A set of intervals $I = \{ I_1, I_2, \dots, I_n \}$ forms a simple chain if, for every $2\leq i \leq n-1$, interval $I_i$ overlaps only with $I_{i-1}$ and $I_{i+1}$. We show that a deterministic memoryless one-directional revoking algorithm achieves a competitive ratio of $2(1 - 1/\sqrt{e}) \approx 0.786$ on the simple chain in the random order model, hence performs worse than the basic greedy algorithm without revoking that has a competitive ratio of $(1 - 1/e^2) \approx 0.864$, but better than any deterministic revoking algorithm in the adversarial model that has a competitive ratio of at most $0.75$. The proof of the latter also leads to a lower bound of $n/4$ for the advice complexity.

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