A bag holds two coins. One is fair. The other has a head on both sides…

Step 1. Two heads. Two-headed coin: $\dfrac{1}{2} \times 1 = \dfrac{1}{2}$. Fair coin: $\dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{1}{8}$. The sum is $\dfrac{5}{8} = 0.625$.

Step 2. Which coin, after two heads. $\dfrac{1/2}{5/8} = \dfrac{4}{5} = 0.8$.

Step 3. Which coin, after one head. The paths to a head on the first toss are $\dfrac{1}{2} \times 1 = \dfrac{1}{2}$ and $\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$. So the posterior is $\dfrac{1/2}{3/4} = \dfrac{2}{3} \approx 0.6667$.