Monty hall problem with leftmost goat selection.

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

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.

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