I am trying to solve the Monty Hall Problem with $3$ doors ($2$ goats, $1$ prize) with a simulation of the rules
- You initially pick a door randomly from the 3 possible doors.
- Monty hall does not know anything about the prize door and the goats door.
- He opens a random door except the one you have chosen ( The door he opens may even be the prize).
- Gives you the option to switch.
- Should switching be preferred.
My simulations give the $\mathbb{P}$robability as $\dfrac{2}{3}$ in this case. The event is $\mathbb{P}(C_2|G)$, where $C_i$ is the probability when the prize is behind $i$, $G$ is the event that a goat is revealed behind the third door.
Even if the 2nd condition is changed to Monty hall every time shows a different door from the chosen one but always the goat door, even then the $\mathbb{P}$robability is $\dfrac{2}{3}=\dfrac{1}{1+\frac{1}{2}}$
are my two probabilities correct
I do not understand the question, so going by your title and intro, I will provide the intuition behind the solution of the Monty Hall Problem. If this does not answer your question, please comment that down below.
Let us say we are playing the "Lets Make a Deal" Game. I am Monty and you are the player.
We have three doors to choose from: A, B and C.
As you have already explained the rules of the game, let us just jump into the solution, shall we?
Solution
Let us assume that A and C have the goats and B has the car.
Now, if you pick C, then I am force to show you a goat, that meaning I have to show you door A. If you switch, then you win the car!
If you picked door A, then I will be forced to show you door C; If you switch, you win the car again.
The probability of losing comes when you pick the right door. If you pick B, then I will show you either A or B. After showing you a goat, if you switch, you lose.
However, we just came up with a conclusion: Out of three games, you win two times, and you lose once. That being said:
P(Winning) = $2/3$
P(Losing) = $1/3$