The Prisoner's Dilemma

Assign a red light a value of 0, a green light a value of 1, and a yellow light a value of 2. Now consider the sum of the values of all of the lights in arithmetic modulo 3. Under this assignment, pushing any button increases the value of exactly three lights by 1 mod 3 (since $2+1\equiv 0$ mod 3), and therefore does not change the sum of the values mod 3 at all. Since all red has a total value of 0 mod 3 and all green has a total value of 13 $\equiv 1$ mod 3, the puzzle cannot be solved.

In fact, this condition on the sum of the values of the lights is the only obstruction to solving the puzzle. Note that for any given light $L$, pushing the button four positions clockwise from it, then the one three positions clockwise, then three positions counterclockwise, and finally four positions counterclockwise results in light $L$ staying the same and all twelve other lights advancing one color.

Extending this idea, start from any light $S$. Find the first light clockwise from $S$ that does not match $S$, call it $L$, and execute the above maneuver once or twice as needed so that now all lights from $S$ to $L$, inclusive, are the same color (since all of the lights from $S$ up to but not including $L$ change in unison, but $L$ stays the same).

By repeating the strategy of the previous paragraph, you can make all of the lights the same color. But if the initial position had a total value of 1 mod 3, the monochrome position reached must be all green, because that is the only monochrome position with a total value of 1 mod 3.