Solution: Five Settings, No Formula

Retrieval check answer. 1 − (0.35 + 0.25 + 0.15) = 1 − 0.75 = 0.25 — referral is 25% of traffic. Same rule as Lesson 12’s missing die-face probability: whatever’s left has to make the total hit exactly 1.

Part 1 — Likelihoods (p³(1−p), for 3 successes and 1 failure out of 4 trials):

(Multiplying every row by the shared constant C(4,3)=4 would give 0.0036, 0.0756, 0.25, 0.4116, 0.2916 — four times each number above. It changes nothing after normalization, exactly Lesson 16’s point: a hypothesis-independent constant cancels.)

Part 2 — Unnormalized posterior (prior weight × likelihood):

Sum = 0.000045 + 0.002835 + 0.031250 + 0.020580 + 0.007290 = 0.062000.

Part 3 — Normalized posterior:

(Sum ≈ 1.0000, confirming the normalization.)

Part 4 — Posterior mean: ≈ 0.604.

0.1×0.0007 + 0.3×0.0457 + 0.5×0.5040 + 0.7×0.3319 + 0.9×0.1176
= 0.00007 + 0.01372 + 0.25202 + 0.23235 + 0.10582
≈ 0.60398 ≈ 0.604

The prior put over half its weight on p = 0.5. After seeing 3-out-of-4 (a raw 75% rate), most of the posterior’s mass shifted onto p = 0.7 (33%) and stayed substantial at p = 0.9 (12%), while p=0.5 still holds a plurality (50%) — that single grid point had the largest prior weight and a reasonably good likelihood, so it wasn’t cheap to dislodge. The posterior mean, 0.604, sits well above the prior’s implied center but well below the raw 0.75 rate — exactly the same “tug-of-war between prior weight and data” you’ve seen since Lesson 14, just computed point-by-point on a discrete grid instead of through a beta formula.

Why this had no closed form. A beta distribution is a smooth curve over [0,1]; this prior is five isolated spikes at specific settings, with nothing at all in between. No choice of Beta(α,β) parameters can represent “exactly these five values, nothing else” — the shapes are fundamentally different objects. Grid approximation doesn’t care: it re-derives the posterior from Bayes’ rule directly, one candidate value at a time, for any prior shape at all — bimodal, five isolated spikes, a hand-sketched curve, anything.

The trade-off, honestly. What you just did — five rows of arithmetic — scales badly if you wanted a smooth posterior over all of [0,1] instead of five discrete settings: that takes hundreds or thousands of grid points, each one requiring its own likelihood computation. That’s exactly the “honest workhorse” Stage 5 builds properly. For now: conjugate updates are a shortcut that only exists for a few lucky prior-likelihood pairings; grid approximation is the general method that always works, at the cost of doing the sum by brute force instead of algebra.

Where this goes: you’ve now summarized posteriors with means. Next lesson asks a sharper question about summarizing them: mean, median, and mode aren’t interchangeable “best guesses” — each one is the answer to a different question about how you’re being scored for being wrong.