6. [4] When flipped, a coin has a probability $p$ of landing heads. When flipped twice, it is twice as likely to land on the same side both times as it is to land on each side once. What is the larger possible value of $p$ ?
$2 p^{2}-2 p+1$. The probability that the coin will land on each side once is $p(p-1)+(p-1) p=$ $2 p(1-p)=2 p-2 p^{2}$. We are told that it is twice as likely to land on the same side both times, so $2 p^{2}-2 p+1=2\left(2 p-2 p^{2}\right)=4 p-4 p^{2}$. Solving, we get $6 p^{2}-6 p+1=0$, so $p=\frac{3 \pm \sqrt{3}}{6}$. The larger solution is $p=\frac{3+\sqrt{3}}{6}$.