# 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. – Ronald Becerra Feb 2 '20 at 22:49
• – MarianD Mar 25 at 6:07

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 ? – Talespin_Kit Sep 7 '18 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. – Don Walpola Sep 7 '18 at 13:18
• note that ${\binom {n}{k}}={\frac {n!}{k!(n-k)!}}$, where $!$ here is the factorial operator: en.wikipedia.org/wiki/Factorial – StatsStudent Mar 1 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 (0 means a goat, 1 means a car):

Constellation No. 1 No. 2 No. 3
1st 0 0 1
2nd 0 1 0
3rd 0 1 1
4th 1 1 1
5th 1 0 0
6th 1 0 1
7th 1 1 0
8th 1 1 1

But to take in consideration the number of goats (2) and cars (1), we end up only with 3 possibilities: The car is in № 3, or in № 2, or in № 1 (goats are always in remaining doors):

Constellation No. 1 No. 2 No. 3
1st 0 0 1
2nd 0 1 0
3rd 1 0 0