Suppose instead of the normal Monty Hall scenario in which we have two empty doors and a car residing behind the third, we instead have three types of objects. One is a car, one is a hard drive, and the third is empty. Does the typical Monty Hall paradox of 2/3 chance of obtaining the car by switching versus a 1/3 chance of obtaining the car by staying apply in this particular case?
Let us assume that the contestant makes an initial pick at his/her discretion (random) and the host proceeds to ALWAYS open the empty door. Let's say that when the contestant's initial pick corresponds to the empty door and the remaining two doors hold the car and hard drive, the host will still open the empty door corresponding to the initial pick. The contestant than chooses between the remaining two doors.
It was my opinion that the action of opening the door with nothing behind it in this three object scenario doesn't inform the contestant such that they can update the posterior probability using Bayes Theorem in any meaningful way. I was convinced that this is where it differed from the original Monty Hall problem and that there will always be a 50/50 chance of success regardless of whether you stay or switch in a three object scenario where the host opens the empty door.
My coworkers, and even the person who posed the question, are insistent that the original rules apply and, given that the host opens an empty door, your chances of obtaining the car by switching are 2/3 and by staying are 1/3. Who is correct?
The original question was posed exactly as seen below:
Here’s the situation: You’re on a gameshow! There are three doors. Behind door #1 is a red Ferrari. Behind door #2 are piles of HDDs needing analysis. Behind door #3 there is nothing. You can select one door to open. The audience is going crazy. You have just pointed to and selected a door, and are about to open it.
Now before you open your door, the host (let’s pretend it’s Richard Dawson, RIP) opens one door himself, the empty one, and shows you and the audience it’s empty.
Then he asks you this question: “Do you want to keep your original selection or switch to the other door?”. The audience is going bananas. Here’s my question to you:
Assuming you’d like to have the Ferrari – What do you do (keep original or switch), and why? (you must explain the logic).
Best Answer
What are the possible outcomes?
(1) Picked car and stuck with it. (Win)
(2) Picked car and switched to hard drives. (Lose)
(3) Picked hard drives and stuck with it. (Lose)
(4) Picked hard drives and switched to car. (Win)
(5) Picked empty door and switched to car. (Win)
(6) Picked empty door and switched to harddrives. (Lose)
The initial picks are all equally likely. If we pick a door with something behind it initially, then we have a $50\%$ chance of winning if we switch. If we pick the empty door, we are forced to switch (by sense, if not by rule), and we again have a $50\%$ chance of winning. Thus, unlike the original Monty Hall problem, it is precisely as it seems: a $50\%$ chance of winning, regardless of strategy.