Recovering a coin bias from a posterior odds ratio

Two biased coins have heads probabilities $p_1$ and $p_2$, both strictly between $0$ and $1$, with $p_1+p_2=1$ and $p_2>p_1$. You pick one of the two coins at random (each with probability $\tfrac12$) and toss it twice, getting heads both times. Given this, coin 2 is now four times as likely as coin 1 to be the one you picked. Find $p_2$.

$p_2=\tfrac23$. With equal priors the posterior odds equal the likelihood ratio, so $p_2^2=4p_1^2=4(1-p_2)^2$. Taking the positive root, $p_2=2(1-p_2)$, so $p_2=\tfrac23$.