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Monty Hall problem (wikipedia.org)
14 points by Maro on Feb 22, 2010 | hide | past | favorite | 27 comments

I always find this the most interesting part of it: "A restated version of Selvin's problem appeared in Marilyn vos Savant's Ask Marilyn question-and-answer column of Parade in September 1990 (vos Savant 1990). Though vos Savant gave the correct answer that switching would win two-thirds of the time, she estimates the magazine received 10,000 letters including close to 1,000 signed by PhDs, many on letterheads of mathematics and science departments, declaring that her solution was wrong (Tierney 1991). " Especially the science PhD's (I have seen quotes from some of the letters..).

Science PhD's are on roughly the same ground as those poor MDs who didn't know what a positive test result does not imply high chances of having a rare disease.

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.

Similar problems arise in contract bridge, where players have a name for it--"restricted choice". I've actually always wondered if mathematicians have a name for it. Anyone know?

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.

It's similar in Skat and Doppelkopf. (Though even more devilish in Doppelkopf since you do not know where your partner is at the beginning of the game.)

There is a certain subtlety to this problem which many people overlook. It is that whether the host <em>always</em> gives the option to switch. Only if the host always gives you the option to switch, there is a benefit in switching. If the host acts randomly(say deciding on a coin flip whether to give the option to switch), then the benefit of switching is nullified. Also, if the host is malicious (gives the option to switch only if you have the door with the prize), then of course switching decreases your chances of winning. I believe that the reason several "PhDs" were confused over the answer has got to do with the fact that they might have assumed a host which acts randomly.

Not quite true.

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.

As you said, "average the overall probabilities between a regular game and a switch-style game", which is 1/3 for a game with no-switching and 2/3 for a game with always-switching. Averaging gives you a probability of 1/2. So if you switch, you win 50 % of the time. But so do you if you do not switch. So there is no benefit in switching, which was basically my point.

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.

You're absolutely right and my argument fails, but it doesn't invalidate the point I wanted to get at. In the long run, your coin flipping method does cause the game to behave as though there were only two doors, but in any given game you'll know whether or not the host is going to choose a door.

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.

Another similar problem from http://www.inference.phy.cam.ac.uk/mackay/itila/

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?

Solution: Host doesn't know what is behind the doors either. Contestant then has a true 1/3 chance. QED

I guess that defeats the game though.

...is the wrong answer.

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.

I know... that's why I said it.

This is actually an interesting variation of the problem. If a random door opens but by chance happens to not have the prize, then the result is different.

But the GP is wrong. If a random door opens, then the odds are 1/2, just like the intuitionistic answer predicts.

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.)

That doesn't make any sense. The intentions of the host prior to opening the second door shouldn't have any impact on the odds of picking a winning door.

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.

The intentions of the host pass you information that you can use to choose the correct door. The information is this: "There's a 2/3 chance I was forced to reveal this door; ergo, there's a 2/3 chance the other door is the correct one."

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.

Search for "forgetful Monty Hall" (or so) if you want to know more about this variation.

A recent paper with a thorough analysis:


I've heard secondhand of this being used as an interview question.

Sounds like a really bad idea to me.

I can see it now. "Behind one of these doors is a job offer with your name on it! The other two go to janitorial supplies, and the networking closet. You just might get lucky!"

why? for a job that involves work with probability modelling it would certainly be relevant.

1. It's notorious for clashing with many people's intuitions and being very confusing. Many smart people who are experts in probability theory answer it incorrectly at first and then it takes them a while to convince themselves they were wrong.

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.

In an interview scenario, I'd give the problem and the correct solution, and just ask the interviewee to tell me 'why?'.

thanks for the detailed response, seems very reasonable.

Not very fair, though, if even maths professors get it wrong.

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.

Wasn't this in the movie 21?

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