Under a Beta(1, 1) prior a strategy wins 7 of its first 10 trades. Give the probability that the next trade wins, the probability that the next two both win, and explain why the second is larger than the square of the first.

Under a Beta(1, 1) prior a strategy wins 7 of its first 10 trades. Give the probability that the next trade wins, the probability that the next two both win, and explain why the second is larger than the square of the first.

Approach: Update to the posterior, then integrate the relevant power of p against it rather than treating future trades as independent with a fixed rate.

2/3. The Beta(1, 1) prior updates to Beta(1 + 7, 1 + 3) = Beta(8, 4), and the posterior predictive probability of a win is the posterior mean 8/12 = 2/3. For two wins in a row the predictive probability is E[p^2] = a*(a + 1)/((a + b)*(a + b + 1)) = (8*9)/(12*13) = 6/13 = 0.4615, which exceeds (2/3)^2 = 4/9 = 0.4444. Future trades are conditionally independent given p and only exchangeable once p is integrated out, so a first win raises the posterior for p and therefore raises the chance of a second. The gap 6/13 - 4/9 = 0.01709 is exactly the posterior variance of p, 8*4/(12^2*13) = 0.01709. That is the general identity E[p^2] - E[p]^2. Treating the posterior mean as a known win rate understates the probability of streaks in either direction.

Follow-up: Under the same posterior, what is the probability of at least 6 wins in the next 10 trades, and how does it differ from a Binomial(10, 2/3) calculation?

Key concepts: posterior predictive, exchangeability, conditional independence, beta prior.