# Find probabilities of winning in a modified version of Monty Hall Problem

So I modified the original version of the Monty Hall problem and allow there to have 4 doors; 1 car and 3 goats behind the doors. I will choose one door, Monty, who knows where the car is, randomly reveals one of the other 3 doors to show a goat.

In a case which I didn't switch doors, my chance of winning is clearly $\frac{1}{4}$.

Suppose I select Door #1, Monty randomly reveals a goat from one of the remaining doors, I make a switch. In this case, I believe I can calculate my probabilities of winning this way:

Let $S:\{$ Succeed if I switch$\}$, and let $D_j: \{$ Car behind the Door $j$, where $j=\{1,2,3,4\}$ $\\}$

Since $P(D_j)=\frac{1}{4}$ ,

Then, \begin{align} P(S)&=P(S|D_{ 1 })\times \frac { 1 }{ 4 } +P(S|D_{ 2 })\times \frac { 1 }{ 4 } +P(S|D_{ 3 })\times \frac { 1 }{ 4 } +P(S|D_{ 4 })\times \frac { 1 }{ 4 } \\ \qquad &=0+(\frac { 1 }{ 2 } \cdot \frac { 1 }{ 4 } )+(\frac { 1 }{ 2 } \cdot \frac { 1 }{ 4 } )+(\frac { 1 }{ 2 } \cdot \frac { 1 }{ 4 } )\\ \qquad &=\frac { 3 }{ 8 } \end{align}

Now, in another strategy, say I do this: I select Door #1. Then,

1. If Monty opens Door #2, I switch to Door #3.
2. If Monty opens Door #3, I switch to Door #4.
3. If Monty opens Door #4, I remain with my initial first choice of Door #1.

Of course, all that Monty reveals are goats. I want to find out the probability of winning with this strategy.

Let $A_j$ be the $j$th event in the above numbered list, where $j=\{1,2,3\}$.

Then to find the probability of winning with this new strategy, I find:

$$P(S)=P(S|A_1)P(A_1)+P(S|A_2)P(A_2)+P(S|A_3)P(A_3)$$

I am not sure, but I rationalise the chance of happening for the events $P(A_1)=P(A_2)=P(A_3)=\frac{1}{3}$. With 4 doors and I have chosen 1 door, Monty left with 3 doors. So I thought it has to be $\frac{1}{3}$.

After here, I'm very confuse with how I should proceed on. $$P(S|A_1) = P(S|A_2) =\frac { P(A_1\cap S)}{ P(A) } =\frac { P(A_2\cap S)}{ P(A) }= \frac { \frac { 1 }{ 3 } \times \frac { 3 }{ 8 } }{ \frac { 1 }{ 3 } } =\frac { 3 }{ 8 }$$

In this equation, I assumed independence between $A_j$ and $S$, even though I don't know if they really are. Also, I am reusing the probability of $P(S)$ from the earlier strategy, which I am still pondering if it even make sense to do this.

Anyway, so if I proceed from above, I will get: \begin{align*} P(S)&=P(S|A_1)P(A_1)+P(S|A_2)P(A_2)+P(S|A_3)P(A_3) \\ &=\frac { 3 }{ 8 }\cdot\frac { 1 }{ 3 } +\frac { 3 }{ 8 }\cdot\frac { 1 }{ 3 } +\frac { 1 }{ 4 }\cdot\frac { 1 }{ 3 } =\frac { 1 }{ 3 } \end{align*}

Finally, my probability of winning with this new strategy would be $\frac{1}{3}$. But I am not certain at all if this is correct or not.

Is what I have done and thought correct? What should I do to find out the probabilities of the different strategies?

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If you are adding conditional probabilities for the purposes of using additivity, you need to keep the conditioning event the same. – guy Mar 13 '12 at 17:07
Your last calculation does not give the probability of obtaining the car. You need to multiply each term by $1/3$ (by the probability of the condition, which is $1/3$ in each case). – David Mitra Mar 13 '12 at 17:15
@DavidMitra Do you mean that I have to do this? $$\frac { 3 }{ 8 }\cdot\frac { 1 }{ 3 } +\frac { 3 }{ 8 }\cdot\frac { 1 }{ 3 } +\frac { 1 }{ 4 }\cdot\frac { 1 }{ 3 } =\frac { 1 }{ 3 }$$ And so $\frac{1}{3}$ is the probability of winning with the last strategy? But I thought I have included it in $P(S|A) =\frac { P(A\cap S)}{ P(A) } = \frac { \frac { 1 }{ 3 } \times \frac { 3 }{ 8 } \quad }{ \frac { 1 }{ 3 } } =\frac { 3 }{ 8 }$ where the $\frac{1}{3}$ got cancelled off? – xenon Mar 13 '12 at 18:09
@xEnOn Maybe I didn't understand what you were trying to do. But if you wanted to calculate $P(S)$ in a different way: $$P(S)= P(S\cap A_1)+P(S\cap A_2)+P(S\cap A_3)=P(S|A_1)P(A_1)+P(S|A_2)P(A_2)+P(S|A_3)P(A_3),$$ where the $A_i$ are your bulleted alternatives. (You used this formula correctly in your first calculation of $P(S)$.) – David Mitra Mar 13 '12 at 18:14
@DavidMitra ohh..yeah, thanks. I should have done that. I cleaned up my question a little with this amendment. Hope this makes my question clearer and easier to understand. Thanks! – xenon Mar 13 '12 at 19:11

This is not exactly correct. In fact, you are in doubt when you say that $S$ and $A_1$ are independent, and they are not.

What follows is some elementary considerations to show you why this is not the case. Note, that this is NOT a mathematician's approach (at least, till the very last paragraph), but might help you to see the picture a little better so that you don't make mistakes in your reasoning.

First, we must be clear about the probability space we are talking about. This space consists of all possible elementary outcomes that are of interest to us (those are the points of the space, each assigned with some probability). In our case an event is $(d,m,y)$, where $d=1..4$ is the door with the car behind it, $m=2..4$, $m\neq d$, is the Monty's choice, and $y=1..4$, $y\neq m$, is your final choice. Note, that things like "your winning" are not elementary events (points in our space) but rather subsets. For example, "you win" is an event $W$ that consists of all points $(d,m,y)$ such that $d=y$.

Now, let us partition our space into four equally probable subspaces: $D_i$ (the same way you did before). Then,

$$D_1=\left\{(1,2,1),(1,2,3),(1,2,4),(1,3,1),(1,3,2),(1,3,4),(1,4,1),(1,4,2),(1,4,3)\right\}$$

$$D_2=\left\{(2,3,1),(2,3,2),(2,3,4),(2,4,1),(2,4,2),(2,4,3)\right\}$$

etc. Note that even though $D_1$ has 9 points when all the other $D_i$'s have only 6, they are all equally probable with probability 1/4 regardless of Monty's or your strategy. What your strategy does change is the probability of each point within a subset.

Now, since Monty always uses the same strategy to pick a door with a goat at random which is different from the one you chose, then we can further separate each $D_i$ into 3 or 2 subsets $D_{ij}$ containing all points $(i,j,...)$. The probability of $D_{1j}$, $j=2..4$, is $\frac{1}{12}$, while the probability of $D_{ij}$, $i=2..4$, $j=2..4$, $j\neq i$, is $\frac{1}{8}$ (they all sum up to 1, as expected).

If you use your original strategy, then the probability of any point ending with 1 is 0 while other points within $D_{ij}$ are equally probable. You win if $i=y$, i.e. with the probability $6/16=3/8$.

Using your second strategy, $m$ determines your choice $y$, therefore, there is only one point with a positive probability within each $D_{ij}$ and the probability of this outcome just equals the probability of $D_{ij}$. So, we have $(1,2,3)$, $(1,3,4)$ and $(1,4,1)$ with probability $\frac{1}{12}$, and $(2,3,4)$, $(2,4,1)$, $(3,2,3)$, $(3,4,1)$, $(4,2,3)$ and $(4,3,4)$ with probability $\frac{1}{8}$. This gives the probability of winning $\frac{1}{3}$.

So, $P(S)=\frac{1}{3}$, $P(A_1)=\frac{1}{3}$, but $P(A_1\cap S)=\frac{1}{8}\neq\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$. So they are NOT independent.

You got the probability of $P(A_1\cap S)=\frac{1}{8}$ right, but your reasoning was wrong: what you said was 1) they are independent, 2) $P(A_1)=\frac{1}{3}$ (by symmetry), 3) let us use $P(S)=\frac{3}{8}$ from the previous setup. Only the second point is correct.

To calculate $P(S\cap A_1)$ correctly, you might have said something like this: $S\cap A_1$ happens only if the car is behind door #3 ( $P\{d=3\}=\frac{1}{4}$ ) and, conditional on this, Monty have chosen the door #2 ( $P\{m=2|d=3\}=\frac{1}{2}$ ), therefore, $P(S\cap A_1)=P\{d=3,m=2\}=P\{m=2|d=3\}\cdot P\{d=3\}=\frac{1}{8}$. That would be a correct way of deriving that probability.

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Thanks! This is really a good way to look at the problem! One thing that I don't understand is the part which you mentioned that "If you use your original strategy...You win if $i=y$, i.e. with the probability $6/16=3/8$." It appears to me that in $D_1$, after removing all the elements which have $y=1$, we still have 6 elements: $$(1,2,3),(1,2,4),(1,3,2),(1,3,4),(1,4,2),(1,4,3)$$ For $D_2, D_3, D_4$, after removing the elements with $y=1$, they left with 4 elements. So $4\times3=12$. Then $12+6=18$. Wouldn't the probability be $6/18=1/3$? – xenon Mar 14 '12 at 7:22
No, and I specifically mentioned that all $D_i$'s have the same probability but different number of points. This is exactly what makes, for example, $(1,2,3)$ less probable than, say, $(2,3,4)$. Think differently: they both require that the car is behind the specific door (probability 1/4), and both require that you choose the specific door (1/2, according to your strategy in the first case), BUT the former also requires that conditional on $d=1$ Monty chooses door #2 (probability 1/3) while the latter requires that conditional on $d=2$ Monty picks door #3 (probability 1/2). – Vadim Mar 15 '12 at 6:15
So, you have: $Pr\{D_1\}=1/4$, $D_1$ contains 9 points. According to Monty's strategy, $D_{12}$, $D_{13}$ and $D_{14}$ are all equally probable. Probability is $1/12$. Similarly, $D_{23}$ and $D_{24}$ are equally probable, but the probability of each is $1/8$. Now, according to your strategy, within each $D_{ij}$ points ending with 1 have 0 probability, while all other points are equally probable. Therefore, e.g. $Pr\{(1,2,3)\}=1/24$ while $Pr\{(2,3,4)\}=1/16$. Overall, you get 6*1/16=3/8. – Vadim Mar 15 '12 at 6:23