I'm bad at probability...Follow

You should always change your answer. The game show is set up to make it appear as though changing or staying have the same odds, and since most people have a bias towards sticking with their first answer (they'd rather be wrong on their first pick, than to move from a right answer to a wrong one), most people will make the wrong choice.


The key to understanding this is that the game host picks which of the other two doors to open. It is not random and he will never open the one with the prize. If he randomly opened one of the other two doors, with a chance that he'd reveal the car (which presumably would mean you'd lose automatically), but instead revealed a goat *then* it would be a 50/50 chance between the two doors remaining. But since he opens the door without the car if one of the other two has the car behind it, it changes the probability.


When you make your selection, you have a 1 in 3 chance that the door you pick has the car. Which means that the other 2 doors collectively have a 2 in 3 chance of containing the car. Showing you the goat behind one of those two doors does not change the odds that one of them contains the car. It only guarantees that if there is a car behind one of those two, it must be behind the door he didn't reveal. Which means that the odds of the car being behind the door you picked remains at 1 in 3, while the odds of the car being behind the door you didn't pick and the host didn't reveal is 2 in 3.


You will win twice as often if you change your selection.


Oh. To answer your hidden question: Yes. That is correct. Because by revealing all the doors you didn't select but one, that one door takes on the probability of holding the prize equal to all the doors. Another way to look at it is that if any of those doors held the prize, it will be behind the one door he didn't show you. Thus, the odds of the prize being behind your door is always 1/n (where n is the total number of doors), but the odds of it being behind the one unselected and unrevealed door will be n-1/n.

Edited, Dec 20th 2012 12:22pm by gbaji

Yeah, I don't get this.

If there are 3 doors and 1 car, the probability of me picking the car up front is 1/3. If it is known that the host will always open whichever door remains unpicked and contains a goat, then it is also known that when prompted to change, there will exist 1 door with a goat, and 1 door with a car. Where does the assumption that the one I originally picked probably does not contain the car come from? Or in other words... two scenarios:

Scenario 1:

1. Show a person 3 doors, 1 of which has a car behind it.
2. Prompt the person to pick one. Then remove whichever remaining door does NOT have the car.
3. Prompt the person to change their pick between the remaining two, if desired.

Scenario 2:

1. Show a person 3 doors, 1 of which has a car behind it.
2. Remove one of the doors which does NOT contain the car.
3. Prompt the person to pick one of the remaining doors

In either case, the chance of selecting the cup with the ball comes down to the final choice between the two remaining cups whether that choice is the initial selection or the second chance, and the original pick seems to have no bearing on the result.

So in the end, you know that one of the two remaining cups has a ball. Whether you change your pick or not, the probability that the cup you pick has a ball is 1/2.

Deja Vu.

http://ffxi.allakhazam.com/forum.html?forum=28&mid=1207003510151407370&h=50&p=2#60
http://ffxi.allakhazam.com/forum.html?forum=28&mid=1207003510151407370&h=50&p=2#72


Edit: I just remember this post cause it was a big arguement with a select few failing to see the advantage of changing your pick. Basically it boils down to what the probability that your first choice was wrong. By changing, you are betting that your first choice was wrong, which is the statistical advantage.

Edit 2: I just realized I have a custom title...

Edited, Dec 20th 2012 4:41pm by TirithRR

Edited, Dec 20th 2012 4:49pm by TirithRR

Another way of looking at it is that you're really being given a choice between picking one door, which must contain a car for you to win, or two doors either one of which can contain a car for you to win. The odds of the car being behind one of the two doors you didn't pick is twice as likely as the car being behind just the one door you picked.


It's a bit of a brain teaser because at first glance it appears like you have two doors and one of them has a car, so it should be 50/50. It's not though. Mythbusters also did a bit on this, and showed the logic and the math behind it, and then just in case people weren't convinced, they rigged a game show and ran a big series of rounds, with one of them always staying and one of them always changing. They also tested the psychological aspect of the game, which is that people's tendencies are to not switch, and sure enough something like 90% of their test group would not switch their answer when given the chance.

That's it I'm going to make Joph an admin.

I can't actually do that.

Here's another way of showing it. With 3 doors, there are 9 possible situations.

 
Door You Picked First:   1   1   1   2   2   2   3   3   3 
        Door With Car:   1   2   3   1   2   3   1   2   3 

As you can see, in only 3 of those 9 situations was your original door choice correct. So you'd have 1/3 odds of winning if you stayed. In 6 of those 9 situations, you'd win the car if you change doors, for 2/3 odds of winning.



Or another illustration of the same math:

Step 1                            Step 2 
                                  Change = 0 cars won 
You picked the car door   (1/3) < 
                                  Stay   = 1 car won 
 
                                  Change = 1 car won 
You picked a non-car door (2/3) < 
                                  Stay   = 0 cars won 

So the "Stay" choices add up to [(1/3) x 1] + [(2/3) x 0] = 1/3 odds of winning a car.
The "Change" choices add up to [(1/3) x 0] + [(2/3) x 1] = 2/3 odds of winning a car.

The intuitive way to think of it is that Monty's door reveals are not random. He has knowledge of where the car is, and so by leaving a particular door closed, he is in fact giving you extra information about that door. Well, sometimes he is:

A) If you picked the car door correctly at first (1/3 of the time), Monty is just revealing any random doors. So he's giving you NO information in that case.

B) If you picked an empty door at first (2/3 of the time), Monty can't just open random doors. He is narrowing down all the remaining choices down to the exact door HE KNOWS has a car behind it. So 2/3 of the time, Monty's reveal is giving you the information of which car has the door.



This concept becomes more clear when you envision 300 doors. The fact that Monty opened all those 298 other doors and left just that one effing specific door gives you a pretty good hint that it's the winning door. That "hint" would only be a red herring in the 1/300 chance that you picked correctly at first.


Edited, Dec 20th 2012 7:16pm by trickybeck

Wait, if the host is going to open an "incorrect" door each time, isn't there really just one choice between 2 doors? The first one being completely meaningless?

Yeah, I get that. I just don't see why the probably doesn't reset the second time when you're asked to choose between 2 doors. I mean if you "forgot" which door you chose the first time, the odds of choosing the right answer would be 50/50.

Kinda get it with lotto tickets...

You want the numbers you chose or the numbers I chose? One of them is the winning ticket.

Edited, Dec 20th 2012 6:01pm by someproteinguy

Correct. Because if you "forgot" which door you picked, it's the same as if the host eliminated one of the doors before you picked. In which case you have a 50/50 chance. But you don't forget. You know that you picked one out of three doors randomly, and that you had a 1 in three chance that the door you picked had a car behind it. Since the host will always eliminate one of the remaining 2 doors, and that door will always be one that does not have a car behind it, the remaining door has a 2/3 chance of having a car behind it, while the one you picked only has a 1/3 chance.


The reason this is the case is because the hosts choice is not random. If there's a car behind one of the two other doors, he will not eliminate that one. So if you pick door number 1 and the car is behind door number 2, he'll eliminate door 3. If it's behind door 3, he'll eliminate door 2. In effect, this means that if either door 2 or door 3 has the car, it will always be the door not eliminated and thus the door you are given the opportunity to pick in the second part will always contain the car if either one of those two doors did.

Imagine if instead of revealing a door without a car behind it, the host instead just said: "You can keep your one door, or you can choose to open both of these doors and if either one has a car, you win". You'd take that offer every single time, right? Well, that's exactly what he's doing when he eliminates one of the other two doors and ask if you want to switch for the third. It's the same thing. If either one of those two doors has the car behind it, you win by switching.

Edited, Dec 20th 2012 6:23pm by gbaji

Yeah, thinking about it I bet my youngest would have picked the right door before me. We do it all the time to help them learn.

Which one is the red crayon? *points to a pile of like 50 crayons*

No sweetie that's yellow.

Here... *grabs the red crayon along with the yellow crayon and holds them up*

Which one is the red crayon?

Good job, that's right!



Wonder why it seems so counter-intuitive in the 'door' situation? General distrust of someone like a game-show host who may be "trying to trick you?" You know, replace the host with someone trustworthy opening the door, and see if the answer changes.

Edit: (also I realize my example isn't exactly the same thing, but it helped with the understanding )

Edited, Dec 20th 2012 6:53pm by someproteinguy

Leave it to you to misinterpret completely valid numbers.

If you pick the first door and never change your mind, your odds of winning are 17%.
If you pick the first door and change your mind, your odds of winning double (~33%).

In either single case, your odds of not winning at all are at least 66% (83% if you pick door 1 and never change). The winning scenarios are mutually exclusive, since one cannot win by sticking with their original pick AND changing their pick. The 50% non-winning attempts is an artifact of the test, and does not imply that your actual odds of winning are 50%. The point of the test is to illustrate that your chances of winning do in fact double by changing your pick when given the opportunity.

Edited, Dec 21st 2012 4:47am by BrownDuck

Any time I want a new iPad, I just find one of those Flash banner ads where you punch the monkey.