Question

Suppose Monty Hall wants to make his game show more interesting. In the new version, there are 7 doors and 2 cars behind the doors. The other 5 doors have goats behind them. A contestant picks a door. Monty Hall opens 3 of the doors with goats behind them (never the door that the contestant picked or a door with a car). The contestant then has the choice of switching doors.find the probability that the contestant wins a car if he/she switches doors. Is it better to switch or stick with the original guess?

*ran a computer simulation and found that the probability when switching is about 2/7*

Almost certain the simulation is correct, but I could be wrong.

Answer 1

In the original choice, the probability of winning the car is 2/7

After 3 doors are opened, the probability that the car is behind the door the contestant picked increases to 2/4, but he never knew which doors would be opened. It might be that his initial guess is wrong and the probability of losing was 5/7.

Now we have two cases with change of door:

1. Initially car was behind door, selected new door and there is car behind door = 2/4*1/3 = 1/6

2. Initially car was not behind door, selected new door and there is car behind door = 2/4*2/3 = 1/3

Total probability that switching door will result in winning a car = 1/6 + 1/3 = 1/2

It is better to switch as there is a higher probability of winning the car by switching doors rather staying with the one selected.