Assignment 5 part 1 - fraction of winning times

Assignment 5 part 1 - fraction of winning times

by Cong Lu -
Number of replies: 5

Hello,

I'm wondering what is the expected result of  fraction of times the contestant winning the cars.

I tested play.many.monty.halls(100, TRUE), the result is 0.31; and 0.14 for play.many.monty.halls(100, FALSE).

My results looks much different with the example. I'm not sure whether my calculation wrong or the result should be a range.

Thanks,

Cong

In reply to Cong Lu

Re: Assignment 5 part 1 - fraction of winning times

by Erik Spence -
As per the articles linked at the top of the Assignment page, the result for switching is 2/3 and not switching is 1/3. Your answers should be near these values.
In reply to Cong Lu

Re: Assignment 5 part 1 - fraction of winning times

by Boyan Konstantinov Tsankov -
Law of large numbers means that you should get close to 66% if you switch and 33% if you keep +- some error (but definitely not 66 -> 31 like you have).
In reply to Boyan Konstantinov Tsankov

Re: Assignment 5 part 1 - fraction of winning times

by Anukrati Nigam -
I am getting similar results as Cong Lu where play.monty.hall(100, TRUE) is 0.52 and play.monty.hall(100, FALSE) is 0.14. Could you please expand on the switching bit? My understanding was if the switch is yes, then monty will not pick the already opened door and will pick one from the remaining doors. How does the switching becomes 2/3?
In reply to Anukrati Nigam

Re: Assignment 5 part 1 - fraction of winning times

by Boyan Konstantinov Tsankov -
Monty doesn't switch the door, the player does. You can imagine that if you are given a choice of three doors (1 with a car and 2 with goats), you have a 1/3 chance of guessing the door with the car on the first try. Therefore, the remaining 2 doors have a combined 2/3 chance of having the car. Now, Monty opens one of the remaining 2 doors with a goat (that is NOT the door you chose, nor the door with the car). Now, all of the sudden, since Monty has shown you one of the doors with the goat, the original 2/3 probability of the doors you didn't choose having the car gets "condensed" onto the door that you didn't pick. This is why choosing to switch doors will yield you 66% chance of winning, and choosing to keep your original door will yield you the original 1/3 chance of having chosen the correct door. Numberphile has a great video on Youtube that explains this problem.