Monty Hall problem
en.wikipedia.org
en.wikipedia.org
Statistics can be non-intuitive because common sense only ever approximates it at best. You've got to either always run the math or build some good heuristics for when to expect that common sense just failed you.
The most common example is this: say you and your partner have nine cards combined in a suit, and you're missing four cards: the queen, the jack, and two unimportant low cards.
Bridge is played like hearts or spades, where one person leads a card and everyone else has to follow suit. It's your lead. You want to find the queen and the jack. If you guess right, you win the contract (= win a lot of points). If you guess wrong, you lose the contract.
Your opponents want to hide the queen and the jack. Assume that your opponents, if one had both missing high cards, are devious enough to randomly play one or the other.
You play the ace, and one defender drops the queen. What are the odds they have the jack? Hint: they're not 50/50.
If the host gives you the option to switch based on a coin flip, that flip will be independent of the state of the game and just average the overall probabilities between a regular game and a switch-style game.
Now, allowing the host to act totally randomly will remove the advantage. In particular, the host needs to randomly choose which door to open from all three (including your own). If he finds the goat, you reset and play again perhaps, if he doesn't find the goat then you've got a 50/50 chance between the remaining doors. The probabilities are simple because each door was treated exchangeably and there could be no possible information flow that is mutual to, entangled with, the location of the goat.
EDIT: In both the versions of the game, I am assuming an omniscience host who always opens the door with a goat behind, if he does open a door at all.
That means that, conditional on a good flip, you may still be able to act to your advantage since there's information transfer from the host to you.
You visit a family whose three children are all at the local school. You don't know anything about the sexes of the children. While walking clumsily round the home, you stumble through one of the three unlabeled bedroom doors that you know belong, one each, to the three children, and find that the bedroom contains girlie stuff in sufficient quantities to convince you that the child who lives in that bedroom is a girl. Later, you sneak a look at a letter addressed to the parents, which reads `From the Headmaster: we are sending this letter to all parents who have male children at the school to inform them about the following boyish matters'.
These two sources of evidence establish that at least one of the three children is a girl, and that at least one of the children is a boy. What are the probabilities that there are (a) two girls and one boy; (b) two boys and one girl?
I guess that defeats the game though.
If the host happened to pick the door with the car, you wouldn't get the option of switching. And the host does not reveal what is behind the door you chose.
There's a 1/3 chance you picked right, a 1/3 chance you picked wrong and a wrong door was revealed, and a 1/3 chance you picked wrong and the right door was revealed. Throw the last third out of your sample space and you get that the odds are 1/2.
This is what that forced Marilyn to admit the odds can be 1/2 on some readings of the problem, which was ambiguously phrased the first time it was printed. (But I think most people read the problem for the 1/3 solution and still came up with 1/2.)
I could be wrong, but I think of it this way: You have a choice between picking one door, or picking two doors. If you choose to pick two doors, the host will happen to open one of them to reveal the non-prize. Irrespective of the differences between the wording of my problem and the MH problem, how are the actions different between the two?
Here's the flaw in your logic:
There's a 1/3 chance you picked right, and a 2/3 chance you picked wrong. In the 2/3 chance you picked wrong, there is a 1/2 chance of the correct door being revealed. No matter if the correct door is shown or not (in this case, it isn't shown), it doesn't change your original odds.
If the host is choosing randomly, no information is passed.
With regards to your last paragraph, the odds of the host picking wrong are either 1/2 or 0 depending on the intentions of the host. I'm a tiny bit annoyed that you think you're pointing out a "flaw" in my logic when you missed something so simple, but whatever.
If you don't like my explanation, direct your attention to the OP's link to Wikipedia, where there are many more variations and explanations that you might like.
On the other hand at this point it seems everybody should have heard about it already, so it might work as a test for curiousness.
2. On the other hand, if you read about it and understood it once, it's trivial, and you can give a correct answer confidently while still being very bad at probability in general. So you'd be selecting for knowing a piece of trivia rather than any real knowledge or experience.
3. Interviewers are not flawless mechanisms, they're people. They have biases that are different to control for even when they're aware of them. You want to steer clear of questions which a) make it easy for the interviewer to feel smugly superior to the candidate; b) the answer to which looks very easy and natural once you know it well, and it's difficult to remember how non-intuitive it had been before. This question fails badly on both counts.