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Published equation contexts

pat least one of k=1−(1−p)k,pall k succeed=pkp_{\text{at least one of } k} = 1 - (1-p)^k, \qquad p_{\text{all } k \text{ succeed}} = p^k

Why this formula appears here

The distinction matters because the two metrics behave very differently even under the most favourable possible assumption — that trials are statistically independent. Write p for the probability that a single trial on a given task succeeds. Under independence, pat least one of k=1−(1−p)k,pall k succeed=pkp_{\text{at least one of } k} = 1 - (1-p)^k, \qquad p_{\text{all } k \text{ succeed}} = p^k. The first quantity rises quickly toward one as k grows, which is the intuitive but misleading sense in which “just try again” seems to fix unreliability. The second quantity falls quickly toward zero, and it is the quantity that matters if a task must be completed correctly on a specific occasion rather than merely being solvable in principle across several attempts.

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pat least one of kp_{\text{at least one of } k}

Symbol p_at least one of k

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

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pall k succeedp_{\text{all } k \text{ succeed}}

Symbol p_all k succeed

pap_all k succeed is one of the signed contributions combined to compute the quantity on the left.

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Published contexts (1)

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pat least one of k=1−(1−p)k,pall k succeed=pk.p_{\text{at least one of } k} = 1 - (1-p)^k, \qquad p_{\text{all } k \text{ succeed}} = p^k.

Equation 2 · AI Agents & Systems

Measuring AI Agent Architectures: Evidence, Benchmarks, and Uncertainty

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

The distinction matters because the two metrics behave very differently even under the most favourable possible assumption — that trials are statistically independent. Write p for the probability that a single trial on a given task succeeds. Under independence, pat least one of k=1−(1−p)k,pall k succeed=pkp_{\text{at least one of } k} = 1 - (1-p)^k, \qquad p_{\text{all } k \text{ succeed}} = p^k. The first quantity rises quickly toward one as k grows, which is the intuitive but misleading sense in which “just try again” seems to fix unreliability. The second quantity falls quickly toward zero, and it is the quantity that matters if a task must be completed correctly on a specific occasion rather than merely being solvable in principle across several attempts.

Meanings in this article

  • pp: the write.
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