There are three doors, behind each is either a car, or a goat. Unlike the original Monty Hall problem, there are 8, equally likely possibilites for the setup of the doors. The possibility sample space can be represented by all possible 3-element long sequences using a 0 (a goat), or a 1 (a car). The sequence 011 represents a goat in the left door, and a car in the remaining two. Also unlike the original problem, Monty will randomly choose one of the two doors you did not pick. If Monty opens a door with a car, you can not get the car from that door. This means that there are possibilities (besides 000), where you can not win anything. There is also the possibility of 111, where you have a 100% chance of winning a car.
The original door you pick is always the left one. I took the liberty of creating a table with all the possible outcomes, given which door is opened by Monty.
In the picture, Dm, and Dr represent Monty opening the middle and right doors, repectively. Depending on what is behind the door Monty opens, 1/2 of the sequences are elimated from the sample space of door setups you have. However, all remaining sequences are still equally likely.
My question is this: what is the probability of you winning any giving game? Does it boil down to 50%
Also, if you played repeately, what is the expected value for the number of cars you are expected to win in any given game?
What is the value for the number of games you have to play to have ~100% of winning one game? 8?