When Does Redundant Reasoning Pay? A Measurable Condition for Error-Tolerant Reasoning in Language Models

Language models are often made more reliable by having them re-derive their own intermediate results. This redundancy can repair a corrupted step, but it also costs accuracy when nothing goes wrong, so it sometimes helps and sometimes hurts, and it is unclear in advance which. We model a reasoning chain as a sequence of noisy steps in which both a brittle procedure (sharp, which states each result once) and a redundant one (flat, which re-derives the previous result before using it) can repair a corrupted step. The model has three parameters, each measured directly on the language model: the two procedures' per-step recovery rates $c_s$ and $c_f$, found by injecting errors into correct reasoning traces, and the per-step accuracy cost of redundancy $\kappa$. We prove that the redundant procedure can overtake the brittle one if and only if $c_f>\kappa c_s$, and give in closed form the per-step error rate beyond which it does. The usual assumption that a brittle chain never recovers ($c_s=0$) makes this condition trivially true, yet on six of our seven task families the brittle procedure repairs 10% to 63% of corrupted steps unaided. On nine settings spanning three models and two chain lengths, eight of them modular arithmetic, predictions from the measured parameters alone, with nothing fitted to the crossover, match the observed crossover with $R^2=0.839$ (mean absolute error $0.021$). The condition also agrees with all six full noise sweeps, including a pre-registered prediction of no crossover that held. In practice this is a check to run before spending compute on redundancy: a few short measurements say whether it can pay off at all, a 64-problem screen flags tasks where it cannot because the model's errors recur at every step, and the same parameters bound the sample size needed to detect the gain. Code: https://anonymous.4open.science/r/reasoning-noise-budget-DF74.