# Variation of the Monty Hall Problem.

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).

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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.

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"If we pick a prize door initially, then we have a 50% chance of winning." That seems to assume that we must decide at random whether to switch or not. –  Henning Makholm Jun 6 '12 at 11:24
@HenningMakholm: Apologies for the misleading phrasing. Does my edit allay that? –  Cameron Buie Jun 6 '12 at 11:50
Perhaps we're just confused at what constitutes a "prize". My intuitive understanding was that the car is the only prize. If you find nothing you go away emptyhanded, and if we find "piles of HDDs needing analysis", that means that we lose twice over: not only do we not get a car, but we get loaded with extra work analysing the disks. So in my understanding, if we pick a prize door initially, then we have 0% chance of winning if we switch. –  Henning Makholm Jun 6 '12 at 12:14
@HenningMakholm: Ah! How did I miss that they needed analysis? I will correct for that. –  Cameron Buie Jun 6 '12 at 12:20

It it's a rule of the game that the door being opened is fixed in advance (such that you don't get a chance to switch if you happen to pick that door), then there is no advantage to switching.

The advantage from switching in the usual scenario comes from the fact that if you happen to pick a non-prize door, the host is forced to reveal information about where the other non-prize door, such that the strategy of always switching will consitently make you win if and only if you initially didn't pick the prize door.

In your variant the risk of picking a non-prize door and not getting a chance to switch neutralizes that advantage. The difference between the two games only show up in those situations where switching in the original game would have benefited you; and losing that opportunity shows up as a lower probability of winning.

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Brandon specifies that if you pick the empty door, you don't necessarily lose; rather the host lets you choose between the remaining two doors. Either way, switching doesn't improve your odds, for the reasons you give. –  Jonas Kibelbek Jun 6 '12 at 1:56
@Jonas: Right; I didn't notice that. –  Henning Makholm Jun 6 '12 at 11:24

It is 1/2 instead of 1/3 because the host is not forced to reveal the other door that is the hard drive when you pick the empty door. So you don't have that two out of three chance that the door with the car is revealed as the remaining door.

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This is the best I could come up with:

The Monty Hall scenario still applies if you don't pick the empty door.

If you don't pick the empty door then {Pick, Empty, Other Prize} is equal to {Empty, Empty, Prize} since you don't know if Pick = Car and you want the Car.

If you pick the empty door and this is revealed to you, then the game degenerates in a "heads-or-tails" with 50% chance of winning.

${2\over{3!}} = {2\over{6}} = 33.3\%$ you pick the empty door, $66.6\%$ you don't.

So your chances to win, considering all the possible outcomes as equiprobable and the fact that you always swap if you didn't pick the empty door (like the classic Monty Hall), is:

$33.3\% \cdot 50.0\% + 66.6\% \cdot 66.6\% \simeq 61\%$

Does it makes sense to you?

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The way I see it, there's nothing sacred about the Ferrari and the hard disk. Changing by choice or because original door was empty shouldn't make it biased towards the car. eg. if you had a Beamer instead of the HD, there's no reason that a hold or a switch should tilt the scale towards one car or the other.

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This might be better as a comment. Maybe you can add some more mathematical rationale to improve it. Regards –  Amzoti Mar 28 '13 at 19:16

Your initial guess has a 1/3 prob of success. After the host has revealed the empty door, you have a 1/2 chance of success. So it's always better to change. You have a 2/3 chance if the third door was opened or 1/2 chance if your original door was opened.

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