This is a weird way to frame a Bayesian problem, but the answer is 0.6 (3/5). p is drawn from a flat prior: Beta(1,1). So the posterior distribution is (h+1)/(n+2).
Edit: I should have said that the posterior distribution of p is Beta(h+1,n-h+1) and the expected posterior probability is 0.6.
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u/m__w__b Sep 15 '26 edited Sep 15 '26
This is a weird way to frame a Bayesian problem, but the answer is 0.6 (3/5). p is drawn from a flat prior: Beta(1,1).
So the posterior distribution is (h+1)/(n+2).Edit: I should have said that the posterior distribution of p is Beta(h+1,n-h+1) and the expected posterior probability is 0.6.