Consider the following 7-door version of the Monty Hall problem. There are 7

doors, behind one of which there is a car (which you want), and behind the rest of which

there are goats (which you don't want). Initially, all possibilities are equally likely for

where the car is. You choose a door. Monty Hall then opens 3 goat doors, and offers

you the option of switching to any of the remaining 3 doors.

Assume that Monty Hall knows which door has the car, will always open 3 goat doors

and offer the option of switching, and that Monty chooses with equal probabilities from

all his choices of which goat doors to open. Should you switch? What is your probability of success if you switch to one of the remaining 3 doors?

Yes, you should switch.

Your probability of success if you switch to one of the remaining doors is 2/7.

Step-by-step explanation:

There are 7 doors.

You choose one door, then your probability of success (winning the car) is one out of seven: 1/7

And you have 6 out of seven chances to not get the car. Your probability of fail is 6/7.

This is, there is 1/7 chances the car is behind the door you opened and 6/7 chances the car is behind one of the other doors.

Each of the 6 doors that you did not open has 1/7 chances of being the door behind which the car is.

When Monty Hall opens 3 goat doors, the 6/7 chances goe to the 3 doors that he did not open. Then, each of the 3 remaining (still closed) doors will have 1/3 of 6/7 chances: (1/3) * (6/7) = 2/7.

Remember, your door, which is closed too, has 1/7 probability of success. Each of the other doors that are still closed have 2/7 probability of success.

Hence, you will have the double probabilities of success if you switch.