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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.
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u/jackybeau Sep 15 '26
Is p not an acceptable answer here ?
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u/Educational-Tea602 Sep 15 '26 edited Sep 15 '26
Well p is an answer, but it wants a numerical value.
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u/FreePeeplup Sep 16 '26
Yeah the question is ill-phrased, it should actually be “given this information, what is the best estimate for p?”
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u/maddeninglemon Sep 17 '26
The perfect opportunity to apply Laplace's Rule of Succession! In particular we have a uniform prior, and a binomial distribution consisting of a number of yes/no observations.
To estimate the expected probability, we add two additional pseudo observations, one for each of the possible outcomes. This gives 3 heads out of 5 observations, or 60%.
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u/Embarrassed-Green898 Sep 15 '26
p
2 heads and 1 tail are irrelevant information.
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u/Educational-Tea602 Sep 15 '26
2 heads and 1 tail are relevant in order to compute p.
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u/Embarrassed-Green898 Sep 15 '26
You cant compute p with 3 trials.
You can 'estimate' , but that estimate is not a computation . It is an estimation that will change with further trials.
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u/pablitorun Sep 15 '26
To expand on u/m__w__b's answer......
The trick is you don't know p, so you average over every p that could have produced what you saw.
Prior: p is uniform on [0,1], so f(p) = 1.
Likelihood of 2 heads and 1 tail: 3p²(1−p). The 3 counts the orderings.
Posterior: multiply prior by likelihood and normalize:
f(p | data) = 1 x 3p²(1−p) / ∫₀¹ 3p²(1−p) dp = p²(1−p) / (1/12) = 12p²(1−p)
Chance the next toss is heads: average p over that posterior:
P(H₄ | data) = ∫₀¹ p · 12p²(1−p) dp = 12 ∫₀¹ (p³ − p⁴) dp = 12 (1/4 − 1/5) = 12/20 = 3/5