Switch doors or not?

[RagnaToad]

I present to you this situation:

You're in a game show. There are three doors. Two of them have a goat behind them, one has the car you want... You can only choose one door. You choose one, and the game show host opens one of the doors you didn't pick, showing a goat behind it . (Note that he will always open a door with a goat behind it and always ask you the following question.) Now he asks you "Would you like to switch doors?", meaning you can switch from the door you originally picked to the door that's left.

The question is: What is the best choice, mathematically speaking?
Switching or not switching? Or do you think it doesn't matter?

[M16]

We were given this question in my physics orientation class. When I was given the solution, I refused to believe. Even after seeing the math, I just don't buy it.

[Sasquatch]

There's no reason to switch doors. If you pick one of the three, and it's not one of the goats (could be the other, or the car), you have a 50/50 shot of it being either.

There are two doors left. You have one of them. One contains a prize, and the other doesn't. Statistically, you have a 50% chance of having the door with the prize in it. If you switch doors, you still have the same 50% chance.

But who wouldn't want a goat, really?

[Furore]

Mathematically I myself would choose to change my guess.
I'm going by some basic probability I did in an earlier maths class but the door you first choose would remain a 1 in 3 possibility while the remaining unopened door would have less possibility of being the wrong door due to less possible outcomes present. I'd imagine if you change doors you'd win something along the lines of 2 of every 3 games, even if there's still a decent chance of it going either way.
Am I right, or should I have done some more advanced maths?

[Fate]

Silver is right.

You see, there are only three possible choices and outcomes in this situation. Let's label the doors as 1, 2, and 3. And let's say that door 1 has the car. Here are the outcomes:

1: You pick door 1 to start with. The person then opens one of the other two doors. If you then switch, you lose. However, if you keep the same door, you win.
2: You pick door 2 to start with. The person then opens the door with the goat. If you switch, you win. However, if you keep the same door, you lose.
3: You pick door 3 to start with. The person then opens the door with the goat. If you switch, you win. However, if you keep the same door, you lose.

As you can see, only one of the choices require that you keep the same choice in order to win, while two of them require that you switch in order to win. So the probability of winning without switching is 1/3, whereas the probability of winning with switching is 2/3. Therefore, for better chances of winning, switch.

[Sasquatch]

That's interesting, but that would be looking at it from the beginning, not on-the-spot chances.

At the beginning, you have a 1/3 chance of picking the right door, and a 2/3 chance of picking the wrong door.

Because there are three doors.

If you have one door picked, and the number of doors go down to two, you have a 1/2 chance. Period. It no longer matters how many doors there used to be. There are two doors, one of each, and you have a 50/50 shot of having chosen the correct one.

The switching makes sense when you explain it, but mathematically, it doesn't hold up.

[RagnaToad]

It certainly does hold up mathematically speaking. Think of it like this:
You choose a door and have 1/3 chance that it's the car. Meaning that when you switch later one, you have a 1/3 chance to switch from the car to a goat.That means you have the rest of the 3/3 -> 2/3 to switch to the car.

Or with a simple [Number of favourable possibilities] divided by [Total number of possibilities] for switching:

Car -> switching means Goat
Goat A -> switching automatically means Car (other goat is eliminated)
Goat B -> switching automatically means Car (other goat is eliminated)

2 out of 3 results of switching get you the car.

NOTE: Eliminating 1 door BEFORE choosing would obviously alter chances from 1/3 to 1/2 of choosing the right one (assuming you want the car ).

[Fate]

Yes, that's right. For example, imagine that there are 500 doors instead of just 3. When you pick one, you have a 1/500 chance of picking the one with the car. Then, the person opens up 498 doors that do not contain the car. By then, switching would mean that you have a 499/500 chance of winning, whereas not switching, you'd still only have a 1/500 chance of winning.

[Che]

Why not switch? Because the door being opened is irrelevant if you weren't going to pick it anyway. The only relevant doors are the two remaining ones, and if you switch doors and it turns out to be a goat then how pissed are you going to be at probability?

[Che]

How do you know you better switch?

At first you have a 33% chance of getting it right. When they open a door to reveal a goat, you then have a 50% chance no matter which you choose. They were going to open the door in the first place, meaning you didn't ever really have a 66% chance to get it right, you only ever had the 50% chance.

You can argue probability all day but the fact of it is, there is only one right answer. You either win, or don't.

[Che]

Wrong. You pick a goat, it doesn't matter which one it is. You're either getting a goat or a car.

50%.

[Pete]

Che is making sense though.

By automatically eliminating one of the options, which is a goat, it does not alter the fact that the remaining two options are on the table. The only thing is that the odds change.

It's kind of similar to what they taught way back in SAT class. If you're given 5 options, and have to guess, your chances of a right answer will increase with every definitive answer you can eliminate.

If I had 5 choices, I'd have a 20% chance of guessing correctly, but when I can eliminate one choice, it jumps to 25%, and then to 33% if I narrow it down to three.

When I start out, I have three choices. Thats a 33% chance of getting the car, which is pretty much a fact. However, once one of the goats is eliminated, it wouldn't matter which one you picked because there is still a 1 in 2 chance that you'd get the car. If you're always shown an incorrect option, and then given the choice to decide again, you'll always wind up with a 50% chance of winning.

I can't really agree with the notion that by not switching, you increase your chances of picking the wrong option. If you're always shown that one of the options is not a car, you still have a 50% chance of getting the car anyway, because by default it has to be one or the other. At first, yes, there is a 1/3 chance, but when a choice is clearly eliminated, you have the same chance to win the car as not winning it.

I wouldn't mind a goat though

[Fate]

No, Rags is right. Take a look at this statistics wheel that I found. http://math.ucsd.edu/~crypto/Monty/images/wheel.jpg

In the wheel, the very middle circle represents the door in which the car is behind. It can either be door 1, door 2, or door 3, right? Now, take a look at the second circle. It represents the door that the player can choose. Once again, the player can either choose door 1, door 2, or door 3. In this case, the probability that the player will immediately choose the correct one is 1 out of 3, or 1/3. Finally, take a look at the last circle. This circle represents the door the can be opened. As you can see, the blue spaces represents the door that has the car behind it. So if the player was right from the very beginning, then the person will open either of the other two doors in which there is a goat behind it. If this was the case, then switching would be a bad option, and if you did do this, you would lose. Now, let's assume that from the start, the player chose an incorrect door with a goat behind it. As you can see, if that happens, there would only be one incorrect door left, and that is the door in which the person will open. If this is the case, switching would be the best option, and you would win if you did. Take a look at the amount of red spaces, which signify an incorrect door. There are four of them. Therefore, from the start, you have a 2 out of 3, or 2/3 chance of choosing a wrong door. And if you do choose a wrong door, then switching would mean that you would win the car, right? So, you have a 2/3 chance of winning if you switch. Also take notice as how there are twice as many red spaces as there are blue spaces. This makes sense because the odds of winning if you switch is double the odds of winning if you don't switch: they are 2/3 and 1/3 respectively.

[Che]

I'm not actually arguing, I know that counting the initial picked door, 2/3 is correct. What Pete and I are saying is after you pick, you essentially have a 50-50 chance. Which would be the chance you get no matter what (since they will always take a door away). The game show just makes it more interesting to the audience by adding the fact that they will take away one of three doors. (Actually it's a really bad game show, considering a game show probably doesn't want to give away 50-50 chances of winning a car every game).

You are correct in saying that it is 2/3 starting with the initial pick of one of three doors. It's just like those text problems in math books where you're like "WTF? This shit never happens."

The fact that you can even get 2/3 out of this is the same reason you have a 1/3 chance at the beginning if they didn't take the door away. Why the F would they take away a free door that wasn't the real one? I'd be the happiest game show person ever.

[Fate]

Yes, the Monty Hall Paradox is real. I think that's where this thread came from. And to say what a famous contestant said, it does matter which door you pick. However, the only possibility where it wouldn't matter "if Monty Hall does not know which door has the car behind it."

[Che]

Oh okay. I see it now. I had to write it all out on my own. I was incorrect in assuming its always 50% chance of getting the car because they'd take away the door, and you stated

(Note that he will always open a door with a goat behind it and always ask you the following question.)

So I'm thinking...Oh, well it's no problem. I wasn't understanding what you meant when you said:

Or else you'd have 1/2 chance from the beginning if you knew they would eliminate a goat automatically...

I was like: "Uh, of course they know they are going to take away a door with a goat, they do it every game." But I see what you are saying if we label A as Car, B as Goat 1 and C as Goat 2.

If you choose A they will take away B or C.
If they took away B, you can stay with choice A or C, and if you stayed with A, you're totally ****ed on winning the car.

If they took away C, you can stay with choice A or B, and if you stay withA, you're totally ****ed on winning the car.

0% chance of winning in that scenario.

I was fooled by the

Do not be fooled by the revealing of the door with the goat behind it. It does not affect the chance of you having picked the right option of the 3, it just increases the chance that the car is in the remaining door you didn't pick.

Because I thought he'd always take away the goat, which I thought meant you automatically won the car if you picked it instead of the chance you could throw it away by picking it and staying with it.

[M16]

I ran through this experiment with my friends today. We had the rule where my friend had to switch his answer every time as the first experiment, and then again where he didn't change doors. When he switched doors, he was correct 80% of the time. When he kept his answer, he was only correct 30% of the time. QED as far as I'm concerned.

[Darkwave]

I think that it is luck that you pick the right door because if you are meant to have the correct door it would open. If you are meant to have the goat it will open on a goat. Depends how lucky you are.

[Alpha]

Ahh the Monty Hall problem aye? We got taught this on the first day of Year 13 Statistics. I would not believe it initially, but I kept thinking about it, and did my own little experiment. To be honest, I don't really get it, it appears to defy logic, but at the same time, does not defy logic. I don't know, I'll leave you guys to explain it. I just know that it's true, and I'd definitely switch doors. You'd be silly not to, unless you wanted a goat. I don't drive, so I suppose I could do with a goat actually. Hmm.

[GHOSTED]

To be honest, I could careless, lol. I woundt switch doors, Im fine with a goat....wait..you do get to keep the goat....right?

[Che]

To be honest, I could careless, lol. I woundt switch doors, Im fine with a goat....wait..you do get to keep the goat....right?

Only if you promise to ass **** it on a daily basis.

[GHOSTED]

Only if you promise to ass **** it on a daily basis.

We can use the F word on here? lol. Some one needs to link me what we can and cant say, lol. I promise to ass**** it on a daily basis...before I use it as T-rex food....