Monty Hall Problem Wrong Reasoning: By Considering All Possibilities

Question

Monty Hall Problem: https://en.wikipedia.org/wiki/Monty_Hall_problem

Here is a wrong reasoning. I am unable to find the flaw in it.

Given below I have attached three images which give us the three initial possibilities of the items behind the doors. Assume that "C" represents a car, and "G" represents a goat. The three compartments in each rounded rectangle are the three doors. The topmost node of each image represents an initial starting position (which can be seen only by the game show host). The colour in red represents the door chosen. So if "C" is in door number 2, and "C" is in red, then door 2 is chosen by the player. The color in green represents the door that the host chooses to show. If the player does not choose to switch, then the number of possibilities that the player wins are 16 among 32, so the probability that player wins is $\frac{1}{2}$ independent of the startegy! What is incorrect here?

Edit: I am looking for a counting argument for the question (and not a probabilistic argument). At the end the argument should be just like $\frac{no.~of~good~possibilities}{total~no.~of~possibilities}$.

Possibility 1:

Possibility 2:

Possibility 3:

Answer 1

The issue is that the two outcomes of winning or losing are not equally likely due to the conditions of the problem. Conditional probability helps avoid this situation, and is what the Monty Hall problem is about.

Answer 2

Using your diagrames:

This is 'Possibility 1'. First, we choose doors randomly with probability $\frac{1}{3}$ (as you can see on the image). Then the host reveals a goat: If we chose doors with car, he can do it in two ways, so there is $\frac{1}{2}$ probability of which door with goat he chooses. In other cases, he has no choice but to pick second doors with goat, hence there is probability $1$. In the last row, you have $4$ scenarios. Let's say that in each of them we switch doors. In the first and second scenario (where we picked car at the start) when we switch we lose: the probability of that is $\frac{1}{3}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{1}{2} = \frac{1}{3}$. In the third and fourth scenario, when we switch doors we win: the probability of that is $\frac{1}{3}\cdot 1+\frac{1}{3}\cdot 1=\frac{2}{3}$. Notice that the same thing happens in your other two 'Possibilities'. I hope this clears why your argument was wrong: not each scenario has equal 'weight'.