Solution: Four Times the Trials, Half the Spread

Retrieval check answer. P(pass) = 0.9×0.2 + 0.25×0.8 = 0.18 + 0.20 = 0.38. P(qualified | passed) = 0.18 / 0.38 ≈ 0.474 — about 47%, not 90%. The 90% is P(pass | qualified), a completely different conditional (Lesson 8’s transposed-conditional trap), and a mediocre false-hire rate (25%) plus a low base rate of qualified candidates (20%) drags it down hard.

Part 1 — the shortcut fails. At n = 20, mean p = 0.33:

variance(20) ≈ 0.33 × 0.67 / 20 = 0.2211 / 20 = 0.011055
SD(20) ≈ √0.011055 ≈ 0.1051

Doubling to n = 40 (same p = 0.33):

variance(40) ≈ 0.2211 / 40 = 0.0055275
SD(40) ≈ √0.0055275 ≈ 0.0743

0.0743 is not half of 0.1051 — half would be 0.0526. The actual ratio is 0.0743 / 0.1051 ≈ 0.707, i.e. 1/√2. Doubling the trial count shrinks the SD by a factor of 1/√2 ≈ 0.71, not 1/2. Your colleague’s “20 more trials” leaves the posterior visibly wider than they think it is — check it against the real number and the shortcut falls apart immediately.

Part 2 — solving properly. variance ∝ 1/n means SD ∝ 1/√n. To cut SD in half, you need √n' to be double √n — which means n' itself must be 4× n, not 2× (√(4n) = 2√n). At n = 20, that’s n' = 80:

variance(80) ≈ 0.2211 / 80 = 0.00276375
SD(80) ≈ √0.00276375 ≈ 0.0526   — exactly half of 0.1051, as required.

Part 3 — the numeric answer: 60 more trials. Current n = 20; target n' = 80; the difference is 80 − 20 = 60 additional trials. (Using the exact beta-variance formula with the +1 term instead of the approximation shifts this to roughly 63 rather than 60 — same conclusion, same order of magnitude, and irrelevant to the point being made.)

Why the “squared in the denominator” instinct is half-right and half-wrong. Variance genuinely does have n (not n²) in the denominator — the colleague’s shortcut silently assumed the wrong power. Because variance ∝ 1/n rather than 1/n², the standard deviation — the more interpretable, same-units quantity you actually care about — scales as 1/√n, the square root softening the relationship. Any process where uncertainty is measured by standard deviation and driven by independent-trial accumulation has this shape: going from “good” to “twice as good” costs four times the data, not twice. This is the same arithmetic, generalized, that made a single new observation barely move a large posterior back in Lesson 16 (movement ≈ 1/(α+β+1)) — both facts come from the same 1/n-type scaling underneath.

The pattern:

Where this goes: you’ve now watched a posterior’s uncertainty shrink as data accumulates. Next lesson asks a related but distinct question — not “how uncertain am I about p?” but “given everything I currently believe about p, what do I actually expect to see on the next trial?” That’s the posterior predictive distribution, and it has its own classic wrong shortcut waiting.