A Scoring Set That Covers the Statistic's Support Has Power Exactly Zero: A Pre-Data Screen, and the Null Distribution of a Ladder with a Free Exponent

A test whose scoring set covers the support of its test statistic has power exactly zero and returns a p-value of exactly 1, whatever figure it reports. This paper states that rule, proves it in one line, and gives the screen: compute the scoring set's measure under the null before reporting any result, which costs one loop and no data. Three constructions are worked through. A test that scores a word count n against φ·round(n/φ) at a tolerance of one word cannot fail, because |x - round(x)| never exceeds 1/2 for any real x, so the deviation never exceeds φ/2 = 0.80902; 200,000 of 200,000 random word counts pass, and the quantity compared against the count is a function of the count, so the comparison has no second argument. A test that assigns a meaning to each digital root cannot fail either, since the digital root of every positive integer lies in {1, ..., 9}. The continuous case is solved in closed form: a geometric ladder with ratio b and a free integer exponent places a random target within relative tolerance τ of some rung with probability ln[(1 + τ)/(1 - τ)]/ln b. For b = φ that band runs from -21.385% to +27.202%, so every real target sits near some rung, the median miss is 11.973%, a reported 2.2% agreement is a one-in-eleven coincidence and a 3.01% agreement is one in eight. Coverage scales as 1/ln b, so the smallest ratio gives the densest rungs and the weakest test: at τ = 0.022 the coverage is 0.0915 for b = φ against 0.0191 for b = 10. The corrected procedure follows: fix the scoring set before seeing the data, keep its null measure strictly below 1, and state the null distribution. Run on the same word counts, the corrected test returns p = 0.27.