Monty Hall's problem argument

Question

Let us take 3 doors A, B, C. Now, let us say car is in A.

Now if player selects A, host opens B, on switching player loses.

If player selects A, host opens C, on switching player loses.

If player selects B, host opens C, on switching player wins.

If player selects C, host opens B, on switching player wins.

So 50% winning prob instead of 66%. Although when I use total prob theorem, I do end up getting 66%. Now, can somebody explain to me the reason why the above mentioned cases are not equally likely?

Answer 1

On the assumption that the player chooses $A,B$, and $C$ with equal probability $\frac13$, and that if the player chooses $A$, the host opens $B$ and $C$ with equal probability $\frac12$, the actual probability of your first event is $\frac13\cdot\frac12=\frac16$. The same goes for your second event. Each of your other two events, however, has probability $\frac13$.

As long as the host has a positive probability of opening each of doors $B$ and $C$ when the player chooses $A$, the probabilities of your first two events will be less than $\frac13$.

Answer 2

The case "Player select $A$ and host opens $B$" happens in one time on six and so for the case "Player select $A$ and host opens $C$".

Answer 3

The four outcomes you have described do not have equal probabilities.

They have probabilities $1/6, 1/6, 1/3, 1/3$ respectively (the player selects a door randomly, and if they choose A the host randomly chooses between B and C, which is why the first two outcomes are $\frac{1}{3} \cdot \frac{1}{2}$), so the probability of winning is $1/3 + 1/3 = 2/3$.