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The tech-broification of American science has officially begun

The Trump administration unveiled the first "Genesis Mission" grants on Thursday, directing $5 billion toward hundreds of AI-driven science projects in an effort the White House has described as "comparable in urgency and ambition to the Manhattan Project." At roughly the same time, Trump's science adviser Michael Kratsios was on Capitol Hill selling lawmakers on […]

2026-07-24 原文 →
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The best config in your bake-off didn't win. Selection did.

Best-of-K eval selection bias: pick the highest-scoring config from K candidates on one eval set and that observed score is biased up. It reports the expected maximum of K noisy estimates, which beats the field mean whenever K exceeds one. The bias appears even when all K configs are truly equal, grows with K, and shrinks with n. Here is the version that bites you. Your bake-off ran a batch of prompts against one eval set, the top one came out ahead, and you shipped it. In production it does worse. That drop reads like bad luck, or drift, or a bad week. It is none of those. It is a number you could have computed before you shipped, and it gets larger the more candidates you tried. I ran a small script to make the gap concrete. Eight configs, one hundred eval items, and here is the catch: I made all eight configs truly identical , every one a fair coin at 50%. There is no real best. Nothing to tune. Then I let selection pick a winner anyway: config 0: 47/100 = 47.0% config 1: 50/100 = 50.0% config 2: 52/100 = 52.0% config 3: 52/100 = 52.0% config 4: 50/100 = 50.0% config 5: 50/100 = 50.0% config 6: 54/100 = 54.0% <- selected winner (argmax) config 7: 44/100 = 44.0% config 6: 54.0% (k=54 n=100 SE=4.98) config 2: 52.0% (k=52 n=100 SE=5.00) RANK: INDISTINGUISHABLE - gap 2.00 pp against 7.06 pooled SE = 0.28 SE < 2.0. Ranking "config 6" above "config 2" is NOT allowed. Held-out the winner on a fresh 100 items: 48/100 = 48.0%. Config 6 wins the bake-off at 54.0%. Then I asked the same eval-guard I use in the McNemar and rule-of-three pieces to rank config 6 against the runner-up. It refused: the gap is 0.28 SE, far under the two-SE bar, so INDISTINGUISHABLE . The ranker would not call config 6 the best. Selection did. And on a fresh held-out set the 54.0% falls back to 48.0%, toward the true 50% it was always going to be. TL;DR Picking the best of K configs by observed pass rate reports the expected maximum of K noisy estimates. The max of K exceeds the field mean, strict

2026-07-23 原文 →
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Utility companies promise to spare us from AI’s energy bill

In the face of backlash to concerns the AI boom will increase consumer electricity bills, the largest utility companies and data center developers in the US are now promising to do something about it. The Wall Street Journal reports that nearly 200 organizations have signed President Donald Trump's "rate payer protection pledge" that's meant to […]

2026-07-22 原文 →
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Zero failures isn't zero risk: the rule of three for evals

The rule of three for evals says zero failures in N runs is a count, not a rate. With 0 failures in N independent runs, the exact 95% upper bound on the true failure rate is 1 - 0.05^(1/N) , which 3/N approximates. After 100 clean runs you still cannot rule out a 2.95% rate, about 1 in 34. Here is the reading that bites you. Your eval harness runs the agent 100 times, prints "0 failures," and the tile goes green. Someone screenshots it into the launch thread. The unspoken translation is "the failure rate is zero." It is not what the data says. I wrote a small script to make the gap concrete, so I ran a real gate over 200 deterministic agent runs first, counted honestly, and got the dashboard everyone trusts: gate: spend<=budget over 200 deterministic agent runs observed failures: 0 (distinct scenarios: 200) naive point rate : 0.00% binomial SE: 0.00 pp naive 95% CI : [0.00%, 0.00%] <- zero width: false certainty Look at the standard error. For a zero count the binomial SE is sqrt(0*1/200) , which is exactly 0, so the naive 95% interval collapses to [0.00%, 0.00%] . A zero-width confidence interval. The math is telling you it is completely certain, from 200 samples, that the true rate is precisely zero. That is obviously wrong, and it is the exact shape of every "all green" board I have ever trusted too much. TL;DR "0 failures in N runs" is an observed count, not a rate. The naive binomial SE of a zero count is 0, which is why a green board looks like proof and isn't. The honest number is the one-sided upper bound. With 0 failures in N runs, the 95% upper confidence limit on the true failure rate is 1 - 0.05^(1/N) . The rule of three, 3/N , approximates it and rounds the risk slightly up. At N=30 the bound is 9.50% (about 1 in 11). At N=100 it is 2.95% (1 in 34). At N=1000 it is 0.30% (1 in 334). Zero failures in 30 runs is compatible with a 1-in-11 true failure rate. To rule out a 0.1% rate at 95% you need about 2995 clean runs, not 50. "We ran it fifty times" and "

2026-07-22 原文 →