Question

A variance of Monty Hall Problem. Let's say I am playing a game where there are 6 doors, there is a car behind 2 doors and there are goats behind 4 doors. I don't know what is behind the doors but I want a car. Let's say I pick a door at random, so Initially, my chances of winning a car are 1/3. But before I open my door to see if I won, the host of the game opens a door revealing a goat and asks me if I want to change doors. If I stick to my initial choice, the probability of winning a car is still 1/3, but what is the probability of winning a car if I change doors given this new evidence?

Answer 1

We have 2 events here :

A - Car behind door selected

B - Goat behind door opened

P(A) = 2/6 = 1/3

Now, event B happens. Now we need to find the probability of car behind the door we selected given that event B happened.

Which is P(A|B) = P(B|A) * P(A) / P(B)

P(B|A), is the probability of event B happening provided that the car is actually behind the door we selected. So, if we selected door 1 for instance and car is behind that door, then the host has 4 options to select from. Suppose he selected door 2.

So, P(B|A) = 1/4

P(B) is the probability of goat behind the door openened. So, we need to think of the ways this can happen.

1. You selected the door with a car behind it and he opened door 2. = 1/3 * 1/4

2. You selected door 2 and he opened door 2. Obviously this is 0.

3. You selected a door with goat behind it ( other than door 2), and he opened door 2. = 3/6 * 1/3

P(B) = 1/3*1/4 + 3/6*1/3 = 1/4

P(A|B) = 1/3 * 1/4 / 1/4 = 1/3

Hence, the probability of car behind the 4 unopened doors in now 2/3 .