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Equation 3 · Part 2 · Measuring Tool Protocols and the Model Context Protocol: Evidence, Benchmarks, and Uncertainty

Symbol p^k

Pr⁡[all k attempts succeed]=pk.\Pr[\text{all } k \text{ attempts succeed}] = p^{k}.
pkp^{k}

What this part means

pkp^k is an input to the expression that computes the quantity on the left.

Its job in the formula

pkp^k is an input to the expression that computes the quantity on the left.

The passage around this formula

That gap is large enough to be worth writing down explicitly. If a task’s outcome were an independent Bernoulli draw with a fixed success probability p equal to the reported average, the probability that all k independent attempts at the same task succeed would be Pr⁡[all k attempts succeed]=pk\Pr[\text{all } k \text{ attempts succeed}] = p^{k}. At p = 0.60 and k = 8 , that model predicts roughly 1.7%. The reported figure — under 25% — sits well above that naive prediction, and the direction of the gap is informative on its own: it is only possible if outcomes are not independent draws from one fixed probability, but rather reflect a task population that splits into instances the agent reliably solves and instances it reliably does not, with the…

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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Sources cited in the article section

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