Imagine you’re on a game show, and you’re given the choice of three doors. Behind one door is a briefcase containing one million dollars, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to pick door #2?”.
The host is not attempting to trick you. He does the same thing every week.
Is it to your advantage to switch your choice of doors?
Yes, it’s better to switch.
Here’s a better game show question: The Lady, or The Tiger?
Switching takes your chances from 1/3 to 2/3, so yes, switching is good.
[quote]pookie wrote:
Switching takes your chances from 1/3 to 2/3, so yes, switching is good.
[/quote]
I still don’t get how this is true…
You have a 1/3 chance. A choice is eliminated. Do you now not have a 1/2 chance?
does it matter if one incorrect door is eliminated?
each remaining door has a 1/2 chance of being right. so if you keep the original choice your odds are 1/2, and if you switch your odds are 1/2.
No, because you most likely picked the wrong door on your first guess.
Wikipedia does a pretty good job of explaining this.
[quote]Beowolf wrote:
pookie wrote:
Switching takes your chances from 1/3 to 2/3, so yes, switching is good.
I still don’t get how this is true…
You have a 1/3 chance. A choice is eliminated. Do you now not have a 1/2 chance?[/quote]
The key here is that you know, regardless of your original pick, that the host is going to open up a door with a goat.
To start out, you have a 1/3 chance of picking the car, and a 2/3 chance of picking a goat. If you pick a goat, then the host shows a goat (which you know he will), and then you switch doors, then your odds are 2/3.
Basically to win you want to pick a goat with your first choice. You know if you pick a goat the host will show the other goat, and the remaining door must have the car. Your chances of picking a goat are 2/3.
[quote]Beowolf wrote:
pookie wrote:
Switching takes your chances from 1/3 to 2/3, so yes, switching is good.
I still don’t get how this is true…
You have a 1/3 chance. A choice is eliminated. Do you now not have a 1/2 chance?[/quote]
Let’s say there are 100 doors, and one door has a prize.
You pick a door and the host then opens 98 doors, revealing them all to be empty. Two doors are left: Your initial pick and another one.
Would you switch? Do you still think that switching is useless because you’ve got a 50/50 chance?
The problem with evaluating that problem is that people do not take into account the added information that the host opening a door adds to the situation.
Another way to look at it.
You pick a door. 1/3 chance of a prize.
The 2 remaining doors have 2/3 chance at the prize.
The host opens one of them and asks you if you want to switch. You are basically asked if you want to keep your 1/3 chance or go for the 2/3 one.
[quote]pookie wrote:
The host opens one of them and asks you if you want to switch. You are basically asked if you want to keep your 1/3 chance or go for the 2/3 one.
[/quote]
But don’t both have 1/2 chance at that point since you know one has the car and one has the goat?
[quote]elano wrote:
pookie wrote:
The host opens one of them and asks you if you want to switch. You are basically asked if you want to keep your 1/3 chance or go for the 2/3 one.
But don’t both have 1/2 chance at that point since you know one has the car and one has the goat?[/quote]
No. The door you picked had 1 in 3 chances of being the right one, and removing one door after you choose does not alter those odds. Hence, while there are 2 remaining closed doors, yours still has 1/3 chance of winning and the other one has 2/3 chance of winning.
If you had 3 doors and picked one, and the host then asked you if you’d prefer to get the best prize from the remaining two doors instead, would you switch? The formulation would be different, but it is the exact same thing being offered here.
…Why in the hell does the other one have a 2/3 chance of being a car?
If you picked a car the guy would still show you one of the goats so…
What?
Its nothing like if he asked you if you wanted the best prize. Don’t you have the same chance of being right with your first pick as with a switch!?
I remembered immediately that the answer is not 1/2. But I took a probability course about 2 years ago so my memory is a bit fuzzy, can’t remember the explanation.
All I remember was getting this question on the final exam and writing:
“1/3 + 1/3 = 2/3”
and getting it right 
[quote]Beowolf wrote:
…Why in the hell does the other one have a 2/3 chance of being a car?[/quote]
Because yours only has 1/3.
Yes, but you only had 1 chance in 3 of picking the car. So there are 2 chances in 3 of the other door being the car.
Of course, switching will cause you to lose the car 1/3 of the time. But since you have no way of knowing in advance whether you did pick the car or not, the best choice is to play the odds and switch.
[quote]What?
Its nothing like if he asked you if you wanted the best prize. Don’t you have the same chance of being right with your first pick as with a switch!?[/quote]
It is exactly the same thing. You pick a door. I then ask you if you’d prefer to get the best prize from the remaining two doors. Basically, you get to pick two doors instead of one, but because the host has already opened one of the “bad” doors, it makes it look as if the remaining doors have 50/50 chance at the prize. They don’t.
Look at my “100 doors” example I gave a few messages back. Your initial door has 1/100 chance of being the right one and the remaining door (After the other 98 are opened) has 99/100 chance of being the right one. Surely you don’t see the 100 door version as a 50/50 proposition? How could you, by picking a random door from 100, expect that door to have the same odds of being right as the door the host leaves closed? If you can grasp that version, then the 3 door version is exactly the same, but with less doors.
The statisticians here are playing sleight of hand by claiming that there are still 3 doors in play, which is only true if you have a half-brain who is still contemplating choosing the door that has been opened to reveal a goat.
This is really no different than there being 3 doors and the host removes a door prior to you making your initial selection. Either way, the real choice is still between only 2 doors. It’s no longer a question of probabilities of 3 choices once the information has changed.
DB
[quote]dollarbill44 wrote:
This is really no different than there being 3 doors and the host removes a door prior to you making your initial selection. Either way, the real choice is still between only 2 doors. It’s no longer a question of probabilities of 3 choices once the information has changed.[/quote]
Not so because when you make your choice, there are three doors in play, not two. That later one door is opened does not change the fact that when you chose, you picked one out of three.
You’re also ignoring the stipulations that the host always opens a door and never reveals the car. It’s very nice of you to note that the information has changed, but then completely ignore that information and treat the problem as if nothing had happened before you got to the two door choice.
I guess with 100 doors, you’d still keep your door given the opportunity to switch after the host has opened 98 of the non-winning ones? Picking one door out of 100 gives you a 50% at the prize? I don’t think so…
They go over this in the movie 21.
If you just don’t believe that it’s true, try it yourself.
1.) Pick a door.
2.) Roll a dice. If you get a 1 or a 2, Door #1 holds the prize. If you get a 3 or a 4, Door #2 holds the prize. If you get a 5 or a 6, Door #3 holds the prize
3.) Eliminate a wrong door from one of the two doors you did not select initially.
4.) Switch your selection to the other remaining door.
5.) Repeat how ever many times you wish, and write down your winning %.
6.) Repeat 1.) through 5.) except this time, do not switch doors in step 4.)
7.) Write down your winning % from step 6.)
8.) Compare the winning % from when you switched doors to when you did not. The more trials you do, the closer you should get to .67 winning % when you switch and .33 winning % when you don’t.
For example, I put together an Excel spreadsheet that randomly repeats this 5000 times and calculates your winning %.
I ran it 10 times, and got winning %'s of
67.8%
66.34%
66.04%
66.76%
66.74%
66.08%
66.52%
66.16%
65.18%
67.00%
So after 50,000 trials with random initial picks, random winning doors, and randomly eliminating a wrong door that you didn’t initially pick, switching your choice averaged out to a winning % of 66.46%.
This isn’t voodoo.
[quote]tGunslinger wrote:
This isn’t voodoo.[/quote]
No, but for some reason this is always a difficult concept for people to understand.
[quote]malonetd wrote:
Wikipedia does a pretty good job of explaining this.
http://en.wikipedia.org/wiki/Monty_Hall_problem[/quote]
Holy shit. Whoever made that page has way too much time on their hands.
Anyway, I’d tell Bob Barker to go fuck some off’s, throw a few of them hot babes in my brand new Cavalier convertible and drive that right off the damn stage. If I’m on a game show, fuck if I’m not winning something.