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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?

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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.

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Car could be in door $3$ if we had chosen door $1$ –  Jean-Sébastien Oct 14 '12 at 5:34
Quick question, why does $Pr(C_1|M_2) = 0.5$ for choosing door 1? –  David Faux Oct 14 '12 at 8:10
You can have either CGG or GGC. In both cases, Monty won't choose door $1$ because you've chosen it. So he defaults to two. Both of these scenarios are equally likely. –  EuYu Oct 14 '12 at 9:55
@EuYu: "Interestingly, if we always keep the decision to switch, then the probability is indeed the same at $\frac{2}{3}$". Can you prove this using total probability i.e. across all the above mentioned configurations. –  RIchard Williams Jan 23 '13 at 10:20
@prasenjit I'm not sure what you mean by "total probability". Why is it not enough to calculate the probability for each configuration? –  EuYu Jan 23 '13 at 15:35

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