A nice problem for you to solve :D

[MusicJam19 737]

You are in a TV show. You have to pick a door, and behind it is what you will get - there are 3 doors - behind 2 is a goat and behind 1 is a car. You pick a door. The host opens a different door, to show a goat. You now have a chance to switch the door you chose to the last door, or you can stick to the door you chose. What do you do and why? (Note: it must be reasoned with a mathematical answer, not just "because sticking is the best".

 

Clue:

 

Use probability!

 

 

Post your answers below!

Edited March 31, 2013 by MusicJam19

[spartan343 8]

Pick any other door. I will either end up with a goat or a car either way. And for the mathmatical bit...50/50

Edited March 31, 2013 by spartan343

[GoldRock 6,165]

Right, so...

 

- You first pick a door; you would have a 1/3 chance of choosing the car first time.

- Once one of the doors with the goats is opened, it's still a 2/3 chance that you would have picked the other goat first time; even though the chances have become 50/50 for selecting between the car or the remaining goat if you were to select a door at this stage, you've already taken your pick when the chance of choosing the car was 1/3.

- Therefore, I'd switch - it's more than likely that the door you picked holds the other goat (2/3) than the car (1/3).

 

Am I right? :)

Edited March 31, 2013 by GoldRock

[Frost.Angel 221]

no, no problems

[MONKBLOB 33]

Switch. Each door has a 1/3 chance of having the car, until the host reveals one door which has a goat. So this door now has no chance, but the original door you picked still has a 1/3 chance ltherefore the unchosen door has a 2\3 chance. Damn i'm good!

 

 

That was right, wasn't it?

 

[MusicJam19 737]

Right, so...

 

- You first pick a door; you would have a 1/3 chance of choosing the car first time.

- Once one of the doors with the goats is opened, it's still a 2/3 chance that you would have picked the other goat first time; even though the chances have become 50/50 for selecting between the car or the remaining goat if you were to select a door at this stage, you've already taken your pick when the chance of choosing the car was 1/3.

- Therefore, I'd switch - it's more than likely that the door you picked holds the other goat (2/3) than the car (1/3).

 

Am I right? :)

Switch. Each door has a 1/3 chance of having the car, until the host reveals one door which has a goat. So this door now has no chance, but the original door you picked still has a 1/3 chance ltherefore the unchosen door has a 2\3 chance. Damn i'm good!

 

 

That was right, wasn't it?

 

Yes, 2/3 is the answer. Look up the monty hall problem and see how many world famous mathematicians got it wrong and said 50/50

[MusicJam19 737]

http://en.wikipedia.org/wiki/Monty_Hall_problem

[MusicJam19 737]

Pick any other door. I will either end up with a goat or a car either way. And for the mathmatical bit...50/50

Nope

Edited March 31, 2013 by MusicJam19

[monkeyburn 13,581]

p(a) > p(B) - 3 = a<b = 3 ^ p(a) = p(B) 1 # 7.234 b(p) = Door number one...However, one component is missing. Unless intentional. Then it would probably be door number 2.

 

"This isn't for a school assignment is it?"

[MONKBLOB 33]

[dennyboy1505 4]

Just walk around the door and look whats behind before you pick it. WHaha you no said there is a wall.

[MONKBLOB 33]

Just walk around the door and look whats behind before you pick it. WHaha you no said there is a wall.

The goats would eat you before you had the chance to see what's there. And yes, they are carnivorous goats...

[GoldRock 6,165]

:o I didn't Google my answer! How could you?! :lol:

[Dihos 2]

This problem is confusing

[MusicJam19 737]

^ Exactly

[dennyboy1505 4]

The goats would eat you before you had the chance to see what's there. And yes, they are carnivorous goats...

OMG GOATS, Soooo scared. :blink:

[Ein_Necro 84]

This is in no Way Nice at all. :lol: . And I'm on a TV Show xD.

Edited March 31, 2013 by Ein_Necro