Monty Hall Problem

Let's look at the Monty Hall problem. A pretty standard Probability problem. 

Let's look at the problem -

The problem is inspired from the Game Show "Let's Make A Deal", hosted by Monty Hall, after whom the problem is named. The rules of the game are - You are given 3 doors, out of which behind 2 are a goat, and behind the third one, lies a car. You are to choose one of the 3 doors, then Mr Monty opens one of the remaining door behind which lies a goat. Now, you are given a choice to either maintain your choice, or to switch. Probabilistically, which is better given that you would want a car.

Thinking casually would probably make you feel that both would probably give a 50-50 chance of winning a car, but mathematically, that is evidently not the case.

Mathematically Thinking

Let us say we have 3 doors, A, B and C. Say, we choose door A. 

P(Car ∈ A) =  ¹⁄₃  

P(Car ∈ B) =  ¹⁄₃  

P(Car ∈ C) =  ¹⁄₃

Because we have 2 Goats and 1 Car behind the doors.

Say, the host opens door B, and shows a goat. "Logically" thinking would say that we have one goat and one car, and choosing one of them among two doors give an equal probability of 0.5 for each. But, the items were divided uniformly before the door was opened, but need not be uniform now. So, the uniform distribution is not required.

We need to get it into the initial condition to use the uniform condition.

Let's get into it.

P(Car | Switch) = P(Car ∈ C | Goat ∈ B) = P(Goat ∈ A) =   ²⁄₃

Basically, to get a car on switching, we would need to have initially chosen a goat before the switch, and that is uniformly distributed. So, we can say that the Probability of getting a car if we switch is 0.67.

So, we should switch.

Bayes Theorem Method

We can also look at it with The Bayes Theorem.

P(A | B) = ^{P(A, B)}⁄_P(B)

Here, take 

A = Choosing a Car after we switch

B = Having a goat in one of the doors not chosen by us

P(A') = P(Choosing a car initially) = 1/3, so P(A) = 2/3

By how our set up is, given any 2 doors, there will always be a door with a goat behind it. If not, then that would imply there being more than one door with a car behind it, or a door without anything behind it.

Thus, P(B | A) = P(B | A') = 1

 P(B) = P(B | A) P(A) + P(B | A') P(A') [Total Probability Theorem]

P(A | B) = P(B | A) P(A) / [P(B | A) P(A) + P(B | A') P(A')]

P(A | B) = 2/3

Which is the result we got earlier.

Alternations to the Problem

An alternative has been presented where there are 100 doors, and behind one door is a car, and behind the rest are goats. The participant chooses one door, and the host opens 98 doors behind which there are goats.

Should we switch?

Use the same idea as before. Get the problem to the initial uniformity.

P(Car | Switch) = P(Car ∈ Door 100 | Goat ∈ Door{2, ..., 99}) = P(Goat ∈ Door 1) =   ⁹⁹⁄₁₀₀

So, it is significantly better to switch.