Re-tossing tails only in pairs

You toss four fair coins together. After seeing the result you may re-toss any two coins that currently show tails (two at a time; a lone tail cannot be re-tossed by itself), and you may repeat this as long as at least two tails are showing. Following the rule of re-tossing a pair of tails whenever you can, what is the expected total number of individual coin tosses, counting the initial four, until no further move is possible?

$\tfrac{176}{27}$. Let $e_k$ be the expected extra tosses with $k$ heads showing; $e_3=e_4=0$. Each pair re-toss costs 2 and adds 0, 1 or 2 heads with probabilities $\tfrac14,\tfrac12,\tfrac14$, giving $e_2=\tfrac83$, $e_1=\tfrac{40}{9}$, $e_0=\tfrac{176}{27}$. Averaging over the binomial first toss, $4+\tfrac{1}{16}e_0+\tfrac{4}{16}e_1+\tfrac{6}{16}e_2=\tfrac{176}{27}$.