Fallacy in conditional probability solution for the Monty Hall Problem

Question

I know there are various established explanations for the problem. But here is my approach:

Just to be specific, the exact problem is as follows:
There are $3$ doors. Behind one of them is a car and behind the other two is nothing. You choose the first door. Now someone tells you that behind the door $3$ there is nothing. What is the probability that the car is behind door $2$, now.

So I define the events as:
$E:$ Car is behind door $2$.
$E1:$ There's nothing behind door $3$
Now I need to find: $P(E|E1)$.
So using the definition of conditional probability: $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$ We get $P(E|E1)=\large\frac{P(E\cap E1)}{P(E1)}$.
Now $E\cap E1$ is the event that Car is behind door $2$ and there's nothing behind door $3$. Now, the set $E=\{(\times,C,\times)\}$ and $E1=\{(C,\times,\times),(\times,C,\times)\}$. So $E\cap E1=\{(\times,C,\times)\}$. $$P(E\cap E1)=\frac{1}{3}$$ Similarly denominator $= \large\frac{2}{3}$.

But this gives the answer to be $\large\frac{1}{2}$.

I don't think there's any problem with the solution. If so, how does my problem statement differ with that of the Monty Hall Problem?

Answer 1

Your E1 is the event that there's nothing behind door 3. This is not the same as the event that someone tells you that there's nothing behind door 3.

Answer 2

Your solution is okay because the problem statement is not the same as Monty Hall's. The thing about Monty Hall's problem statement is that the presenter knows which door contains the prize and that, regardless of whether the door you initially selected contains contains the prize or not, the presenter will choose a door that you did not select and that does not contain a prize and reveal this to you.

Notice that such a choice is always possible.

If you initially picked the door with the prize (probability $1/3$), then the presenter has freedom in his choice of revelation, because both of the other two doors contain nothing. However, regardless of that, if you then change doors you will pick a door with no prize.

Now, if you initially picked a door without the prize (probability $2/3$), then the presenter must reveal to you the only other door that does not contain the prize. In this case, if you switch, the other door necessarily contains the prize.

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Say there were $n\geqslant 3$ doors, only one of which has the prize. Let $I$ be the event that the door you choose initially has the prize. Let $S$ be the event that the door you switch to (after the presenter's revelation) has the prize.

$$\begin{align} \Bbb P(S) &= \Bbb P(S|I)\cdot \Bbb P(I) + \Bbb P(S|I^C)\cdot \Bbb P(I^C) \\&= 0 \cdot \frac1n + \frac1{n-2}\cdot \frac{n-1}n = \frac{n-1}{n(n-2)} \end{align}$$

Notice that with $n=3$ you recover $\Bbb P(S) = \frac23$.