Example. Counterexample: Equal one-shot success, unequal successful coverage [ftip-007E]
Example. Counterexample: Equal one-shot success, unequal successful coverage [ftip-007E]
Let the evaluation law be uniform on two instances \(x_1,x_2\). Consider protocols \(P\) and \(P'\) with successful-support probabilities
\[ \bigl (p_P(x_1),p_P(x_2)\bigr ) =\left (\frac 12,\frac 12\right ), \qquad \bigl (p_{P'}(x_1),p_{P'}(x_2)\bigr ) =(1,0). \]Both protocols have mean one-shot success \(1/2\). At threshold \(\alpha =1/2\), however, Definition [ftip-005V] gives
\[ \operatorname {Cov}_{1/2}(P)=1, \qquad \operatorname {Cov}_{1/2}(P')=\frac 12. \]Thus average success does not determine successful-support coverage. The example is finite and uses the same task law and threshold for both protocols; it does not compare their training costs or establish acquisition.