Two cars, one goat

[Xist]

My cousin posted an article about Marilyn vos Savant on Facebook. Twenty-five years ago, she "solved" the Monty Hall Problem, and ten thousand people wrote in to tell her that she was wrong. It was simply that, if presented with three doors, told that two doors each have a goat, but the third has a car, and told to choose one, what is the chance of correctly choosing the door hiding the car if you change your mind after shown the third door?

The writer said that you would think that you have a fifty-fifty chance regardless, but vos Savant argued that it would be two-thirds.

Many paragraphs later, I understood that, if shown the car, your chances of choosing the car increase to 100% if you chose another door, but change your mind--presumably to the open one showing the car. If you do not choose the car after seeing it, there is a 100% chance that you are an idiot.

Seriously, I do not use that term lightly. There is only one car, you are shown the car, so why would you say "No thanks, the one and only car is behind another door?"

So, you are correct in thinking that you have a 50-50 chance whether you change your mind or not, assuming that the third door does not have the car.
http://priceonomics.com/the-time-eve...ign=pockethits

[RedDevil]

if you are shown the car hell yes, you know where it is.
Was that what you meant?

Marylin was right of course, but somehow it is hard to grasp.

We had a game show over here with had (I think) the same feature.
Three doors, only one had a prize in it.

You have to choose one; the chance it has the prize is 1/3.
While the door you chose stays closed, the game show host opens one of the two other doors.
If you have accidentally chosen the right door then he can open either one, but if not he must open the one remaining door that has no prize; the prize is behind the other door.

So if your choice was wrong at first, switching means you get the prize and not switching means you don't.
If your first choice was right switching means you won't get the prize, not switching means you keep it.
As the chance your first choice was right was only 1/3, you double your chances to 2/3 by switching.

The essence is this.
If all choosing was purely random the chance would stay 1/3.
But the game show host does not do a blind pull. He knows where the prize is and has to choose a door that has no prize, doubling the chance that the prize is behind the remaining door - if it weren't he would have opened that door.

But the vast majority of people will never grasp that for some reason.

Knowing this, if the prize is not that big and you don't want 1/3 of a chance to be ridiculed by your neighbours (you had the prize but you threw it away!) stay with your first choice.
If the prize is big or your neighbours are smart enough, choose the other door.

At one time the game show host accidentally opened the door with the prize. That made for a 100% chance

Oh, in the case of the Lincoln Continental and the goats the goat may get better MPG.
I'd still want the Lincoln, just to sell it and buy an electric goat.

[Gasoline Fumes]

If I understand the problem correctly, it has to be a 50-50 chance. Three doors. Pick one. A door you didn't pick is opened, revealing a goat. Two doors remain, a car and a goat. You now get to switch or stay. How is that different than picking from two doors? Why would the original choice matter? And is a Lincoln Continental better than a goat?

Edit:
Nevermind, now I get it!

[niky]

Actually... the door opened by the host ALWAYS has a goat behind it. That's the game show format. The host CANNOT pick the winner for you. That's a given.

In which case, you really do improve your chances by changing your mind.

And statistics actually backs this up, strangely.

Monty Hall problem - Wikipedia, the free encyclopedia

You have a 2 in 3 chance of winning by changing your mind.

[RedDevil]

Originally Posted by Gasoline Fumes

Two doors remain, a car and a goat. You now get to switch or stay. How is that different than picking from two doors? Why would the original choice matter? And is a Lincoln Continental better than a goat?

Before he opens the door the chance is 66.7% that the car is behind one of the doors you did not choose.
When he opens one of those doors the chance the car is behind the other door is still 66.7 %, unless he accidentally opens the door with the car in which case you'll choose that door.

Two doors remain, the one you chose at first with a 33.3% that it has the car, and the door the host did not (could not) open.

As your door still has that same 33.3% chance of having a car behind it (the car did not move) the remaining door has that 66.7% chance.

Swap and sell the Lincoln for a herd of goats.

[Xist]

That table shows all six possible outcomes, which are very limited, but the only way to have a 2/3s chance is if there is a possibility for the host to show you the car, and for you to choose that door once you know, but why would they ever do that? Are all of their contestants drunks?

That would make it interesting...

[RedDevil]

You are one of many who don't get it.
Don't take it personal - many smart people don't get it.
Maybe you just don't want to believe there can be a 2/3 chance.
Something makes you read, yet refute, the truth.

You throw in an assumption "the only way to have a 2/3s chance is if there is a possibility for the host to show you the car"...
What is that based on?

The point is that the host shows you where the car is not, while he knows where it is.
His choice of which door to open is NOT random.
That changes the odds.

If you choose A and the car is behind B he'll show C. Swap and you get a car.
If you choose A and the car is behind C he'll show B. Swap and you get a car.
Only if the car is behind door A (1/3 chance) can he choose which one to open as none has a car. If you swap you don't get the car.

[freebeard]

My cousin posted an article about Marilyn vos Savant on Facebook.

Why is this about your cousin and Facespook?

The aticle was in New York Times in 1991:

Behind Monty Hall's Doors: Puzzle, Debate and Answer?
By JOHN TIERNEY,
Published: July 21, 1991

[Fat Charlie]

Step one- you pick a door, whichever one you pick is meaningless.
Step two- the host shows you one door that doesn't have the prize.
Step three- you decide which of the two doors the host didn't open has the prize.

Call it "switching" or just flip a coin at that moment, it's 50/50. Your initial "choice" just narrows down the host's choice of doors, and the host's choice is the only piece of information you have.

[RedDevil]

Hi Fat Charlie, read the article Freebeard linked to, or Wikipedia.
It is deceitful, as if designed to put you on the wrong leg.

Remember, the car does not move.
Whatever you choose first has 1/3 chance of being right, no matter what happens next.
The only other possibility is the third door, the one the host did not open. So that has the combined chance of both remaining doors, as one of them was taken out by the host; not at random but deliberately revealing (the) one without a car.

If the host did open one wrong door before you had to make your first choice, then it would be 50/50 between the remaining doors.
But he does so after you've chosen, which means the chance the car is behind one of the remaining doors is 2/3.
And it is still 2/3 when he opens one of them.

It all looks so simple that you may be tempted to not think it through.
So simple, no need to read all this... right?
And it is simple - but you can easily make a wrong turn in thinking it out.

Read the article Freebeard linked to.
Once you get it, you cannot believe you ever missed it.
Swapping gives you a 2/3 chance of winning, trust me.