Let $f(x)=1-x, g(x)=\frac{1}{x}$. Then for any $x, f(f(x))=x$ and $g(g(x))=x$. Furthermore, $f(g(x))=1-\frac{1}{x}, g(f(g(x)))=\frac{x}{x-1}, f(g(f(g(x))))=\frac{1}{1-x}, g(f(g(f(g(x)))))=1-x=f(x)$, so for all $n, x_{n}$ is one of $x, \frac{1}{x}, 1-\frac{1}{x}, \frac{x}{x-1}, \frac{1}{1-x}, 1-x$, and we can understand the coin flipping procedure as moving either left or right with equal probability along this cycle of values. For most $x$, all six of these values are distinct. In this case, suppose that we move right $R$ times and left $2012-R$ times between $x_{0}$ and $x_{2012}$. For $x_{2012}=x_{0}$, we need to have that $R-2012+R \equiv 0(\bmod 6)$, or $R \equiv 1(\bmod 3)$. The number of possible ways to return to $x_{0}$ is then $a=\binom{2012}{1}+\binom{2012}{4}+\cdots+\binom{2012}{2011}$. Let $b=\binom{2012}{0}+\binom{2012}{3}+\cdots+\binom{2012}{2010}=\binom{2012}{2}+\binom{2012}{5}+\cdots+\binom{2012}{2012}$. Then we have $a+2 b=2^{2012}$ and that $b=\frac{(1+1)^{2012}+(1+\omega)^{2012}+\left(1+\omega^{2}\right)^{2012}}{3}$, where $\omega$ is a primitive third root of unity. It can be seen that $1+\omega$ is a primitive sixth root of unity and $1+\omega^{2}$ is its inverse, so $(1+\omega)^{2012}=(1+\omega)^{2}=\omega$, and similarly $\left(1+\omega^{2}\right)^{2012}=\omega^{2}$. Therefore, $b=\frac{2^{2012}-1}{3}$, so $a=2^{2012}-2 b=\frac{2^{2012}+2}{3}$, and our desired probability is then $\frac{a}{2^{2012}}=\frac{2^{2012}+2}{3 \cdot 2^{2012}}=\frac{2^{2011}+1}{3 \cdot 2^{2011}}$. For some $x_{0}$, however, the cycle of values can become degenerate. It could be the case that two adjacent values are equal. Let $y$ be a value that is equal to an adjacent value. Then $y=\frac{1}{y}$ or $y=1-y$, which gives $y \in\left\{-1, \frac{1}{2}\right\}$. Therefore, this only occurs in the cycle of values $-1,2, \frac{1}{2}, \frac{1}{2}, 2,-1$. In this case, note that after 2012 steps we will always end up an even number of steps away from our starting point, and each of the numbers occupies two spaces of opposite parity, so we would need to return to our original location, just as if all six numbers were distinct. Therefore in this case we again have that the probability that $x_{2012}=x_{0}$ is $\frac{2^{2011}+1}{3 \cdot 2^{2011}}$. It is also possible that two numbers two apart on the cycle are equal. For this to be the case, let $y$ be the value such that $f(g(y))=y$. Then $1-\frac{1}{x}=x$, or $x-1=x^{2}$, so $x=\frac{1 \pm i \sqrt{3}}{2}$. Let $\zeta=\frac{1+i \sqrt{3}}{2}$. Then we get that the cycle of values is $\zeta, \bar{\zeta}, \zeta, \bar{\zeta}, \zeta, \bar{\zeta}$, and since at the end we are always an even number of spaces away from our starting location, the probability that $x_{2012}=x_{0}$ is 1. Finally, we need to consider the possibility that two opposite numbers are equal. In this case we have a $y$ such that $f(g(f(y)))=y$, or $\frac{x}{x-1}=x$, so $x=2$. In this case we obtain the same cycle of numbers in the case where two adjacent numbers are equal, and so we again obtain the probability $\frac{2^{2011}+1}{3.2^{2011}}$. Therefore, the only possibilities are $1, \frac{2^{2011}+1}{3 \cdot 2^{2011}}$.