Solution
= Solution
Given the coin, the next toss is independent of its earlier tosses and has that coin's head <probability>. Averaging using the <posterior distribution> gives the <posterior predictive probability>
$$
\boxed{\mathbb P(\text{next head}\mid H)
=\frac4{29}\frac13+\frac9{29}\frac12+\frac{16}{29}\frac23
=\frac{33}{58}.}
$$
Observing two heads raises the <probability> of having chosen a high-bias coin.