# What is the total number of outcomes in the Monty Hall problem?

While calculating the probability of outcomes in the Monty Hall problem (https://en.wikipedia.org/wiki/Monty_Hall_problem) why is the total number of outcomes taken to be three (https://betterexplained.com/articles/understanding-the-monty-hall-problem/)? It should be 8 since every door can have 2 outcomes (goat/car) and there are 3 doors, so the total number of outcomes should be 2^3 = 8. How is the total number of outcomes systematically calculated?

While it is true that there are $2^{3}$ possible configurations for three outcomes, each of which has two possible values, this is not the situation we have here. There is one car, which is to be positioned behind one door. There are thus ${{3}\choose{1}} = {{3}\choose{2}} = 3$ possible configurations. In setting up the game, once the door to place the car behind is selected, the two goats get positioned behind the remaining two doors. Alternatively, the goats may be positioned behind two of the doors, then the car is placed behind the remaining door. The binomial coefficient appears as there are two classes of object, goat and car, and no ordering of the objects within a class. If there were a chance of a car or goat appearing behind any door, with, for instance, three cars being a possible configuration, then there would be $2^{3}$ possible configurations.
• ${3}\choose{1}$ is standard mathematical notation for the combinations / binomial coefficients, and is read "3 choose 1." If my answer addressed the issues in your question, you could accept it to indicate the question has been answered. – Don Walpola Sep 7 '18 at 13:18