1 post tagged “three doors”
It's the door thing in 21. HAHAHA. Statistics yo. ;) Watch the movie, I think it's über cool.
Imagine that you are in a game show. There are three closed doors. Behind one of these doors is a car; behind the other two are goats. You do not know where the car is, but the host does.
The hosts asks you to pick a door and the host opens one of the remaining doors, one he knows doesn't hide the car. If you have already chosen the correct door, the host is equally likely to open either of the two remaining doors.
After the host has shown a goat behind the door that he opens, you are always given the option to switch doors. What is the probability of winning the car if you stay with your first choice? What if you decide to switch?
We will assume that there is a winning door and that the two remaining doors, A and B, both have goats behind them. There are three options:
1. You first choose the door with the car behind it. You are then shown either door A or door B, which reveals a goat. If you change your choice of doors, you lose. If you stay with your original choice, you wins.
2. You first choose door A. You are then shown door B, which has a goat behind it. If you switch to the remaining door, you win the car. Otherwise, you lose.
3. You first choose door B. You are then shown door A, which has a goat behind it. If you switch to the remaining door, you win the car. Otherwise, you lose.
Each of the above three options has a 1/3 probability of occurring, because you are equally likely to begin by choosing any one of the three doors. In two of the above options (#s 2 & 3), you win the car if you switch doors; in only one of the options (#1) do you win if you do not switch doors. When you switch, you win the car twice (the number of favorable outcomes) out of three possible options (the sample space). Thus the probability of winning the car is 2/3 if you switch doors, which means that you should always switch doors.
This result of 2/3 or 66.6% may seem counterintuitive to many of us because we may believe that the probability of winning the car should be 1/2 or "50-50" once the host has shown that the car is not behind door A or door B. Many people reason that since there are two doors left, one of which must conceal the car, the probability of winning must be 50%. This would mean that switching doors would not make a difference. As shown through the three different options, however, this is not the case.
What if there were 1,000 doors? You would have a 1/1,000 chance of picking the correct door. If the host opens 998 doors, all of them with goats behind them, the door that you chose first will still have a 1/1,000 chance of being the one that conceals the car, but the other remaining door will have a 999/1,000 probability of being the door that is concealing the car. Here switching sounds like a pretty good idea.
Reference: http://mathforum.org/dr.math/faq/faq.monty.hall.html
Yeah, I'm cool like that. 8D
