The Monty Hall Problem

[Ramor]

You're on a gameshow, "Let's Make a Deal". There are three doors, Door 1, Door 2, and Door 3. Behind two doors are goats and behind one door is a car. We're assuming that you want the car, and that the host knows where everything is.

You pick door 1. The host, Monty Hall, says, "Let's make a deal. I'm going to show you what's behind Door 2." He does. It's a goat. Then he says, "Now you have a choice. You can either stay with door number 1, or you can switch to door number three, and take whatever's behind whichever door you pick."

So, do you stay, do you switch, and does it matter either way?

A friend keeps bugging me with this stupid ass question (he also typed it) and says that the chance that you get the car *after* you switch is doubled; my argument is that it isn't, it'll stay 50/50, here's why I argue that (yeah I realize I'm probably wrong, but math can kiss my ass lately):

So, you have a 1/3 chance of getting the car initially, after he shows you a goat and you switch, you're supposedly supposed to have a 2/3 chance of getting the car. I don't think so, because if you pick a goat, he'll show you the other one, if you pick the other goat he shows you that, 2 chance for the car; now, if you pick the car, he can show you *either* of the two goats, meaning 4 total scenarios (assuming he'll pick at random, instead of always picking the same goat if you were to pick the car) meaning 2 chances to get the car out of 4 total = 50% chance.

So what do you think, I'm sure I'm wrong (as every website contends it is in fact 2/3) but eh, I don't like mathematicians >=(

[Annihilator]

easy
disregard everything contextual
you have 2 doors
one has a prize
one chance, out of two possibilities
cant get any simpler than that

[Blarg]

I liked this problem when I first saw it a few years ago. I was also confused at first because I thought it was just your run of the mill question, but if you think about it enough it makes sense. I can't remember what made me realise exactly why you change doors, so i cant help you there.

[quannum]

switching is better. youtube vid for you: http://www.youtube.com/watch?v=o2L_2psS9uI

[Ramor]

I've been arguing with him all class about it (yaysub) and he keeps showing me crap on the intarweb, I can see why he considers it 2/3 but I also see reason for it to be 1/2, but like 99% of the internet stuff says it doubles when you switch so bah. Shall watch the youtube when I'm not blocked by school filters :D

[Millael]

The way it works is as follows. There are three doors - A, B, and C. You yourself pick one of these doors.

Say you pick A. A has a 1/3 chance of being the correct door. Now, you can split the doors up into two groups.

Doors you DID pick - A (1/3)
Doors you DID NOT pick - B and C (2/3)

Now, if you look at it this way, the subset of doors you picked is only 1/3 of the doors, while the subset of the doors you did not pick is 2/3 of the doors. Since there are two doors that you did not pick, one of them is guaranteed to have a goat behind it. Revealing one of these doors does absolutely nothing to change the odds that you originally picked the correct door, since you have obtained absolutely zero information pertaining to the door you chose. It is only narrowing down which of the two you did not choose is more likely to be correct.

Say C was revealed to have a goat behind it. You now have the following groupings.

Doors you DID pick - A (1/3)
Doors you DID NOT pick - B and Goat (2/3)

The easiest way to look at is as follows - what results would you get if the car was behind each of the three doors, if you chose A?
Car behind A: A(Car), B(Revealed Goat), C(Goat) Correct With A, C showing
Car behind B: A(Goat), B(Car), C(Revealed Goat) Incorrect With A, B showing
Car behind C: A(Goat), B(Revealed Goat), C(Car) Incorrect With A, C showing

As you can see, there are three possible scenarios, and the fact that the goat is revealed has absolutely no bearing on the odds of the original door being correct. I apologize if this post makes absolutely no sense - I woke up about 45 minutes ago and am currently trying not to fall asleep at work.

[Seyrr]

One thing to keep in mind: the host knows where everything is. This is actually really important, because if you do the problem with the same parameters except the host doesn't know, you get a completely different result.

So, there are two doors with goats. No matter what door you pick, there is a door with a goat behind it that you didn't pick. (If you picked the car, both of the other doors have goats. If you picked a goat, then it's the other goat door.) When the host shows you a door with a goat behind it, he's not telling you anything you didn't already know. It doesn't change your knowledge of the situation one bit.

So, your situation to start with, and that hasn't changed:

There's a 1/3 chance you picked the right door to start. There's a 2/3 chance you picked the wrong door to start. Switching is saying you picked the wrong door to start.

I hope that helps. The key insight is that the host's opening a door doesn't tell you anything new, because YOU know that HE knows where the car is.

[Ryko]

Uh, if there are goats behind 2 doors and he showed you a door with a goat, why would it matter if you didn't choose the door he showed you? EIther way, if you're wrong you end up with a goat anyway and if you are, voila, new car. Nothing is lost or changed by taking the chance.

[Septimus]

Uh, if there are goats behind 2 doors and he showed you a door with a goat, why would it matter if you didn't choose the door he showed you? EIther way, if you're wrong you end up with a goat anyway and if you are, voila, new car. Nothing is lost or changed by taking the chance.

In the end it is still a 50/50 chance, since he was going to open a door to reveal a goat.

[Millael]

Let's try breaking this down a little more.

Of A, B, C, goats are 0, car is 1.

A + B + C = 1
The Monty Hall problem

A + (B+C) = 1
1/3 2/3
In this stage, the person has chosen A as their door. B and C cannot both equal 1. The host knows exactly where the car is. Since he knows this, he shows the player one of B or C, one of which is GUARANTEED to be wrong, thus having zero effect on the original door. The grouping of doors B and C still have a 2/3 chance of being correct, but one of the doors is now known.

A + (0+C) = 1
1/3 2/3

Thus, the unpicked door has a 2/3 chance of being correct. This is possibly one of the most pain in the ass problems to explain to others, because it's such a clunky thing to describe, and people are deadset in their notions that it HAS to be a 50/50 chance. Give me a moment, and I'll try to find a more user-friendly explanation.

EDIT: If you still don't quite get it, there are a number of different explanations here, at http://en.wikipedia.org/wiki/Monty_hall_problem. Or, just contact me on AIM, and I'd be happy to run through enough sets of doors until the numbers themselves prove it.

[Seyrr]

Okay, let's look at it a different way. There are two types of doors, Doors You Chose (DYC) and Doors You Passed (DYP). Let's say that for the sake of example, you choose door 1. Then:

DYC = 1
DYP = 2, 3

There's a 1/3 chance the car is in the DYC group, and a 2/3 chance it's in the DYP group.

When the host reveals a door that has a goat behind it, he's saying "Hey, one of the doors in DYP has a goat!" But before we even made the first choice, we knew he would be able to do that no matter what. So the host's reveal doesn't really change anything. It's still 1/3 that the car is in DYC, and 2/3 that it's in DYP.

What the host's reveal DOES do is tell us that if we say the car is in DYP, which door we should actually say (if he showed 2, we pick 3, and vice versa.)

Again, this is a demonstration that knowing what other people know can influence our strategy. If the host had forgotten which had the goat, and he guessed one to reveal and got lucky, then it would be a 50/50 chance. But since he knows everything, he can pretend to give us information that isn't really useful.

Edit: http://en.wikipedia.org/wiki/Monty_Hall_problem has a very extensive treatment of the problem, including several aids to understanding. It is a 2/3 chance if you switch, people. >.>

[Byrd]

I've had this conversation many, many times, and there's some very good explanations here. as to why your odds are better to switch.

Seyrr and Millael both have very good explanations of why it's better to switch, no point in repeating their discussions. If you don't believe them, there are numerous websites where you can test it yourself. Or you can get a deck of cards and a friend and test it yourself.

Bottom line: if you stick with your original choice, you're gambling that you were right the first time. If you switch, you're gambling that you were wrong, since you get the better of the other two options.

I made good extra credit in a math statistics class with my writeup of this problem.

Memo to all: do not have this conversation with drunk people who are convinced they are right. We nearly had a brawl break out.

[Enkidu]

The easiest way to visualize it is instead of 3 doors, make it 100 doors. One door has a car behind it, the other 99 have goats. You pick one door and the host opens another 98 doors revealing 98 goats, now he asks you if you'd like to keep your first choice (1/100 chance) or switch (1/2 chance).

In essence it's the same problem, just that you have a much greater chance of winning with the 100 door scenario. I saw this on TV a long time ago, not sure if the video explains it that way... 20 minute video DO NOT WANT.

[dietvanillapepsi]

This argument had a big resurgence when Deal or No Deal first aired. That show runs on the same concept. There is really no advantage to switching since your odds are the same for both doors. Like others have said, the host revealing a door for you provides no new information that you can use to your advantage so the odds stay the same.

[dietvanillapepsi]

The easiest way to visualize it is instead of 3 doors, make it 100 doors. One door has a car behind it, the other 99 have goats. You pick one door and the host opens another 98 doors revealing 98 goats, now he asks you if you'd like to keep your first choice (1/100 chance) or switch (1/2 chance).

In essence it's the same problem, just that you have a much greater chance of winning with the 100 door scenario. I saw this on TV a long time ago, not sure if the video explains it that way... 20 minute video DO NOT WANT.

Your math is faulty. If 98 doors are revealed then both doors have a 1/2 chance of being the right one. The "first choice" does not remain 1/100 chance while the "switch" has a dynamic chance. The odds for both change when new information is presented but the odds still remain equal.

[Seyrr]

This argument had a big resurgence when Deal or No Deal first aired. That show runs on the same concept. There is really no advantage to switching since your odds are the same for both doors. Like others have said, the host revealing a door for you provides no new information that you can use to your advantage so the odds stay the same.

No, no, no. Those two statements directly contradict each other. The host does reveal no new info, so your odds stay the same. But those odds are 1/3 that you picked correctly and 2/3 that you picked incorrectly. They are not the same for both doors.

Okay, another example, taken from Wikipedia. Try getting three cards, say an ace of spades (car) and the red kings (goats). Shuffle 'em up and divide them into two groups, the Player Group and the Host Group. The Player Group has one card; it is the door you chose. The Host Group has two cards; they are the other two doors. Now, the host will discard a red king from the Host Group (reveals a goat). How often does the Player Group have the ace of spades?

The host discarding a red king doesn't change where the ace went. It went to the Player Group 1/3 of the time, and to the Host Group 2/3 of the time.

[Millael]

This argument had a big resurgence when Deal or No Deal first aired. That show runs on the same concept. There is really no advantage to switching since your odds are the same for both doors. Like others have said, the host revealing a door for you provides no new information that you can use to your advantage so the odds stay the same.

Dude. You just said "Like others have said", but then you completely contradicted what the rest of us have said. Here, have a visual aid to help... http://en.wikipedia.org/wiki/Image:Monty_tree.svg

[Seyrr]

The easiest way to visualize it is instead of 3 doors, make it 100 doors. One door has a car behind it, the other 99 have goats. You pick one door and the host opens another 98 doors revealing 98 goats, now he asks you if you'd like to keep your first choice (1/100 chance) or switch (1/2 chance).

In essence it's the same problem, just that you have a much greater chance of winning with the 100 door scenario. I saw this on TV a long time ago, not sure if the video explains it that way... 20 minute video DO NOT WANT.

Your math is faulty. If 98 doors are revealed then both doors have a 1/2 chance of being the right one. The "first choice" does not remain 1/100 chance while the "switch" has a dynamic chance. The odds for both change when new information is presented but the odds still remain equal.

The "switch" does not have a dynamic chance. It's always 99/100, since switching is saying "my first choice was wrong". When we switch, we say "fuck door 1, I'll take all of the other 99 doors." That is, "I'll take the 98 doors that have goats, plus the remaining door." This choice is effectively that remaining door, because they don't wanna give you 98 goats. Who knows what you'd do with them, you pervert.

[layoneil]

This argument had a big resurgence when Deal or No Deal first aired. That show runs on the same concept.

it's not anywhere close to the same concept because the host isn't choosing and also the host doesn't know.

and yeah, your understanding of it is totally opposite of what people are saying.

[Byrd]

Your math is faulty. If 98 doors are revealed then both doors have a 1/2 chance of being the right one. The "first choice" does not remain 1/100 chance while the "switch" has a dynamic chance. The odds for both change when new information is presented but the odds still remain equal.

You really think there is a 50% chance you picked the right door the first time out of 100 doors? Because that's what you're saying if you think that you still have a 50% chance after he reveals the 98 other doors as being incorrect.