Solution
Three More From a Fresh Prior
Show the problem again
A new free-throw shooter has an unknown success probability $p$. Before any data, your prior for $p$ is Beta(2, 3). Given $p$, each shot is an independent Bernoulli($p$) trial. You watch her take 5 shots and she makes 4. What is the posterior predictive probability that she makes all of her next 3 shots? Give your answer as a reduced fraction.
Worked solution
Beta is conjugate to the Bernoulli likelihood, so the posterior is Beta(2+4, 3+1) = Beta(6, 4). The predictive probability of three further successes is $E[p^3]$ under Beta(6, 4). Equivalently, after each success the posterior shifts from Beta(a, b) to Beta(a+1, b), so the predictive chain is $\frac{6}{10}\cdot\frac{7}{11}\cdot\frac{8}{12}$. The product is 336/1320 = **14/55**. It is above the plug-in value $(0.6)^3 = 27/125$: ignoring uncertainty about $p$ gets the answer wrong, because $E[p^3] > (E[p])^3$ by Jensen's inequality.
Source: Original problem written for AxiomIQ (standard Beta-Binomial posterior predictive calculation). Statement written for AxiomIQ.