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Equation 12 · How Benchmark Contamination Actually Works in Agentic Evaluation

What does this equation mean?

passdeterministick=pass1for every k.\text{pass}^k_{\text{deterministic}} = \text{pass}^1 \quad \text{for every } k .

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Inputs and operationspass^1 quad for every k
Result or conditionpass^k_deterministic
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This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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kdeterministick_{\text{deterministic}}

Symbol k_deterministic

kdk_deterministic is part of the quantity the equation computes from the expression on the right.

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kk

Symbol k

k is part of the quantity the equation computes from the expression on the right.

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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subscript

subscript

The lower label selects a particular version, component, or indexed member of the quantity. For example, x₀ and xₜ can be values at different positions.

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superscript

superscript

A raised number can be a power. When it is a label or bound, it selects a case or the upper limit of a sum; the formula’s structure distinguishes these uses.

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How to interpret it

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What the article says around this equation

averaged over tasks to produce the benchmark’s headline number. The metric is designed to punish an agent whose competence is real but inconsistent across resampled trials — a stochastic policy with true per-trial success probability p has passk\text{pass}^k ≈\approx pkp^k , which falls quickly as k grows. But a policy whose output on a given task is deterministic — an empty response, always, regardless of sampling — produces the identical transcript on every trial, so passdeterministick=pass1for every k\text{pass}^k_{\text{deterministic}} = \text{pass}^1 \quad \text{for every } k . A degenerate exploit is not merely invisible to pass@1; it is more invisible to passks^k, precisely because passks^k was built to reward consistency, and a fixed, checker-satisfying non-answer is the most consistent…
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averaged over tasks to produce the benchmark’s headline number. The metric is designed to punish an agent whose competence is real but inconsistent across resampled trials — a stochastic policy with true per-trial success probability p has passk\text{pass}^k ≈\approx pkp^k , which falls quickly as k grows. But a policy whose output on a given task is deterministic — an empty response, always, regardless of sampling — produces the identical transcript on every trial, so passdeterministick=pass1for every k\text{pass}^k_{\text{deterministic}} = \text{pass}^1 \quad \text{for every } k . A degenerate exploit is not merely invisible to pass@1; it is more invisible to passks^k, precisely because passks^k was built to reward consistency, and a fixed, checker-satisfying non-answer is the most consistent thing a policy can produce. The metric engineered to catch the difference between competence and luck cannot, on its own, tell competence apart from a policy that never varies because it never tries.

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