A capability that appears suddenly, a benchmark score that leaps — these are among the most misread phenomena in the field. Two small calculations show how much of the drama lives in the ruler rather than the model.
Here is a claim you will meet often: a model was scaled up, and at some size an ability it did not have simply appeared. It is a striking claim, and sometimes true. But before you believe it about any particular benchmark, there are two calculations to run — and they will frequently show that what changed was not the model but the ruler held against it.
Two calculations do the work. The first shows how a smoothly improving model can produce a sharp jump on a benchmark that scores exact matches. The second shows how contamination — test items seen during training — inflates observed accuracy, and how to invert the inflation. Neither requires anything past Chapter 1.
Suppose a model performs a task with \( k \) independent parts — adding two \( k \)-digit numbers, say — and gets each part right with probability \( p \). If the benchmark scores an exact match, awarding credit only when all \( k \) parts are correct, then the measured accuracy is \( p^k \). Now watch what a smooth improvement in \( p \) does to \( p^k \).
Do this one on paper as you read. Pick a per-part accuracy of 0.9 and eight parts. Before computing, write down whether you expect the exact-match score to be closer to 0.9 or closer to 0.5. Nearly everyone writes something in the eighties or nineties, because 0.9 feels high and eight feels small. The answer is 0.43, and the size of that surprise is precisely the size of the illusion this section is about.
A cliff manufactured from a slope
Exact-match accuracy is \( p^k \). As a function of the underlying per-part skill \( p \), this is smooth — but for large \( k \) it is nearly flat at low \( p \) and rises steeply near \( p = 1 \), because raising a number just below one to a high power collapses it, while raising a number near one preserves it. The 50% point sits where \( p^k = \tfrac12 \), i.e. at \( p = 2^{-1/k} \), which increases toward 1 as \( k \) grows.
Concretely, take \( k = 8 \). At \( p = 0.67 \), exact-match accuracy is \( 0.67^8 = e^{8\ln 0.67} = e^{-3.20} = 0.041 \) — about 4%. At \( p = 0.90 \), it is \( 0.90^8 = e^{-0.843} = 0.43 \) — about 43%. A modest, smooth improvement in per-digit skill, from \( 0.67 \) to \( 0.90 \), produces a tenfold leap on the benchmark. The model got steadily better; the benchmark reported a sudden emergence.
Where the cliff appears to fall
Tabulate exact-match accuracy \( p^k \) at \( k = 8 \) as the per-part skill improves:
p = 0.50 → 0.004 · 0.70 → 0.058 · 0.80 → 0.168 · 0.90 → 0.430 · 0.95 → 0.663
Between \( p = 0.5 \) and \( p = 0.7 \) the benchmark barely moves — from under half a percent to under six. Between \( 0.8 \) and \( 0.95 \) it moves from 17% to 66%. An observer watching only this metric would report a capability that lay dormant and then arrived; an observer watching \( p \) would report steady, unremarkable progress. The apparent onset sits near the 50% frontier \( 2^{-1/8} = 0.917 \), and for larger \( k \) it sits closer to 1 and looks sharper still.
An abrupt jump in a thresholded score is not evidence of a discontinuous change in the underlying capability. The per-part skill \( p \) — visible in a smooth metric like per-token log-loss — may be improving perfectly steadily while the exact-match metric snaps upward. Before calling a capability “emergent,” ask whether the ruler is a threshold on many smoothly-improving parts. Much reported emergence survives this test poorly.
A model does \( k \)-digit exact-match addition with per-digit accuracy \( p = 0.98 \). What is the largest \( k \) for which exact-match accuracy stays at or above 50%?
Need \( 0.98^k \ge 0.5 \), i.e. \( k \le \ln(0.5)/\ln(0.98) = (-0.693)/(-0.0202) = 34.3 \). Largest integer \( k = 34 \).
A subtler distortion is contamination: some fraction of the test items appeared in the training data, so the model answers them from memory rather than skill. This inflates the observed score, and a two-line model lets us both quantify and invert the inflation.
Solving for the leak, then flipping it with Bayes
Let a fraction \( c \) of test items be contaminated — answered correctly with probability 1 — while on the clean fraction \( 1-c \) the model's true accuracy is \( p \). The observed accuracy is the mixture
\[ \text{observed} = c\cdot 1 + (1-c)\,p = c + p(1-c). \]Given an observed score and a known (or assumed) clean accuracy, solve for \( c \). Suppose observed \( = 0.70 \) and clean accuracy \( p = 0.60 \): then \( 0.70 = c + 0.60(1-c) = 0.60 + 0.40c \), so \( c = 0.25 \). A quarter of the benchmark leaked, and the observed score overstates true skill by ten points.
Now the more revealing question: given that an item was answered correctly, what is the probability it was contaminated? By Bayes' rule, contaminated items contribute \( c \times 1 \) to the correct-answer mass, out of a total correct mass equal to the observed accuracy:
\[ P(\text{contaminated}\mid\text{correct}) = \frac{c}{\text{observed}} = \frac{0.25}{0.70} = 0.357. \]Over a third of the model's apparent successes are memory, not skill. This is the calculation to run whenever a headline score seems too good — and it is nothing more than the Bayes flip of Chapter 1.
A benchmark shows observed accuracy 0.85; the model's clean accuracy is believed to be 0.75. (a) What contamination fraction \( c \) explains the gap? (b) Given a correct answer, what is the probability it was contaminated?
(a) \( 0.85 = c + 0.75(1-c) = 0.75 + 0.25c \Rightarrow c = 0.40 \). (b) \( P = c/\text{observed} = 0.40/0.85 = 0.47 \).
One more measurement idea belongs here, though Chapter 15 develops it fully: taking the majority vote of several independent attempts. We met the calculation in Chapter 1 — five chains at \( p = 0.6 \) vote correctly with probability \( 0.68 \). The one fact to carry forward is that voting helps only when the single-attempt probability already exceeds one half; below that threshold, the majority amplifies the error instead of correcting it. Voting sharpens a capability that is already there; it does not manufacture one. That distinction — sharpening versus creating — is exactly the metric-versus-capability theme of this chapter in another guise.
A benchmark can mislead in two quite different ways, and this chapter priced both. Exact-match scoring turns a smooth improvement in per-part skill into an apparent cliff, and the 50% frontier \( p = 2^{-1/k} \) tells you where the cliff will seem to fall. Contamination inflates a score by \( c + p(1-c) \), and Bayes' rule reveals how much of the model's success is memory. Both are warnings to measure the ruler before trusting the miracle — a habit that separates careful readers of the field from credulous ones. Chapter 10 closes Part II by descending to the idea beneath all of this: that learning is compression, that simplicity is a probability, and that a single length prior turns Occam's razor from a preference into a theorem.
A · Drills
B · Problems
C · Challenge
Reproduce Derivations 8.1 (the \( p^k \) cliff and its 50% frontier) and 8.2 (the contamination mixture and its Bayes flip) on blank paper. Attempt B-1 and B-2 closed book. You pass when you can produce the \( 0.67 \to 0.90 \) becomes \( 4\% \to 43\% \) figures and the \( c/\text{observed} \) Bayes ratio without notes.