Emergent abilities in large models: Exercises

This is a self-testing page. Try each problem before opening its solution. Difficulty climbs from L1 (recognition) to L5 (mastery). The engine behind almost every answer is the same lens from the parent note, Emergent abilities in large models: a smooth per-token skill $p$ raised to a task length $L$ gives $Acc = p^{L}$ , which looks like a jump but is smooth underneath.

Definition Symbols used everywhere below (earn them once)

$p$ = probability the model gets one token right. A number between $0$ and $1$ . Think "how good per word."
$L$ = number of tokens that all must be right for the task to count as solved (exact-match).
$Acc = p^{L}$ = exact-match accuracy: succeed only if every one of the $L$ tokens is right, and each is an independent coin-flip with success chance $p$ . Multiplying $p$ by itself $L$ times is why we write $p^{L}$ .
$TokenScore = L p$ = expected number of correct tokens (a smooth metric).
$N$ = model scale (parameters).
$D$ = the training dataset — specifically the number of training tokens the model was fitted on (its size / the data distribution it learned from). More data means the model has seen more examples.
$L_{loss} (N, D)$ = training loss, written as a function of both scale $N$ and dataset size $D$ (careful: this " $L_{loss}$ " is the loss, a different symbol from task-length $L$ — the subscript keeps them apart).
$E$ = irreducible entropy: the floor the loss can never drop below, no matter how big the model. It is the intrinsic unpredictability of language itself (even a perfect model faces genuine randomness). The loss gap $g = L_{loss} - E$ is the shrinkable part of the loss.

Level 1 — Recognition

Exercise 1.1

Which of these is the definition of an emergent ability (Wei et al., 2022)? (a) Any ability that improves with scale. (b) An ability absent in small models, present in large ones, and not predictable by extrapolating small-model performance. (c) An ability that only large models can ever have, forever.

Recall Solution

(b). The key phrase is cannot be predicted by extrapolating small-model performance. (a) is too broad — smooth improvement is not emergence. (c) overclaims permanence; emergence is about the shape of the scale curve, not a metaphysical limit.

Exercise 1.2

On the standard picture (smooth loss curve + S-shaped accuracy curve), which quantity is the smooth one and which shows the sharp look?

Recall Solution

Loss is smooth (a power law with no kink). Task accuracy shows the sharp look (flat at chance, then a hockey-stick). Mnemonic: "Smooth Loss, Sharp Look."

The figure below plots both against model scale $N$ (horizontal axis, arbitrary units, running left-to-right from small to large models) on the same vertical axis (a dimensionless "value" from $0$ upward). The black curve is the loss: it slides down smoothly with no bend. The red curve is exact-match accuracy $p^{20}$ : it hugs the bottom (near $0$ ) for a long stretch, then curls up sharply. Same underlying model, two lenses — the red arrow marks the point where accuracy crosses into "looks emergent."

Figure — Emergent abilities in large models

Figure s01 — Horizontal axis: model scale $N$ (arbitrary units). Vertical axis: value (dimensionless). Black = smoothly declining loss; red = exact-match accuracy $p^{20}$ pinned near zero then rising sharply.

Exercise 1.3

Name the two competing explanations for emergence and match each to a one-line summary.

Recall Solution

Capability phase transition: a hard task chains many fragile sub-skills. Let $k$ = the number of serial sub-skills the task requires. If each is right with probability $p$ , the whole chain succeeds with probability $p^{k}$ . This $p^{k}$ explodes (rises fast) once $p$ passes a knee → genuinely sharper skill.
Metric mirage (Schaeffer et al., 2023): the jump is an artifact of a harsh, thresholded metric (exact-match); a smooth metric (edit distance, log-likelihood) reveals smooth progress underneath.

Level 2 — Application

Exercise 2.1

A task needs $L = 10$ tokens all correct. Compute exact-match accuracy $p^{10}$ for $p = 0.5$ , $p = 0.8$ , and $p = 0.99$ . What story do the three numbers tell?

Recall Solution

$p = 0.5 : 0. 5^{10} = 0.000977 \approx 0.10%$ — effectively "can't do it."
$p = 0.8 : 0. 8^{10} = 0.1074 \approx 10.7%$ .
$p = 0.99 : 0.9 9^{10} = 0.9044 \approx 90.4%$ — "solved it."

Story: a per-token skill climbing $0.5 \to 0.99$ (smooth) turns $0.1%$ into $90%$ accuracy. That feels like a switch, but nothing discontinuous happened.

Exercise 2.2

For the same jump ( $p : 0.8 \to 0.99$ ), compute the smooth metric $TokenScore = L p$ with $L = 10$ . How much (in relative terms) does it change, versus how much exact-match changes?

Recall Solution

$TokenScore (0.8) = 10 \times 0.8 = 8.0$
$TokenScore (0.99) = 10 \times 0.99 = 9.9$
Relative change: $9.9/8.0 = 1.2375$ → about a 24% increase.

Exact-match went from $0.107$ to $0.904$ , a factor of $\approx 8.4 \times$ . Same model, same skill gain — the harsh metric magnifies it $\sim 8 \times$ while the smooth metric barely moves. This is the mirage in numbers. Compare with Cross-entropy loss and perplexity, which is likewise a smooth per-token lens.

Exercise 2.3

A 3-digit addition answer is $L = 4$ tokens (three digits + a carry token). Model A has $p = 0.6$ , Model B has $p = 0.95$ . Which reports as "emergent" on exact-match, and by what factor did accuracy grow?

Recall Solution

Model A: $0. 6^{4} = 0.1296 \approx 13.0%$ .
Model B: $0.9 5^{4} = 0.8145 \approx 81.5%$ .
Growth factor: $0.8145/0.1296 = 6.285 \approx 6.3 \times$ .

Model B looks emergent (near-random → strong). The mechanism is the exact-carry chain of Chain-of-thought prompting-style arithmetic: one wrong digit fails the whole answer.

Level 3 — Analysis

Exercise 3.1

Two tasks improve their per-token skill by the same absolute amount $Δ p = 0.1$ (from $p = 0.85$ to $p = 0.95$ ). Task X has $L = 5$ , Task Y has $L = 40$ . Compute the accuracy gain (final minus initial) for each. Which one looks more emergent, and why does that follow from the algebra?

Recall Solution

Task X ( $L = 5$ ): $0.8 5^{5} = 0.4437$ , $0.9 5^{5} = 0.7738$ → gain $= 0.7738 - 0.4437 = 0.3301$ . Task Y ( $L = 40$ ): $0.8 5^{40} = 0.001500$ , $0.9 5^{40} = 0.1285$ → absolute gain $= 0.1285 - 0.001500 = 0.1270$ , but measured as a factor it is huge ( $0.1285/0.001500 \approx 85.7 \times$ ), off a near-zero base.

Why Y looks more emergent: for large $L$ , $p^{L}$ sits crushed near $0$ until $p$ is very close to $1$ , then rises steeply. The slope of $p^{L}$ with respect to $p$ is $L p^{L - 1}$ , which is largest for big $L$ once $p$ nears $1$ . So Task Y's curve has a later, steeper knee — the visual signature of emergence.

Exercise 3.2

Show algebraically why the log of exact-match accuracy is linear in $L$ , and use that to explain why "the larger $L$ , the sharper the transition."

Recall Solution

Start from $Acc = p^{L}$ . Take the natural log: $ln Acc = L ln p .$ Since $0 < p < 1$ , $ln p < 0$ . So $ln Acc$ is a negative number scaled by $L$ . When $p$ is small, $ln p$ is very negative → $Acc \approx 0$ . As $p \to 1$ , $ln p \to 0^{-}$ → $Acc \to 1$ .

Doubling $L$ doubles the depth $ln Acc$ for any fixed $p < 1$ , pushing small- $p$ accuracy even closer to zero while $p \to 1$ still snaps to one. The gap between "pinned at zero" and "snaps to one" widens with $L$ ⇒ a sharper-looking transition.

Why the axis choice changes the "sharpness": the eye judges "sharp" by how fast the plotted height changes. On a linear accuracy axis, the height is $p^{L}$ — for large $L$ this stays glued to the floor and then rockets up, so your eye sees a knee. But if we plot the height as $ln Acc$ against $ln p$ , the relation $ln Acc = L ln p$ is a perfectly straight line of slope $L$ — a line has no knee anywhere. Nothing about the model changed; we only swapped the vertical (and horizontal) ruler for a logarithmic one, which stretches the crushed-near-zero region back open. The "sharp transition" was living in the linear ruler, not in the mathematics. This is the same trick the mirage view uses: pick a lens (metric/axis) and the discontinuity appears or vanishes.

Figure s02 — Horizontal axis: per-token skill $p$ (0 to 1, dimensionless). Vertical axis: exact-match accuracy $p^{L}$ (0 to 1). Black curves are $L = 2, 5, 10$ ; the red curve is $L = 40$ , staying pinned near zero until $p$ is close to 1 — bigger $L$ gives a later, steeper apparent knee.

The figure makes this concrete. The horizontal axis is per-token skill $p$ (from $0$ to $1$ ); the vertical axis is exact-match accuracy $p^{L}$ (from $0$ to $1$ ). Each black curve is a different $L$ (labelled $L = 2, 5, 10$ ); the red curve is $L = 40$ , which stays pinned near the floor and only lifts once $p$ is very close to $1$ — the red arrow marks that "stays near 0 until $p$ close to 1" behaviour. Bigger $L$ ⇒ later, steeper apparent knee, exactly as the algebra predicts.

Exercise 3.3

Parent note's link: $p (N) = exp (- (L_{loss} (N, D) - E))$ where $L_{loss}$ is the training loss (a function of scale $N$ and dataset size $D$ ) and $E$ the irreducible entropy (the loss floor defined in the preamble). If the loss gap $g = L_{loss} - E$ shrinks smoothly from $g = 0.5$ to $g = 0.05$ , and the task needs $L = 20$ tokens, compute exact-match at both ends. Confirm it is a smooth composition despite the visual jump.

Recall Solution

First $p = e^{- g}$ :

$g = 0.5 : p = e^{- 0.5} = 0.6065$ , so $Acc = 0.606 5^{20} = e^{- 0.5 \times 20} = e^{- 10} = 4.54 \times 1 0^{- 5}$ .
$g = 0.05 : p = e^{- 0.05} = 0.9512$ , so $Acc = 0.951 2^{20} = e^{- 0.05 \times 20} = e^{- 1} = 0.3679$ .

Note the clean form $Acc = e^{- Lg}$ . Every function ( $e^{- g}$ , then raising to $L$ ) is continuous and smooth, yet accuracy went from $\sim 5 \times 1 0^{- 5}$ to $0.37$ . Emergence = amplification by the exponent $L$ , not a discontinuity. This ties directly to Compute-optimal training and Scaling laws (Kaplan, Chinchilla), where $g$ shrinks like a smooth power law.

Level 4 — Synthesis

Exercise 4.1

Chain-of-thought prompting adds intermediate reasoning steps. Model it: without CoT the task needs $L = 1$ final token at skill $p = 0.7$ . With CoT it needs a chain of $k = 6$ reasoning tokens each at skill $p$ , but if the chain is right the final answer is essentially certain. (a) For a small model with $p = 0.7$ , does CoT help or hurt? (b) For a large model with $p = 0.97$ , does CoT help? Use $Acc_{no-CoT} = p$ and $Acc_{CoT} = p^{6}$ .

Recall Solution

(a) Small, $p = 0.7$ : no-CoT $= 0.7$ ; CoT $= 0. 7^{6} = 0.1176$ . CoT hurts ( $0.118 < 0.70$ ) — extra fragile steps multiply errors. (b) Large, $p = 0.97$ : no-CoT $= 0.97$ ; CoT $= 0.9 7^{6} = 0.8330$ .

Here CoT is still slightly below raw guessing in this toy setup — the point is the gap collapses as $p$ rises: the CoT penalty shrinks from $0.70 - 0.118 = 0.582$ down to $0.97 - 0.833 = 0.137$ . In the real world CoT's benefit comes because the direct answer's effective $p$ is far lower than the per-step $p$ (hard problems can't be one-shot), so once per-step $p$ is high, decomposition wins. CoT emergence = a composition/crossover effect, matching the parent note.

Exercise 4.2

Design a smooth diagnostic. You suspect an "emergent" benchmark is a metric mirage. You have small models scoring $0%$ exact-match. Propose two smooth metrics from the Connections and predict what each would show if the mirage hypothesis is true.

Recall Solution

Per-token log-likelihood / Cross-entropy loss and perplexity: should decline smoothly across small models — a clean trend you can extrapolate to forecast the large-model jump.
Token edit distance to the gold answer: should shrink smoothly (fewer wrong tokens) even while exact-match sits at $0%$ .

Prediction under mirage: both smooth metrics improve monotonically and predictably, so the large-model "leap" is forecastable — the discontinuity was only in the harsh exact-match lens. This is exactly the red-teaming argument: small-model trends do inform us. Contrast with Inverse scaling and Grokking (delayed generalization), where the underlying dynamics themselves are non-monotone or delayed — those are not pure metric artifacts.

Level 5 — Mastery

Exercise 5.1

Regulators want to name the scale $N^{*}$ where a dangerous capability "turns on." Two tempting definitions of "steepest" are on the table. Using $Acc (p) = p^{L}$ with the per-token skill rising smoothly with scale, $p = p (N)$ : (a) does $p^{L}$ have an inflection point in $p$ (a place where its curvature $\frac{d ^{2}}{d p ^{2}} p^{L}$ changes sign)? (b) On the scale axis $N$ , where is the accuracy curve genuinely steepest, and is that a physical threshold or a convention?

Recall Solution

(a) Curvature in $p$ . First derivative: $\frac{d}{d p} p^{L} = L p^{L - 1}$ . Second derivative: $\frac{d ^{2}}{d p ^{2}} p^{L} = L (L - 1) p^{L - 2} .$ For $L > 1$ and $p \in (0, 1)$ this is strictly positive and never zero — it does not change sign. So $p^{L}$ is convex everywhere on $(0, 1)$ : there is no inflection point in $p$ . The "knee" your eye sees is pure convexity (hug the floor, then curl up toward $p = 1$ ), not a true turning point. The steepest slope in $p$ is at the boundary $p \to 1$ , where $L p^{L - 1} \to L$ .

(b) Steepest on the scale axis $N$ . Because accuracy reaches $N$ only through $p (N)$ , use the chain rule: $\frac{d Acc}{d N} = \frac{d ( p ^{L} )}{d p} \cdot \frac{d p}{d N} = L p^{L - 1} p^{'} (N) .$ This can peak at a finite $N$ — but the location depends entirely on the shape of $p (N)$ (how fast skill grows), not on any intrinsic property of $p^{L}$ . There is no universal $N^{*}$ baked into the mathematics.

Interpretation / answer to the regulators. Since $p^{L}$ has no inflection in $p$ , there is no canonical, model-independent threshold scale. Any reported " $N^{*}$ " is a convention — e.g. "the scale where exact-match first crosses $50%$ ", i.e. solve $p (N^{*})^{L} = 0.5 \Rightarrow p (N^{*}) = 0. 5^{1/ L}$ . It is a chosen line in the sand, not a physical singularity. (On a log-log plot $ln Acc = L ln p$ is a straight line of slope $L$ — no knee at all — confirming the "turn-on" is an axis illusion.) This is the mastery point: "emergence" gives you a convention to define $N^{*}$ , never a discontinuity to discover it.

Exercise 5.2 (capstone)

A dangerous capability needs $L = 15$ tokens correct. Regulators want an early-warning trigger. From two small checkpoints you measure per-token skill $p_{1} = 0.60$ at scale $N_{1}$ and $p_{2} = 0.75$ at scale $N_{2} = 2 N_{1}$ , and you find $p$ grows as $p (N) = 1 - c N^{- γ}$ . (a) Solve for $γ$ . (b) Predict $p$ at $N = 8 N_{1}$ . (c) Compute exact-match at $N_{1}$ , $N_{2}$ , and $8 N_{1}$ and state at which scale the capability "emerges" (cross $50%$ ).

Recall Solution

Write $q (N) = 1 - p (N) = c N^{- γ}$ (the per-token error).

At $N_{1}$ : $q_{1} = 1 - 0.60 = 0.40$ .
At $N_{2} = 2 N_{1}$ : $q_{2} = 1 - 0.75 = 0.25$ .

(a) Ratio: $\frac{q _{2}}{q _{1}} = (\frac{N _{2}}{N _{1}})^{- γ} = 2^{- γ} = \frac{0.25}{0.40} = 0.625$ . So $γ = - lo g_{2} (0.625) = 0.6781$ .

(b) At $N = 8 N_{1} = 2^{3} N_{1}$ : $q = q_{1} \cdot (8)^{- γ} = 0.40 \times 8^{- 0.6781}$ . Now $8^{- 0.6781} = 2^{- 3 γ} = 0.62 5^{3} = 0.2441$ . So $q = 0.40 \times 0.2441 = 0.09766$ , giving $p = 0.9023$ .

(c) Exact-match $= p^{15}$ :

$N_{1}$ : $0.6 0^{15} = 4.702 \times 1 0^{- 4}$ ( $\approx 0.05%$ ).
$N_{2}$ : $0.7 5^{15} = 0.01336$ ( $\approx 1.3%$ ).
$8 N_{1}$ : $0.902 3^{15} = 0.2143$ ( $\approx 21.4%$ ).

None crosses $50%$ yet, but the smooth per-token error law $q \propto N^{- 0.678}$ was fully forecastable from the two small checkpoints — the exact-match "silence" ( $0.05% \to 1.3%$ ) hid a clean, extrapolable trend. That is the whole safety point: measure the smooth quantity, extrapolate the error law, and predict the emergence before it happens.