Question

(a) In the Monty Hall problem with 100 doors, you pick one and Monty opens 98 other doors with goats. What is the probability of winning (assuming you would rather have a car than a goat) if you switch to the remaining door? Explain your answer.

(b) Suppose Monte opens 98 doors without checking for cars. What is the probability that, once the doors are open, changing your choice will not change your chances of winning.

Answer 1

I'll try to make a generalised formula for this problem .

Suppose there are n doors and K are 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 1/n.

Monty then opens k doors, where 0≤ k ≤n−2 (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 n−1/n , times the probability of picking it now, which is 1 / n−k−1. This gives us a total probability of

(n−1/n )⋅(1/n−k−1) = (1/n)⋅(n−1/n−k−1) ≥ 1/n

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

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

Maximum number of doors revealed
If k is at its maximum value of n−2, the probability of picking a car after switching becomes

(1/n)⋅(n− 1)/(n−(n−2)−1) = (1/n)⋅(n−1/1) = n−1/n

So in our case k=98

therefore, probability of winning (assuming you would rather have a car than a goat) if you switch to the remaining door is:

100-1/100 = 99/100 =0.99 :Ans