Monty hall problem with leftmost goat selection.

Question

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?

Answer 1

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.