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Hidden bias forecast

StatisticsHard

A coin has a hidden bias $P (H)$ equal to $\frac{1}{4}$ , $\frac{1}{2}$ or $\frac{3}{4}$ , with prior probabilities $\frac{1}{4}$ , $\frac{1}{2}$ and $\frac{1}{4}$ . Given the bias, tosses are independent. The first $3$ tosses came out tails, tails, heads in this order. What is the probability that each of the next $2$ tosses is heads?

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Hint

The order of the observed tosses fixes the likelihood; then average the squared bias over the posterior.

Solution

The likelihood of tails, tails, heads is $p (1 - p)^{2}$ , equal to $\frac{9}{64}$ , $\frac{8}{64}$ and $\frac{3}{64}$ for the biases in the order listed.
Multiplying by the priors gives $\frac{9}{256}$ , $\frac{16}{256}$ and $\frac{3}{256}$ , so the posterior is $\frac{9}{28}$ , $\frac{16}{28}$ and $\frac{3}{28}$ .
The chance that the next $2$ tosses are heads is the posterior mean of $p^{2}$ , which is $\frac{9 \times \frac{1}{16} + 16 \times \frac{1}{4} + 3 \times \frac{9}{16}}{28}$ .
This equals $\frac{\frac{25}{4}}{28} = \frac{25}{112}$ .

Answer: $\frac{25}{112}$

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