Distributionally robust Markov decision processes provide a principled framework for sequential decision making under model uncertainty. We study how many samples are necessary and sufficient to learn an $\varepsilon$-optimal robust policy under the average-reward criterion. A generative model provides samples from the nominal transition kernel, whereas policy performance is evaluated over $(s,a)$-rectangular total-variation uncertainty sets of radius at most $σ$. Let $H_0$ and $H_σ$ denote the nominal and robust optimal bias spans, respectively. We identify $σH_0$ as the perturbation scale separating high- and low-tolerance regimes. Our matching upper and lower bounds show that, up to logarithmic factors, the minimax total sample complexity is $$ NSA \asymp \frac{SA}{\varepsilon^2}\begin{cases} \min\{H_0,H_σ\}, & \varepsilon\gtrsimσH_0,\\ \min\{H_0,H_σ\}+σH_σ^2, & \varepsilon\lesssimσH_0. \end{cases} $$ Here $S$ and $A$ are the numbers of states and actions, and $N$ is the number of samples per state-action pair. The sample complexity consists of a linear-span term that resembles the nominal AMDP results and a robustness-specific term that appears only in the low-tolerance regime. We attain these rates using reduction-based plug-in procedures that select the reduction---nominal or robust---and its discount factor: a span-informed procedure that makes these choices using known span parameters, and a span-agnostic procedure that calibrates both choices from data.

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