Four Times the Trials, Half the Spread

First, a quick retrieval check — a different concept, new setting.

A hiring test has this property: P(pass | qualified candidate) = 90%. It’s not a perfect test though — P(pass | unqualified candidate) = 25% (some unqualified people get lucky). Suppose 20% of applicants are actually qualified (the base rate). A candidate passes. What’s P(qualified | passed)? Compute it — then compare it to the bare 90% number a recruiter might quote from memory. They’re not the same question.

(Worked out in the solution.)

Lesson 15 gave you the mechanics: Beta(α, β) behaves like α virtual successes and β virtual failures. Lesson 16’s colleague scenario aside, here’s a different, extremely common wrong shortcut about the same beta-binomial machinery — this time about how fast a posterior’s uncertainty shrinks.

Terms (standalone):

Posterior standard deviation (SD): how spread out the posterior distribution is — a small SD means you’re fairly confident about the true rate p; a large SD means substantial remaining uncertainty. It’s the square root of the variance.
Pseudo-count weight: as in Lesson 15, α + β is the total “trial count” behind a Beta(α, β) posterior — real trials plus the prior’s virtual ones. Priors don’t have to be specified with whole-number pseudo-counts: a prior stated as “centered at 33%, with the weight of 20 trials” means α = 0.33 × 20 = 6.6 and β = 0.67 × 20 = 13.4 — fractional pseudo-counts are completely normal once a prior is defined by a centre and a weight rather than raw counts.
The approximation this lesson uses: for a beta posterior with mean p and total weight n (n = α + β), the variance is well approximated by p(1−p) / n. (The exact formula has an extra +1 in the denominator that matters little once n isn’t tiny; this track uses the simpler form throughout, same as the exact-enough convention from Lesson 15’s “how much data to overturn a prior” estimate.)

A colleague’s shortcut. You currently hold the posterior Beta(6.6, 13.4) — built from a prior centered at p = 0.33 with the weight of 20 trials (so n = α + β = 20). Your colleague says: “Variance has the trial count squared in the denominator, so if you want to cut the standard deviation in half, just double the trials — 20 more should do it.”

Part 1 — Check the shortcut directly. Using the approximation variance ≈ p(1−p)/n, compute the posterior’s SD right now (at n = 20), and then compute what the SD would be if the trial count merely doubled to n = 40 (same posterior mean p = 0.33 throughout, as if new data kept arriving at the same underlying rate). Did doubling n actually halve the SD?

Part 2 — Find the trial count that genuinely halves the SD. Using the same approximation, solve for the total trial count n' at which the SD is exactly half of its value at n = 20. (Hint: variance ∝ 1/n, so SD ∝ 1/√n — work out what has to happen to n for √n to double.)

Part 3 — The numeric answer above. How many additional trials beyond the current 20 does that require? That’s the number to submit.