The basic solution to the Monty Hall problem remains the same.
There is a $\tfrac{1}{3}$ chance of having picked the car at first, so there's a $\tfrac{2}{3}$ chance of it being behind another door and picking it after switching.
However, that doesn't mean that everything stays the same. After having picked a door, there are two doors left, a left one and a right one. Monty must open the leftmost door with a goat behind it. If (and only if) the car is behind the left door, he must open the right one. If the car is behind another door (either the right one or the one you picked), he must open the left door.
$\begin{array} {ccc|cc} \begin{array}{c}\text{Door} \\ \text{chosen}\end{array} & \begin{array}{c}\text{Car} \\ \text{is behind}\end{array} & \begin{array}{c}\text{Monty} \\ \text{must open}\end{array} & \begin{array}{c}\text{Stay} \\ \text{prize}\end{array} & \begin{array}{c}\text{Switch} \\ \text{prize}\end{array} \\ \hline 1 & 1 & 2 & \text{Car} & \text{Goat} \\ & 2 & 3 & \text{Goat} & \text{Car} \\ & 3 & 2 & \text{Goat} & \text{Car} \\ \hline 2 & 1 & 3 & \text{Goat} & \text{Car} \\ & 2 & 1 & \text{Car} & \text{Goat} \\ & 3 & 1 & \text{Goat} & \text{Car} \\ \hline 3 & 1 & 2 & \text{Goat} & \text{Car} \\ & 2 & 1 & \text{Goat} & \text{Car} \\ & 3 & 1 & \text{Car} & \text{Goat} \\ \end{array}$
If Monty picks the right door, the car has to be behind the left door and switching is a guaranteed win. If Monty picks the left door, the car is behind your door or the right door with equal probability.
As can be seen in the table above, there's a $\tfrac{1}{3}$ chance of Monty opening the right door with a certain win after switching and a $\tfrac{2}{3}$ chance of him opening the left door with a $\tfrac{1}{2}$ chance after switching.
Your total chance of winning when using the switching strategy is $\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot \frac{1}{2} = \frac{2}{3}$ as we said it would be.
So the left door is the right door.