Awesome Game ShowFollow

You're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door and the host, who knows what's behind the doors, opens another door and reveals a goat - no matter what you have picked he will always reveal a goat. He then says to you "Do you want to switch?"

Should You? Why?

I know what Aegis's Answer would be

Yes, you have 2/3 chance of picking a goat first time around and when he reveals the other goat you win by changing

People round here are far too smart

Switch doors if you want a 2 in 3 chance of winning the car was the correct answer.

Exactly how though?

I understand that you have a 67% chance of picking a goat the first time, and a 50% chance of picking a goat the second time if you repick. But in reality, him revealing one of the goats has no way of changing what is actually behind the first door you chose. (This isn't a cat in a box).

It's like if a teacher asked a question, which you had no clue what the answer was. Multiple choice, A, B, or C. You pick A at random, and the teacher tells you afterwards that C is not the correct answer. You don't gain anything really by changing your answer after that.

Always switch doors unless you have "faith" in your door.

If you want the car, you should switch.

The reason why it works is simple: There's 2 possibilities, on your first random pick you either pick the car 1/3 of the time, or a goat 2/3 of the time.

If you randomly pick the car, then the other goat is out of the picture and you lose by switching, meaning you lose 1/3 of the time.

If you randomly pick the goat, then the other goat is once again out of the picture, but by changing you pick the car, therefore winning 2/3 of the time.

It's only 50% if you don't include the information about the previous doors.

Out of 100 doors any door you pick is almost certainly the wrong door. If you know you're wrong, and he takes away all other choices except one, then you know that must be the right choice.

Or you can try thinking of it this way. Instead of the doors you can actually see all the prizes. You can see which of the 99 are goats and which one is a car. Pick any prize at random (not caring what you win). Most of the time you will pick a goat, because there are 99 of them and only one car, after switching you would have ended up with a car. If you by chance pick the 1 car out of those 99 goats then after switching you would have picked a goat. So since you pick a goat most of the time, by switching you end up picking a car most of the time.

Ok. I get it now.

Thanks, that certainly clears thing up

No you don't, it's Equally as likely that you picked the right door the first time.

a choise of 2 doors, one being right is still 50/50 reguardless of previous choices.

No this is awesomer

click

Math says to @#%^ off

Superior Lulz

With two doors to choose from, your odds are 50/50 only in a _random_ system.

But when you are posed with your second choice (to switch or not), it is not a random system: one door will win 1/3 of the time and the other will win 2/3 of the time. This is, by definition, not random.

You're such a f*cking moron. It's really funny.

Incidentally, this is nothing like how Deal or No Deal works.

it's dead AND alive... or something?

This bit is wrong



The box you chose was chosen at random from 3 boxes thus a 1 in 3 chance of winning. The box that is left was not left unopened randomly, it was left unopened on the basis that the other box contained a goat and so the probability of it containing the car is the sum total of the boxes at the initial point of the game (1 in 3 plus 1 in 3 = 2 in 3).

Tarv your original statement was:

It's not equally as likely that you picked the right door the first time. It's not a 50% chance of one being right.

I don't mean to be rude, but what your trying to arguing what you are saying is pointless. You're now arguing that you mean that out of the two boxes/doors there is a 50% chance one has a car and that one can't be certain which box/door has the car. While both of these statements are true they mean nothing useful. Out of any two objects you have a 50% chance of selecting one, and nearly every situation involving probability lacks perfect certainty. Right, but not relevant.

Edited, Apr 1st 2008 4:58am by Allegory

When I was a kid (ha!), I had a pet goat named Florence. She slept on the end of my bed.

This made my day.

From The Straight Dope:

I take it someone watched 21.

I think the reasoning was bullsh*t, and you now have a 50/50 of winning with either door.

I am not being dense, this is exactly my point. You do *not* have a 50/50 chance. You have a 33.3/66.6 chance.

You only have a 50/50 chance if the car is randomly behind one of the two curtains, but it is not randomly behind one of the curtains. It's twice as likely to be behind the curtain not initially picked.