You want an agent to complete a 10 step task with at least 90% end to end success rate. All steps are independent. What minimum per step accuracy p must each step achieve to meet this target?
End-to-end success is p^N for independent steps. To clear 90% over 10 steps you need the tenth root of 0.9, about 98.9% per step.
Imagine a relay race where 10 runners each have to pass a baton without dropping it. The team finishes cleanly only if every single handoff works. If each runner is 90% reliable, that sounds great, but the chance all ten succeed is 0.9 multiplied by itself ten times, which is only about 35%. To get the whole team to a 90% finish rate, each handoff has to be almost perfect. Working backwards, each runner needs to succeed about 99 times out of 100. The lesson is that long chains punish small per-step mistakes brutally, because the errors multiply instead of adding. That is why people prefer fewer, more reliable steps over many shaky ones.
Concept explanation~2 min read
Everything you need to truly understand this topic: intuition, mechanics, step by step explanation, code, formulas, and worked example. Click to expand.
Concept explanation~2 min read
Everything you need to truly understand this topic: intuition, mechanics, step by step explanation, code, formulas, and worked example. Click to expand.
This question probes whether you internalise the single most important piece of arithmetic in agent engineering: errors across a multi-step trajectory compound multiplicatively, not additively. The scenario gives you a 10-step task, an independence assumption, and a 90% end to end target, then asks for the minimum per-step accuracy.
The setup is deliberately a trap for intuition. A 90% per-step success rate sounds like a strong engineer's number. But over ten independent steps it collapses to roughly 35% end to end. Recovering a 90% whole-task target requires each step to clear nearly 99%. Understanding why, and what it implies for system design, is the difference between a candidate who has shipped agents and one who has only read about them.
Why successes multiply for independent steps
An agent task succeeds only if every step succeeds. There is no partial credit in the model: one failed tool call, one misparsed observation, one bad branch, and the whole trajectory is counted as a miss. When the steps are independent, the joint probability of all of them succeeding is the product of their individual success probabilities. This is the basic multiplication rule for independent events, and it is the entire engine behind cascading failure in agents.
If each of the N steps shares the same success probability p, that product becomes p multiplied by itself N times. So the end to end reliability is a single clean expression in p and N. The shape of that expression matters more than any single number: because p is below one, raising it to a higher power always shrinks the result, and it shrinks faster the more steps you add.
The key word is independent. It means the outcome of one step does not change the probability of any other. That assumption is what lets us collapse N separate probabilities into one power. In a real agent it is rarely exactly true, but it is the right first model: it is simple, it is conservative in the common case, and it isolates the variable that dominates everything else, which is the number of steps. We will return to how realistic the assumption is, but for the arithmetic it gives us exactly the formula we need.
\text{reliability} = p_1 \cdot p_2 \cdots p_N = p^{N} \quad \text{when all } p_i = pSituations where this technique stops working.
2–4 min · Everything important, quickly.
Real products, models, and research that use this idea.
- SWE-bench coding agents show this directly: editing a file, running tests, and fixing failures across many steps means a single bad tool call tanks the whole trajectory's pass rate.
- Anthropic's guidance in Building Effective Agents argues for the simplest viable workflow, which is exactly the step-count lever the p^N math implies.
What an interviewer would ask next. Try answering before peeking at the approach.
QIf steps have different success rates, how does the end to end formula change?
Replace p^N with the product of each step's p_i. The geometric mean of those rates plays the role of the single equivalent p, so the weakest steps dominate the product.
Red flags & common mistakes
The phrases that signal junior thinking. Click to expand.
Red flags & common mistakes
The phrases that signal junior thinking. Click to expand.
Treating per-step errors as additive instead of multiplicative. A 90% step rate does not give 90% over ten steps; it compounds down to roughly 35%.
60 second bullets to scan on the way to the call.
Why end to end success is the product of per-step success rates for independent steps.
Write the reliability formula as p to the power N.
Primary sources. Browse if you want the original framing.
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