What is the probability that, at the end of the game, one card of each color was turned over in each of the three rounds? 
Three players each have a red card, blue card and green card. The players will play a game that consists of three rounds. In each of the three rounds each player randomly turns over one of his/her cards without replacement. What is the probability that, at the end of the game, one card of each color was turned over in each of the three rounds? Express your answer as a common fraction.

I thought the answer would just be $1*(\dfrac{1}{3})^2*1*(\dfrac{1}{2})^2$ since the first person can have any card shown and the next $2$ must match, then the next round the first person can have any card shown and the others must match. The last time everyone matches so the probability is $1$. 
 A: The answer is $\frac{1}{18}$. 
The way to see it is that there are $216$ possible sequences constituting the game (each of players A, B, and C can play any of 6 permutations of cards, e.g. 
$(1,2,3)$ or $(1,3,2)$ or ...
To "win" (meet your conditions of each round being three different colors), player A can start with any card $(3)$.  wlog, assume that card is $R$. Then player B can start with $B$ or $G$ $(2)$ and player C must start with start with the third color $(1)$.
Then on turn 2, player A can still choose either of the remaining cards $(2)$. wlog, assume he chooses the color played by player B on turn 1.  Now player B cannot choose the color player 1 played on turn 1 because that would paint player C into an impossible corner -- he no longer has that third color available.  So player B has only one choice.  Finally, player C also has only one choice.
Of course, in the third turn everything is determined.
So the number of winning sequences is $3 \cdot 2 \cdot 2 = 12$ and the probability of a win is $\frac{12}{216} = \frac{1}{18}$.
A: The probability that the three players choose different colors on the first turn is $\frac{3}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$. Given that they do, consider the generating function for the colors played on the second turn: $(r+b)(b+g)(r+g)=\color{red}1b^2 2g+\color{red}1b^2 r+\color{red}1b g^2+\color{green}2 b g r+\color{red}1b r^2+\color{red}1g^2 r+\color{red}1g r^2$. The probability that the players choose different colors in the second round is therefore $\frac{\color{green}2}{\color{green}2+\color{red}6}=\frac{1}{4}$. Given that they do, which happens with probability $\frac{2}{9}\cdot\frac{1}{4}=\frac{1}{18}$, the probability they choose different colors on the last turn is $1$, so the final probability is $\frac{1}{18}$.
