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Compute end to end success for a 6 hop multi-agent chain at 92 percent per step

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A multi-agent workflow runs as a chain of six agents, each independently correct on its own step 92 percent of the time. Assume independence across hops. Compute the end to end success probability of the full chain. Express as a decimal rounded to 4 places.
TL;DR

End to end success is 0.92^6 = 0.6064. A 92 percent per-step rate compounds to 61 percent over six hops, the canonical multi-agent reliability tax.

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Easy to grasp

Imagine six friends passing a glass of water down a line. Each one is pretty careful and only spills 8 times out of 100 attempts. Sounds safe, right? But each spill is a fresh chance, so by the time the sixth friend gets the glass, the cup has been at risk six separate times. The chance the water survives the whole journey is not 92 percent, it is 92 percent multiplied by itself six times, which lands around 61 percent. Long handoff chains punish each link's failure rate harder than people expect. The fix is fewer hops, a quick re-check at each step, or accepting you will need to redo runs.

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.

Compounding error is the cheapest argument against long multi-agent chains and the single hardest intuition to internalise. The interview question hides a calculation that takes ten seconds and a lesson that reshapes how you design agent systems for the rest of your career.

The arithmetic is 0.92^6 ≈ 0.6064. The lesson is that a per-step rate everyone agrees feels safe (92 percent) collapses into an end to end rate everyone agrees is unacceptable (61 percent) once you chain six agents in series. This is not a quirk of the numbers chosen; it is the geometry of independent multiplication.

The headline: 92 percent per step. 61 percent end to end. The gap is the whole topic.

The arithmetic, walked carefully

Step by step

Independent events multiply. The probability that six independent steps each succeed is the product of their per-step success probabilities. With a uniform per-step rate of p = 0.92:

P(end to end)=p6=0.92×0.92×0.92×0.92×0.92×0.92P(\text{end to end}) = p^6 = 0.92 \times 0.92 \times 0.92 \times 0.92 \times 0.92 \times 0.92

Doing this without a calculator:

  • 0.92^2 = 0.8464
  • 0.92^3 = 0.8464 x 0.92 = 0.7787
  • 0.92^4 = (0.92^2)^2 = 0.8464 x 0.8464 = 0.7164
  • 0.92^6 = 0.92^4 x 0.92^2 = 0.7164 x 0.8464 = 0.6064

Rounded to four places: 0.6064. As a percentage: about 60.6 percent.

Why intuition gets this wrong

People hear 92 percent and anchor to 'almost always works'. The chain has six fresh opportunities to fail, and each one is independent, so the 8 percent failure compounds. A useful sanity check: the failure probability of the chain is 1 - 0.92^6 = 0.3936, or roughly 40 percent. Four runs in ten fail somewhere along the line. That number is the one product owners actually feel.

Where the independence assumption breaks
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Real products, models, and research that use this idea.

  • Devin's agent traces showed long planner-executor chains hitting compounding error past 5 hops, prompting tighter sub-task scoping.
  • AutoGen's max_turns guard exists partly to bound the compounding-error tax on chatty workflows.
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What an interviewer would ask next. Try answering before peeking at the approach.

QIf you add a critic at every hop that catches 70 percent of upstream errors, what is the new end to end success?
A

The effective per-step rate becomes 0.92 + 0.08 x 0.7 = 0.976, so end to end ≈ 0.976^6 ≈ 0.866. Walk the interviewer through why the critic improves the per-step number rather than acting as a multiplier on top of the original chain math.

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Red flags & common mistakes

The phrases that signal junior thinking. Click to expand.

Most common mistake

Adding the failure rates (0.08 x 6 = 0.48) instead of multiplying success rates. Addition gives 52 percent failure; multiplication gives 39 percent failure. The two are not the same.

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60 second bullets to scan on the way to the call.

  • Why independent step probabilities multiply rather than add

  • How to compute 0.92^6 step by step without a calculator

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