An agent pipeline has 10 sequential steps. Each step succeeds independently with probability 0.9. Assuming no recovery mechanism, what is the end to end probability that all 10 steps succeed?
Independent step successes multiply, so a 10-step agent at 0.9 per step ends at 0.9^10, which is about 0.349, roughly 35% end to end success.
Imagine a relay race where ten runners each have to hand off a baton. Each runner drops the baton one time in ten, so each one succeeds nine times out of ten. You might guess the team usually finishes fine. But every single handoff has to work for the baton to cross the line. The chance the whole chain survives is nine-tenths multiplied by itself ten times, not nine-tenths once. That comes out to about one in three. So even though each runner looks reliable on their own, the team finishes cleanly only about a third of the time. Long agent chains feel the same pain: each tool call seems trustworthy, yet stacking ten of them turns a small per step error into a big end to end failure rate.
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 looks like arithmetic, but it is really a test of whether you understand how cascading failure works in multi-step agents. The agent runs ten sequential steps, each succeeds independently with probability 0.9, and there is no recovery. The task succeeds only if every single step succeeds, so we are asking for the probability that an entire chain of ten links holds at once.
The instinct many people have is to answer 0.9, or to average something, or to treat a 90% step rate as proof the agent is basically reliable. All of those are wrong. The correct mental model is a chain: it holds only if every link holds, and the probability that every link holds is the product of the individual link probabilities. A 90% per step rate is the reliability of one link, not the reliability of the whole chain, and conflating the two is exactly the mistake the question is designed to catch.
Getting the number right is the easy part. The valuable part is understanding why the answer is so much lower than intuition expects, what it implies for the per step targets you must hold, and what it forces you to do when you design real agent loops that run for many steps.
Why the probabilities multiply
Each step is an independent trial that either succeeds or fails, a Bernoulli trial with success probability p. The whole task is an AND across all ten of those events: step one AND step two AND step three, all the way to step ten. For independent events combined with AND, the joint probability is the product of the individual probabilities, which is the multiplication rule for independent events. Independence is the key assumption: it lets us say the chance step two succeeds does not change based on whether step one did.
That is why you multiply 0.9 by itself ten times rather than averaging or adding. Averaging would give 0.9, which silently assumes a single representative step rather than ten that all must hold. Adding failure rates would give one minus a linear term, which wildly underestimates the damage and can even exceed one for long chains, which is nonsense for a probability. The product is the only operation consistent with the requirement that all steps succeed together, and it is the operation interviewers expect you to reach for instantly.
The shape of the result is exponential decay in the number of steps. Each additional step multiplies in another factor below one, so the survival probability shrinks geometrically rather than linearly as the chain gets longer. A handy way to feel this: taking the logarithm turns the product into a sum, so log-reliability falls by a fixed amount per step. Equal steps subtract equal log-probability, which is precisely what exponential decay means, and it is why the curve looks gentle at first and then collapses.
P(\text{all succeed}) = \prod_{i=1}^{N} p_i = 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: resolving an issue needs many correct edits and test runs in a row, so a high per-action rate still yields a much lower full-task resolve rate.
- Anthropic's guidance on building effective agents pushes for the fewest steps that work, precisely because each added tool call multiplies another sub-one factor into the success product.
What an interviewer would ask next. Try answering before peeking at the approach.
QIf you add one retry per step, how does the effective per step success rate and the end to end rate change?
A retry succeeds unless both attempts fail, so the effective rate becomes 1 minus the failure squared. For p equal to 0.9 that is 0.99, and raising 0.99 to the tenth recovers about 90% end to end.
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.
Averaging the per step rate or quoting 0.9 as the answer. Independent successes multiply, so the end to end rate is 0.9 raised to the number of steps, not 0.9 itself.
60 second bullets to scan on the way to the call.
State why independent step successes multiply rather than average.
Write the reliability formula as p to the power of N.
Primary sources. Browse if you want the original framing.
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