Monty hall problem extended.

I just learned about the Monty Hall problem and found it quite amazing. So I thought about extending the problem a bit to understand more about it.

In this modification of the Monty Hall Problem, instead of three doors, we have four (or maybe n) doors, one with a car and the other three (or n1) with a goat each (I want the car).

We need to choose any one of the doors. After we have chosen the door, Monty deliberately reveals one of the doors that has a goat and asks us if we wish to change our choice.

So should we switch the door we have chosen, or does it not matter if we switch or stay with our choice?

It would be even better if we knew the probability of winning upon switching given that Monty opens k doors.


I decided to make an answer out of my comment, just for the heck of it.

n doors, k revealed

Suppose we have n doors, with a car behind 1 of them. The probability of choosing the door with the car behind it on your first pick, is 1n.

Monty then opens k doors, where 0kn2 (he has to leave your original door and at least one other door closed).

The probability of picking the car if you choose a different door, is the chance of not having picked the car in the first place, which is n1n, times the probability of picking it now, which is 1nk1. This gives us a total probability of n1n1nk1=1nn1nk11n

No doors revealed
If Monty opens no doors, k=0 and that reduces to 1n, which means your odds remain the same.

At least one door revealed
For all k>0, n1nk1>1 and so the probabilty of picking the car on your second guess is greater than 1n.

Maximum number of doors revealed
If k is at its maximum value of n2, the probability of picking a car after switching becomes 1nn1n(n2)1=1nn11=n1n
For n=3, this is the solution to the original Monty Hall problem.


Source : Link , Question Author : Shaurya Gupta , Answer Author : Community

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