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,
- If Monty opens Door #2, I switch to Door #3.
- If Monty opens Door #3, I switch to Door #4.
- 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?