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

While calculating the probability of outcomes in the Monty Hall problem why is the total number of outcomes taken to be three (see 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?

• If you calculate the outcomes as 2*2*2, you are implying that the content of each door is independent of the others, like if we could have a car in door1 and then a car again in door2. But that's not true. Once the car is in one specific door, the other ones are determined to have a goat. Commented Feb 2, 2020 at 22:49
• Commented Mar 25, 2021 at 6:07
• Commented Aug 31, 2022 at 11:45

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.

Furthermore, the outcomes of interest are which door is actually picked by the contestant. These correspond to whichever door in the actually realized configuration is chosen. So suppose, from left to right, behind the doors were a goat, the car, and a goat; there are three choices a contestant could make - left, middle, or right door - but the contestant only wins if they select the middle door. This is true regardless of the actual configuration of goats and car behind the door - contestants only win when their choice (the guessed configuration) is the same as the game's 'choice' (the car/goat configuration) of where the car is.

• Does writing the numbers in curly braces vertically has any special meaning in statistics ? Commented Sep 7, 2018 at 11:34
• ${3}\choose{1}$ 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. Commented Sep 7, 2018 at 13:18
• note that ${\binom {n}{k}}={\frac {n!}{k!(n-k)!}}$, where $!$ here is the factorial operator: en.wikipedia.org/wiki/Factorial Commented Mar 1, 2021 at 23:14

You are confused with all possible constellations of goats/cars, not taking in consideration the required number of goats (2) and cars (1).

It means that your calculation $$2^3$$ describes this set of possibilities:

Constellation No. 1 No. 2 No. 3
1st goat goat goat
2st goat goat CAR
3nd goat CAR goat
4rd goat CAR CAR
5th CAR goat goat
6th CAR goat CAR
7th CAR CAR goat
8th CAR CAR CAR

But to take in consideration the terms of this television game, i.e. the car is behind one and only one door, we end up only with 3 possibilities from the previous table:

• the 2st constellation – the car is behind № 3 (goats behind remaining doors),

• the 3rd constellation – the car is behind № 2 (goats behind remaining doors),

• the 5th constellation – the car is behind № 1 (goats behind remaining doors):

Constellation No. 1 No. 2 No. 3
2st goat goat CAR
3nd goat CAR goat
5th CAR goat goat