Solution: Don't Just Square the Mean

Retrieval check answer. P(flagged) = 0.70×0.12 + 0.05×0.88 = 0.084 + 0.044 = 0.128. P(buggy | flagged) = 0.084 / 0.128 = 0.656 (about 66%) — Lesson 6’s two-factorizations-of-a-joint trick, on a code-review bot instead of a spam filter.

Part 1 — Beta(4, 3), mean ≈ 0.571. Beta(1,1) + 3 successes, 2 failures → Beta(1+3, 1+2) = Beta(4, 3). Mean = 4 / 7 ≈ 0.5714.

Part 2 — Shortcut’s answer: 0.327. (4/7)² = 16/49 ≈ 0.3265.

Part 3 — Exact answer: 0.357.

P(both succeed) = [4/7] × [5/8] = 0.5714 × 0.625 = 0.35714... ≈ 0.357

0.357 is noticeably higher than the shortcut’s 0.327 — a real gap (about 9% relative), not rounding noise.

Part 4 — why the gap exists. The shortcut pins p at exactly 4/7 and treats the two trials as independent flips of that fixed-rate coin. But p isn’t pinned — Beta(4,3) is a whole distribution of plausible rates, some above 4/7, some below. If the first trial actually succeeds, that success is itself evidence nudging your belief about p upward: the posterior would update to Beta(5,3), mean 5/8 = 0.625, a bit higher than 4/7 ≈ 0.571. So the second trial’s true probability of success — conditional on the first succeeding — is a little better than the plain mean suggests. Squaring the mean silently assumes the first trial teaches you nothing about the second. In reality, because both trials depend on the same unknown p, they’re positively correlated once you’re honest about your uncertainty: Var(p) > 0 always pushes E[p²] above (E[p])², and E[p²] is exactly what “both trials succeed” needs.

Seeing it as a sum, not just a formula. The exact predictive probability is really E_posterior[p²] = ∫ p² · Beta(p; 4,3) dp, computed in closed form above as 20/56 = 5/14 ≈ 0.357. You can approximate that same integral with a coarse discrete sum over a few candidate values of p, weighted by how plausible each is under Beta(4,3) (density ∝ p³(1−p)²):

Weighted average of p (≈0.578) is close to the true mean (0.571) — a coarse 4-point grid, so not exact. Weighted average of p² (≈0.363) is close to the true E[p²] (0.357) — and notice it does not equal the squared weighted-average-of-p (0.578² ≈ 0.334). Even with only four grid points, summing p² × weight gives a different, higher number than squaring the summed p × weight — the same gap, visible as arithmetic instead of calculus. Stage 5 turns this coarse sketch into the real tool (fine grids, many points); for now, the point is that “integrate/sum over the whole posterior” and “plug in one summary number” are genuinely different operations, and they disagree exactly when uncertainty about p is large enough to matter.

The general pattern, for the next k trials all succeeding from a Beta(α,β) posterior:

[α/(α+β)] × [(α+1)/(α+β+1)] × [(α+2)/(α+β+2)] × ... × [(α+k−1)/(α+β+k−1)]

— each factor is “the posterior mean, updated one hypothetical success at a time.” As α+β grows large (a sharp, confident posterior), this converges toward mean^k, because there’s barely any uncertainty left for a hypothetical success to update — another instance of Lesson 17’s “more weight, less movement.”

Where this goes: Stage 2 is done. Stage 3 formalizes something you’ve been doing informally the whole time — treating yesterday’s posterior as today’s prior — and asks precisely when the order evidence arrives in is allowed to not matter.