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Apathy
10-15-2008, 09:36 PM
As long as these threads are so popular now, I figured I'd make one on pretty much one of the most famous math incorporated riddles in the world.

Skip down for the riddle, this is just my rambling because I'm long winded I guess.

It's one of Marilyn vos Savants' for those of you who don't know, she's the holder of the world's highest IQ, or at least she was, and I doubt anyone overtook her since. (She's also married to the guy who invented the artificial heart, who ironically enough, had to go to an italian medical school because he wasn't admitted to any others, in a small part because of his surprisingly low IQ, in the 80's iirc.)

Anyway, riddle time. I first heard this in my stats class last year. It was printed in a column Marilyn writes in 1990, and when she published the answer, mathematicians and statisticians from around the country wrote in explaining to her why she was wrong, or to condemn her for trying to trick modern day Americans.

But she wasn't wrong.

RIDDLE

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, 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?' Is it to your advantage to switch your choice of doors?"

Highlight for answer.
The wrong answer? No.
Any reasonable person will see that the door the host opened has a goat, and assume that door #1, and door #2 both have a 50% chance of having a car.
Unfortunately, this is incorrect. In my opinion, the easiest way to prove the correct answer, is to look at all of the possibilities.

DOOR 1 DOOR 2 DOOR 3 RESULT
GAME 1 AUTO GOAT GOAT Switch and you lose.
GAME 2 GOAT AUTO GOAT Switch and you win.
GAME 3 GOAT GOAT AUTO Switch and you win.

GAME 4 AUTO GOAT GOAT Stay and you win.
GAME 5 GOAT AUTO GOAT Stay and you lose.
GAME 6 GOAT GOAT AUTO Stay and you lose.

When you switch doors, you win 2/3 games. When you stay with what you have, you only win 1/3.

The reason this works is that The Host will always eliminate a door with a goat in it. He can do this because he knows what is behind the doors. So in game number three from above, You Pick Door #1. The Host is forced to eliminate Door 2.

Another way to look at it. When you make your first choice, before the host eliminates any, there is a 1/3 chance of you having the car, and a 2/3 chance of the other doors having it. If the host eliminates one of the other doors, the probability doesn't change, because there's still three doors. You just know one of them now, and know not to pick it. So the door he didn't pick gets all of that 2/3, and your door still has 1/3.

Fuck Yes.

Tyler Durden
10-15-2008, 09:56 PM
This was in the movie 21.

Apathy
10-15-2008, 09:59 PM
Hmm.

I walked out of that movie.

Jebus
10-15-2008, 10:05 PM
Why don't the two fours want any dinner?

Because they already eight!

lost_nvrfound
10-15-2008, 10:48 PM
Fuck. too late to comprehend. I will have to look back at this in the morning.

hshduppsnt
10-15-2008, 11:13 PM
hmm...

I'll have to pass this one along...

funny that I think most people would say no, but basic probability playing out there...

though my favorite is still the box of ropes that take an hour to burn (but at different and non-uniform rates) and you have to find a way to time 45 minutes. What thread has these other riddles you speak of? I must go find...

Endymion
10-16-2008, 03:17 AM
ah, the monty hall problem. yes, you switch. the door the host opens changes nothing because it gives you no information. you pick a door, your chances of being right are 1/3. thus there's a 2/3s chance it's behind one of the other two doors. clearly, one of the two doors you didn't pick has a goat behind it (pigeon hole principle applies). the host opens a door that will have a goat behind it, which does not change the odds that the car has a 2/3s chance of being behind one of the two doors you didn't pick. it does, however, change the relative odds of those two doors -- the goat one definitely doesn't have it, so the remaining door has a 2/3s shot of having the car.

RickyCrack
10-16-2008, 07:08 PM
three friends go to a hotel only to find that there is only one room with three beds so they decide to share the room. they ask the clerk how much the room is and he responds $30. they divide it evenly at $10 each. later the clerk learns that he made a mistake that the room is only $25 so he sends the bellhop to give the three men their $5 back. unable to decide how to split the $5 up evenly they each take $1 and give the bellhop a $2 tip.

so basically each of the three friends spends $9 on the room. 9 x 3 = 27 + 2 = 29. so where did the extra dollar go?

Cock Joke
10-16-2008, 07:13 PM
My brain hurtz.

RickyCrack
10-16-2008, 07:16 PM
I'll give you a hint. think order of operations

Lodat225
10-16-2008, 08:44 PM
three friends go to a hotel only to find that there is only one room with three beds so they decide to share the room. they ask the clerk how much the room is and he responds $30. they divide it evenly at $10 each. later the clerk learns that he made a mistake that the room is only $25 so he sends the bellhop to give the three men their $5 back. unable to decide how to split the $5 up evenly they each take $1 and give the bellhop a $2 tip.

so basically each of the three friends spends $9 on the room. 9 x 3 = 27 + 2 = 29. so where did the extra dollar go?

Don't you, uh, subtract the 2? Why would you add it ..

Apathy
10-16-2008, 09:53 PM
Don't you, uh, subtract the 2? Why would you add it ..

What are you talking about? He's just showing why it doesn't add up to 30.

The amount they each paid (9) times the # of guys (3) = 27$
The amount they tipped the bell boy = 2$
The tip (2$) and their bill (27$) added up = 29$

His question is, if they initially paid 30$, then where is the last dollar?

In reality, all thirty of the original dollars still exist. The clerk has 25, the three guys each have 1, and the bellhop has 2.
25+1+1+1+2 = 30

The reason it looks like they don't is because each man didn't really pay nine dollars. They paid one-third of twenty-five dollars. That's how much the bill was.

Endymion
10-17-2008, 12:44 AM
the point is they each paid nine dollars, and that went to the room AND the tip, not just the room.

Tyler Durden
10-18-2008, 08:54 PM
ah, the monty hall problem. yes, you switch. the door the host opens changes nothing because it gives you no information. you pick a door, your chances of being right are 1/3. thus there's a 2/3s chance it's behind one of the other two doors. clearly, one of the two doors you didn't pick has a goat behind it (pigeon hole principle applies). the host opens a door that will have a goat behind it, which does not change the odds that the car has a 2/3s chance of being behind one of the two doors you didn't pick. it does, however, change the relative odds of those two doors -- the goat one definitely doesn't have it, so the remaining door has a 2/3s shot of having the car.

I don't understand. When you pick a door, you have a 1/3 chance of getting the car. When the host reveals the door with the goat, there is now a 1/2 chance that either of the other door is right. It doesn't change the odds if you switch, there is still a 50% chance even if you left it the way it was.

Cock Joke
10-18-2008, 09:29 PM
I don't understand. When you pick a door, you have a 1/3 chance of getting the car. When the host reveals the door with the goat, there is now a 1/2 chance that either of the other door is right. It doesn't change the odds if you switch, there is still a 50% chance even if you left it the way it was.

Exactly what I was thinking but couldn't put into an explanation. :o

Apathy
10-18-2008, 10:32 PM
Sorry, but you're both wrong. I've explained as well as I can in the first post.

Because the host knows which doors have goats in them, and he always reveals a door with a goat in it, the door you selected still has 1/3, and the other two still have 2/3. You just know that one of the two that has 2/3 is wrong.

It's basically been proven throughout countless simulations and computer programs.

Tyler Durden
10-18-2008, 10:47 PM
After he reveals the goat, he basically gives you a choice to pick between the two doors. 1 or 2. And the chance that either would be right is 50%. You say the odds stay at 1/3, but they don't. They would, if he revealed it but didn't let you pick again, but he IS letting you pick again. So if you're given a choice of picking between the two then both have a 50% chance of containing the car.

Apathy
10-18-2008, 11:54 PM
I know it's difficult to grasp, but you're wrong.

I felt the same way when I first had it explained to me. It took days before I finally twisted my brain around to understanding it.

http://www.youtube.com/watch?v=mhlc7peGlGg
http://montyhallproblem.com/

You just going to have to accept it.

Edit: This might help you understand it better. I found where you're getting stuck.

After he reveals the goat, he basically gives you a choice to pick between the two doors. 1 or 2.
No. The reason why the probabilites don't change is because you still have a choice between three doors. It's just obvious that you're not going to pick the door that he revealed as a goat, because you'd lose. This makes it seem like it's fifty-fifty. But it's still in thirds.

Tyler Durden
10-19-2008, 01:13 AM
So I wrote out the riddle on paper, and I still don't understand. Tell me where I go wrong...

For the scenario, Door 1 contains a goat, Door 2 contains a car and Door 3 contains a goat.

#1 - Choose Door 1, host reveals Door 3
Not Swapping = Goat
Swapping = Car

#2 - Choose Door 2, host reveals Door 1
Not Swapping = Car
Swapping = Goat

#3 - Choose Door 2, host reveals Door 3
Not Swapping = Car
Swapping = Goat

#4 - Choose Door 3, host reveals Door 1
Not Swapping = Goat
Swapping = Car

From the results, swapping will get you a car 2/4 of the time, or 50%. I honestly don't understand how it could be any different.

Endymion
10-19-2008, 01:18 AM
let's put it this way:

i have three boxes. in one is a diamond ring, in the other two nothing. i ask you to put one box on the left side of the room, and the remaining two on the right side of the room. there's a 33% chance the ring is on the left, and a 66% chance it's on the right. additionally, there's a 100% chance that at least one of the two boxes on the right is empty. if i look in both boxes on the right, and then show to you that one of them is empty, i have given you no new information since you already knew that one of them was empty. since i haven't actually done anything, there is still a 33% chance that the ring is on the left and a 66% chance that it's on the right. i now say that you should pick a side of the room you think the ring is on -- which would you pick?

Tyler Durden
10-19-2008, 01:26 AM
But in the problem with the doors, the doors aren't distinguished in any way. In your problem the boxes are distinguished by their location in the room, in the door problem there is nothing to go by. Your problem separates one box, the door problem separates no doors. So picking is harder and more complex.

Endymion
10-19-2008, 01:28 AM
So I wrote out the riddle on paper, and I still don't understand. Tell me where I go wrong...

both 2 and 3 shouldn't be considered separate cases. you pick either door 1, 2, or 3. every single time, the host will show you a goat. you're only right one of those times, so 2 of the times you'll win by switching.

Tyler Durden
10-19-2008, 01:30 AM
They should be considered different cases because when doing an experiment you should measure every possible scenario that could happen.

Endymion
10-19-2008, 02:14 AM
They should be considered different cases because when doing an experiment you should measure every possible scenario that could happen.

you also have to give them the proper weights: you pick 1, 2 or 3 with equal probability. when you pick 2, half the time the host will open 1, half the time open 3. that means of your 4, 1 has a 1/3rd shot, 2 has a 1/6th shot, 3 has a 1/6th shot, and 3 has a 1/3rd shot.

Splinter[PI]
10-19-2008, 02:22 PM
We talked about that thing at school, so that's not really hard for me.

Pranksta
10-19-2008, 08:48 PM
Look at it this way:
There are 3 possibilities: you pick Goat 1, Goat 2, or the Car.
Possibility 1: You pick Goat 1. The host is forced to pick Goat 2. The third one is the car.
Possibility 2: You pick Goat 2. The host is forced to pick Goat 1. The third one is the car.
Possibility 3. You pick the Car. The host picks either of the goats, doesn't matter which one. The third one is the other goat.

Hence, out of the 3 possibilities, in 2 of them you would benefit from switching.

Cock Joke
10-19-2008, 09:27 PM
The Youtube video made it clearer for me. After the host picks the door with the goat, you may think, "He must not have picked that other one because it's got the car." So, you should always switch doors, as the guy mentioned. Switching increases your chances of winning the car. Pretty cool if you think about it!