We've all heard of the famous Monty Hall problem. However, what if Monty always picks the leftmost goat (and the player knows this)? Does this change the problem?
I don't think it does because Monty is always picking a goat door anyway. Does that make sense?
Certainly it changes the problem. This means in particular that Monty can never choose door $3$ (labelled left to right). If Monty chooses door $2$ then we know that the car must be in door $1$. This is information which we never had before.
Edit: A slight mistake from above. I forgot to account for the fact that Monty's decision changes based on the door that we choose. Nevertheless, the game is changed quite a bit due to the asymmetry.
Here are the probability breakdowns. Let $C_i$ be the event that the car is in door $i$ and $M_i$ be the event that Monty opens door $i$.
If we choose door $1$: $$\Pr(C_1|M_2) = 0.5,\ \ \Pr(C_3|M_2)=0.5$$ $$\Pr(C_1|M_3) = 0,\ \ \Pr(C_2|M_3) = 1$$ If we choose door $2$: $$\Pr(C_2|M_1) = 0.5,\ \ \Pr(C_3|M_1) = 0.5$$ $$\Pr(C_1|M_3) = 1,\ \ \Pr(C_2|M_3) = 0$$ If we choose door $3$: $$\Pr(C_2|M_1) = 0.5,\ \ \Pr(C_2|M_1)=0.5$$ $$\Pr(C_1|M_2) = 1,\ \ \Pr(C_3|M_2) = 0$$ So in this asymmetrical game, it depends on the door that Monty chooses. There is no benefit to switching in certain cases, while in others it's a sure win.
Interestingly, if we always keep the decision to switch, then the probability is indeed the same at $\frac{2}{3}$, so the chances of us winning by switching remains $\frac{2}{3}$ but now we have knowledge of when the switch will benefit us and when it will not.