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Coin bag, streak observed

ProbabilityHard

A bag holds $3$ coins whose probabilities of heads are $\frac{1}{4}$ , $\frac{1}{2}$ and $\frac{3}{4}$ . One coin is chosen uniformly at random and flipped $3$ times independently, giving $3$ heads. What is the probability that the next flip of the same coin is heads?

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Hint

Update the weight of each coin by the likelihood of $3$ heads, then average the heads probabilities with the new weights.

Solution

The likelihoods of $3$ heads are $\frac{1}{64}$ , $\frac{8}{64}$ and $\frac{27}{64}$ , so the posterior weights are proportional to $1$ , $8$ , $27$ .
The weights sum to $36$ , so the posterior probabilities are $\frac{1}{36}$ , $\frac{8}{36}$ and $\frac{27}{36}$ .
The next-flip heads probability is $(1 \times \frac{1}{4} + 8 \times \frac{1}{2} + 27 \times \frac{3}{4}) \div 36$ .
The numerator is $0.25 + 4 + 20.25 = 24.5$ , so the result is $\frac{24.5}{36} = \frac{49}{72}$ .
The answer is $\frac{49}{72}$ .

Answer: $\frac{49}{72}$

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