Change.
You have a two in three chance of getting the car. If you don't change you have a 1 in 3 chance.
(dchurch bah, Sat 15 Oct 2011, 18:49, 1 reply)

Please show your working.
This brings back dim memories of Marylin vos Savant explaining the maths behind this on a TV show once, but since she has an IQ of 228 and I was about 7 (years old) at the time, my memory needs refreshing.
( .Yeti., Sat 15 Oct 2011, 18:52, closed)

This is the usual reasoning:
You have a 2 in 3 chance of having picked a door with a cabbage. If this is so, then there's one door with a cabbage and one with a car left. Given that the host has shown you one with a cabbage, the other one will have the car.

You have a 1 in 3 chance of having picked the door with the car. If this is so, then you'll get a cabbage by changing.

Changing gives you a 2/3 chance of getting the car, and staying gives you a 1/3 chance. Thus you should change.

---

But actually it depends why the host picked a door with a cabbage behind it (ie do they know where the car is, and do they want you to win or lose). See en.wikipedia.org/wiki/Monty_Hall_problem
(apeloverage committed the vile act of onanism on, Sat 15 Oct 2011, 19:30, closed)

Easy.
The first door you pick has a one-third chance of having a car behind it. The two remaining doors therefore have a two-thirds chance of having a car behind one of them.

If you could choose both of those doors, I'm sure you'd agree you'd have a two-thirds chance of getting the car - but you don't have to! The host, who knows where the cabbages are, removes one of those two doors from play - meaning the door that neither yourself or the host chose has the car behind it two-thirds of the time.

If the host didn't know where the cabbages were, the odds would be equal, but he'd sometimes open the door with the car behind it.
( Legionary - My god, Brian, there's a message in my alphabets, Sun 16 Oct 2011, 12:03, closed)

But...
Suppose you choose door 1
The host opens door 2
You swap to door 3.

Now the host asks "are you sure?"

Now you have the choice between door 3, or doors 1&2

By the logic of the 'correct' answer, you should always swap and keep swapping. This can't be right, can it?
( Ghoti Fingers, Sun 16 Oct 2011, 16:44, closed)

No, it's not :)
The odds of your first choice being right are always one third, because you made that choice when there were three doors. That doesn't change when the host opens one of the two other doors - and so you should always switch (once), because you can either stick with your original choice (with a one-third chance of being correct) or switch to the only remaining door.

Remember, the key is that the host knows where the car is, so he never opens that door. Without that knowledge, it wouldn't make any difference whether or not you switched, but one third of the time the host would accidentally open the door with the car behind it. His simply asking "Are you sure?" without opening a door doesn't make any difference.
( Legionary - My god, Brian, there's a message in my alphabets, Sun 16 Oct 2011, 18:59, closed)