smc

Let $x$ be the probability of flipping heads. It follows that the probability of flipping tails is $1-x$. The probability of flipping $2$ heads and $2$ tails is equal to the number of ways to flip it times the product of the probability of flipping each coin. $$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{4}{2}\right)x^2(1-x{)}^2& =\frac{1}{6}\\ 6x^2(1-x{)}^2& =\frac{1}{6}\\ x^2(1-x{)}^2& =\frac{1}{36}\\ x(1-x)& =\pm \frac{1}{6}\end{array}$$ As for the desired probability $x$ both $x$ and $1-x$ are nonnegative, we only need to consider the positive root, hence $$\begin{array}{rl}x(1-x)& =\frac{1}{6}\\ 6x^2-6x+1& =0\end{array}$$ Applying the quadratic formula we get that the roots of this equation are $\frac{3\pm \sqrt{3}}{6}$. As the probability of heads is less than $\frac{1}{2}$, we get that the answer is $\boxed{\frac{3-\sqrt{3}}{6}}$.