Question
In: Statistics and Probability
Suppose Monty Hall wants to make his game show more interesting. In the new version, there...
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?
Solutions
Expert Solution
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.
orchestra answered 2 years agoRelated Solutions
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# 1. Monty-Hall Three doors
Recall the Monty-Hall game with three doors, discussed in class.
Run a simulation to check that the probablility of winning
increases to 2/3 if we switch doors at step two.
Set up the experiment two functions "monty_3doors_noswitch" and
"monty_3doors_switch" (these functions will have no input
values):
```{r}
monty_3doors_noswitch <- function(){
}
monty_3doors_switch <- function(){
}
```
Use your two functions and the replicate function to compute the
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