We have two probabilities here:

The probability of a car = 1/3.

The probability of a car given a door with a goat has been opened = 1/2.

BTW, the host will ALWAYS be able to show a goat after the first pick, as there are two goats and the contestant cannot select them both.

Look at the way the question is posed and what you know as FACT and not how the game show is played or what might happen.
From the very moment he first picks you know the contestant has either picked the car or he hasn’t. Regardless of which of the three he has picked there will ALWAYS be a door remaining without a car behind it. Furthermore we know that one door is going to be opened without a car. So from the outset you know that the contestant will always be left with two doors to pick from, one with the car and one without. Effectively, at least the way this question is constructed, this is really only a choice of two doors from the very start.

That's true, but it all assumes that a scenario exists where the host CAN possibly reveal the winner.

But what about this very specific scenario:

3 doors, 1 car 2 goats.

Contestant chooses a door.

Before that door is revealed, the host, without knowing where anything is, opens one of the 2 remaining doors. Behind it happens to be a goat.

Should the contestant change his initial pick?

Truly you have a dizzying intellect. :D

I though of Schrodingers cat too, but this appears to be Monty's goat.:D

Ok, I should rephrase that. It doesn't matter if the host knows what door to open. It only matters that he opens a goat door.

In your example, yes you should change your pick. Because:
1- You probably guessed wrong at first
2- Once the goat door is opened, there's only one door left; you now have the opportunity to make your probably wrong guess a probably right guess.

Seems right to me. Kind of an interesting twist that Len picked up on, because if you read the Wiki and other explanations, they seem to suggest that it is important that the host know, even in the above scenario.

I like Len's 1000 door scenario. Let's pretend the host doesn't know, and could open the winning door accidentally.

Pick a door (0.1% chance it's a winner).

Now the host opens 998 doors, and they all happen to be losers.

So, do you change your pick? Are the odds greater that you accidently pick the winner, or are they greater that the hose accidentally picks 998 losers?

The answer (I think) is keep your pick. The odds change as picks are made.

The odds you get it right in 1 out of 1000 picks are 1/1000

The odds the host can't get it right in 998 picks out of 999 are 998/(999x998x997x996...) or really really small.

So if the host has had 998 tries out of 999 to get the winner and can't get a winner, I'm betting I got real lucky with my guess and will keep it.

The only time that the contestant would have a 33% probability of picking the correct door is if he had to pick one from the three and it was opened at that point. But that isn’t the case as we know that one of the incorrect options will be disregarded, which makes the probability of picking the correct door 50% from the outset.

To visualize this, suppose I sit at my computer and write either a number 1, 2, or 3 on a piece of paper and put it next to my keyboard. I then ask one of you to guess the number and post it on the forum. Once you post it I will tell you which of the remaining numbers is NOT the answer that I have written down.

THIS IS THE POINT THAT YOU NEED TO CALCULATE THE PROBABILITY THAT YOU HAVE THE RIGHT ANSWER!!!

Now I ask you if you want to change your guess.

What is the probability if you change your mind?

It makes no difference what number I have written, you just know it is either the one you picked or its the remaining one.

60 pelicans take your offer. 30 will change their answer, 30 will keep their answer.

You've written down number 1.

20 pelicans tell you they've chosen number 1. You tell them that number 2 is not the answer. 10 stay with 1, 10 change to 3.

20 pelicans tell you they've chosen number 2. You tell them that number 3 is not the answer. 10 stay with 2, 10 change to 1.

20 pelicans tell you they've chosen number 3. You tell them that number 2 is not the answer. 10 stay with 3, 10 change to 1.

So of the 30 that stuck to their guns, 10 win and 20 lose.
Of the 30 that changed their answer, 20 win and 10 lose.

Have I missed some iteration?

Can we get on to some conditional probability stuff already?

Whether they or right or wrong is one thing, but just before they know the answer they all have a 50% probability of being correct.

Yes.

20 pelicans tell you they've chosen number 1. You tell them that number 3 is not the answer. 10 stay with 1, 10 change to 2 .

I think.

Okay then...

180 pelicans take your offer. 90 will change their answer, 90 will keep their answer.

You've written down number 1.

60 pelicans tell you they've chosen number 1. You tell 30 that number 2 is not the answer. 15 stay with 1, 15 change to 3. You tell the other 30 that number 3 is not the answer. 15 stay with 1, 15 change to 2.

60 pelicans tell you they've chosen number 2. You tell them that number 3 is not the answer. 30 stay with 2, 30 change to 1.

60 pelicans tell you they've chosen number 3. You tell them that number 2 is not the answer. 30 stay with 3, 30 change to 1.

So of the 90 that stuck to their guns, 30 win and 60 lose.
Of the 90 that changed their answer, 60 win and 30 lose.

The percentages haven't changed.

Not true - just because there are two choices doesn't mean they are two equal choices.

Say I have a deck of cards, and ask you to pull out the ace of spades, without looking. You choose your card.

I then look at the remaining 51 cards, keep one to myself and turn the other 50 over to show the ace of spades is not among them. I ask if you'd like to keep your card or take mine instead.

Even though you now have a choice of only two cards, do you think you have a 50% chance that keeping your card is the right call?

Absolutely correct.

There are two cards, one is the Ace of Spades. If I hold one card and you hold the other I have a 50% chance of holding the Ace.


EDIT:

I was a bit quick to jump on this and I'm reconsidering. I think with increased numbers this gets to be a completely different kettle of fish...

Warm up the crow pie

I'll try one more scenario.

I offer you a big-ass jar full of red jellybeans, save one which is black. "Close your eyes and pick the black one. Keep your fist closed once you have yours", says I.

So, you close your eyes, pick one, close your fist and wonder if you've picked the sole black jellybean.

Now, I say "Tell ya what, aerkuld, I am now going to take out handful after handful of these jellybeans, pick thru every last one, and eventually pick the one I want. I will even show you every jellybean except the one I pick. Mind you, I won't keep my eyes closed".

Once I've done that, I say "Okay, aerkuld ole buddy ole pal, I'll bet you that I have the black jelly bean and you don't. In fact I'll bet you a dollar. If I win, you owe me a dollar. If I lose, I'll pay you ten dollars".

If this is a bet you'd take, I will gladly fly you out to sunny California for a fun weekend of Pick the Black Jellybean.

Edit:
Seeing your edit makes me feel a little silly for driving the point home with a sledge hammer.

And you're welcome to come out to sunny California anytime you'd like. We won't even have to play Pick the Black Jellybean :D

Amail - I see your point and to be fair I have been equally vocal on the opposing view.

I'll let you know next time I come back to the Golden State & I'll give you some notice so you can get the jelly beans in.

:)

You take the other door. The 52 card deck analogy is the best means to have people understand what happens as the remaining cards (doors) are revealed.
To prove it to some co-workers (years ago), I even wrote a C program that generated a random "door" - 1, 2, or 3 - as the winner, and also randomized the contestant's door choice and whether they swapped.

Re: Probability Challenge - Can You Solve It?
 

2 doors, 1 correct choice
1 divided by 2 = 1/2 = 50%

Problem solved