Don't Just Square the Mean

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

A joint probability factors two ways (Lesson 6): P(A,B) = P(A|B)·P(B) = P(B|A)·P(A). A code-review bot flags 8% of all pull requests as “risky.” Among genuinely buggy PRs, the bot flags 70% of them. Among non-buggy PRs, it flags 5%. Overall, 12% of PRs turn out to be genuinely buggy. What’s P(flagged)? Then, what’s P(buggy | flagged)? Two quick multiplications and a division — Lesson 6’s derivation, applied fresh.

(Worked out in the solution.)

You’ve built posteriors, watched them sharpen, and measured how fast they shrink. Now a different question: given everything your posterior currently believes about the true rate p, what should you expect to happen on future trials?

Terms (standalone):

Posterior predictive distribution: the probability of an observable future outcome, averaged over everything the posterior currently believes about p — not computed by pretending p is pinned at one value.
The tempting shortcut: plug in the posterior mean as if it were p, then compute an ordinary probability from there. Under this shortcut, “the next two trials both succeed” gets estimated as mean².

Your posterior, one step past where a bare example would stop. You start from a uniform prior, Beta(1, 1), and observe 3 successes out of 5 trials. First derive the posterior yourself (Lesson 15’s addition-of-counts rule), then use it — this puzzle doesn’t hand you the posterior pre-built.

Part 1 — Derive the posterior. What Beta(α, β) results from Beta(1,1) plus 3 successes and 2 failures? What’s its mean?

Part 2 — The shortcut’s answer. Using that mean, what does “square the mean” predict for P(next two trials both succeed)?

Part 3 — The exact answer (the numeric answer above). The honest calculation asks the question one trial at a time: the first trial succeeds with probability equal to the current posterior mean; given that success, the posterior itself updates (exactly like Lesson 15’s rule) before you ask about the second trial. So:

P(next two both succeed) = [α / (α+β)] × [(α+1) / (α+β+1)]

Compute this exactly for your Part-1 posterior, to 3 decimal places, and compare it to Part 2’s shortcut answer. Which one is bigger, and by roughly how much?

Part 4 — Why the gap exists. In one or two sentences: why does treating the two future trials as independent (at a fixed, pinned-down p = the mean) systematically get this wrong? What does a success on the first trial do to your belief about the second, that squaring the mean silently assumes doesn’t happen?