Which Door to Open? (II)
(See Question)
So, have you decided whether to stick to the door that you had initially chosen, or make a switch? Or perhaps you’re thinking that since 1 door is being opened, the probability of opening the door with a car behind would be 1/2 for each of the remaining 2 doors? Then think again!
Although we can make use of conditional probability to solve this problem, I will illustrate how to solve by using diagrams and lay-man explanation.
In the 3 situations below, the car is always in Door A, and we select different initial doors in each of the 3 different situations shown.
Situation 1:
- Car is in Door A.
- You select Door A initially.
- Game host opens Door B to show you the car isn’t in B, and asks if you want to switch your choice.
- You decides to switch from Door A to C.
- Game host opens Door C, and you don’t win the car.
Situation 2:
- Car is in Door A.
- You select Door B initially.
- Game host opens Door C to show you the car isn’t in C, and asks if you want to switch your choice.
- You decides to switch from Door B to A.
- Game host opens Door A, and you win the car!!!
Situation 3:
- Car is in Door A.
- You select Door C initially.
- Game host opens Door B to show you the car isn’t in B, and asks if you want to switch your choice.
- You decides to switch from Door C to A.
- Game host opens Door A, and you win the car!!!
So, from the illustration, you could see that the player switched in all three situations, and got to win the car in 2 out of 3. So, the probability of winning the car if you switch is 2/3, while the probability of winning the car if you stick to your initial choice is only 1/3.
So, switching is always a better option when you see it in a mathematical perspective. Get it?
This problem is actually called the "Monty Hall Problem", and it could be solved by using Bayes’ Theorem easily.
Extension: If you’re still interested until this point, do work this problem out.
Now, imagine there are 4 doors instead of 3, and the car is hidden behind one of the doors. The player chooses a door. The game host then opens some other door that is a loser. The player then switchs to another door. The game host then opens another as-yet-unopened losing door, different from the player’s current choice. Now, there are only two unopened doors left: the player’s current choice and another one. The player decides to switch again. What is his probability of winning the car?


































