\lambda }$ with the smallest variance.

Step 3: the data. With $S=10$: $${\left(\frac{4}{5}\right)}^{10}\approx \mathrm{0.107.}$$

Compare. The maximum likelihood estimate plugs in $\hat{\lambda }=\frac{10}{5}=2$ and gives $e^{-2}\approx 0.135$; it is biased. One can check unbiasedness directly: $\mathbb{E}\left[(4/5{)}^S\right]=e^{-5\lambda (1-4/5)}=e^{-\lambda }$, using the generating function of the Poisson($5\lambda$) total.