Uh, what if you picked the car the first time?

Quote from Exilement

lol wut

Quote from BloodFruit

lol wut.

shuddap. im dumb.

Quote from muttonhead

shuddap. im dumb.

lol wut.

Quote from Cereal Man

Uh, what if you picked the car the first time?

Congratulations, you had a 1/3 chance of doing that.

You have a 2/3 chance of picking one of the goats.

That means switching every time will get you the car 2/3 times, while staying gets you it 1/3 of the time.

So if you picked it the first time, you're pretty much screwed, because mathematically you should switch, every time.

What? No. Just no. After he opens one door to show 1 goat, it's no longer a chance out of three, it's out of two. Get it? Got it? After one goat is eliminated, and you are choosing between door a and door b, there is NO door c. Therefore, it's a 50/50 chance at the last. Before the revealing of one goat and after are two seperate problems, not one.

Quote from Sawc

What? No. Just no. After he opens one door to show 1 goat, it's no longer a chance out of three, it's out of two. Get it? Got it? After one goat is eliminated, and you are choosing between door a and door b, there is NO door c. Therefore, it's a 50/50 chance at the last. Before the revealing of one goat and after are two seperate problems, not one.

That is what most people think. However you are wrong. Your peanut sized brain apparently cannot comprehend the riddle.

You can ignore the 3rd door, yes, but you can't ignore your first choice.

In the simplest form of this problem, you can assume this: if you pick the car, switching gets you a goat. if you pick one of the 2 goats, switching gets you the car.

Remember this:

Options (Switching)

Pick the car first, switch, you lose

Pick goat A first, switch, you win

Pick goat B first, switch, you win

Options (Staying)

Pick the car first, stay, you win

Pick goat A first, switch, you lose

Pick goat B first, switch, you lose



6 outcomes, that means each outcome has a 1 in 6 chance of occuring. There are 3 that cause you to win, and 3 that cause you to lose. You could say that this would constitute a 50/50 chance of winning, since 3/6 outcomes will cause you to win.

However, you only have 2 options (switch or stay). That means each option gives you 3 possible outcomes. 3 allow you to win, 3 allow you to lose. That means there are only two possible ways to divide those 6 outcomes between the 2 options:

One option has 3 possible chances to win, the other has 3 possible chances to lose.

OR

One option has 2 possible chances to win and 1 to lose, and the other has 1 possible chance to win and 2 to lose.

The first situation would be the only possibility for this to give you a 50/50 chance. However, that's simply a hypothetical. It's not how the problem goes at all.

There's no possible way for this to be reduced to a 50/50 chance, and there's no possible way for me to explain it any further.

Quote from muttonhead

Your peanut sized brain

Irony .

This also might help:

still dont grasp the concept

Geez Exilement, you're doing a lot of work.

Quote from Eccentrix

I'd rather ride the goat

Are you thinking what I think you're thinking? *wink wink nudge nudge*

idk *wink wink nudge nudge*

I remember seeing a newspaper article about this a while back. I'll see what I can find.

*edit*

Proof of 2/3:

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