The problem with the Unfinished Game is that it does NOT distinguish properly between what Jeff Atwood intended it to be (hence ending with a 2/3 probability) and the "nosy neighbor" scenario.

It comes to the same downfall as Monty Hall/Fall: intent is KEY because it shifts the probabilities.

In Monty Fall, the probabilities shift to 50/50 because Monty is not adding information (more information is coming in, but that information is as likely to drop us out of the scenario a (he accidentally opens the door we chose, or he accidentally opens the door with the car behind it) as it is to guide us. Thus, the problem domain has contracted but no new information has been imparted on the existing choices.

In the Unfinished Game scenario, we are supposed to realize that the information offered is NOT random, but as a result of a pointed question ("Is one of your children a girl") but not TOO pointed question ("Is your oldest child a girl"). If it were random, you end up with the "nosy neighbor" scenario, wherein the extra information contracts the problem set (you are as likely to have seen a boy walking across as a girl) instead of ellucidating the remaining domain.

Which brings me back to the first paragraph of Atwood's piece, where he scolds us for telling him his formulation of the Unfinished Game problem was incredibly poorly stated. Sorry, but it was. He does not say if this person pointedly told us he has a girl at random or with some surrounding information, and supposes that we must infer that there was some surrounding context to that information.

Personally, as a parent, I answered "~1". If you are a parent and tell someone that "one of my children is a girl" you are saying that the other one is not. Simple human nature. If I have two girls, I'm going to say I have two girls, not one. The only possibilities then are:

1. The other child is a boy
2. The other child is a hermaphrodite of some sort
3. The person is being deliberately coy about their children's genders

Since 2 and 3 are generally very small probabilities, that leaves just about a 100% chance that the other child is a boy.

Of course, Atwood's scenario posits that (3) is the dominant probability, which then overwhelms the odds of the situation, and further that there is not a REASON that they are being deliberately coy (for instance, some odd societal stature loss from having two girls).

For the record, a better statement of the "Unfinished game" problem would be:

In a random gathering of people, we asked everyone with two children to stand up and everyone else to sit down. We then asked those standing to only remain standing if one or more of their children was a girl. One woman was standing at this point. What are the odds that she had a girl and a boy?

The problem, though, is that making the problem clearer starts sending warning signals to the quizee. Not sure how to best pose this scenario in a "natural" manner without either causing critical ambiguity or tipping the quizee off that something unintuitive is afoot.

@Tobermory:

"Monty Fall is the same as Monty Hall once the chance of accidentally revealing the car is gone."

"Intent" was bad choice of words on my part. Since Monty is not a random event generator, but has actual decision-making powers, his actions enhance instead of simply restrict the problem scope.

Lay out the possibilities end to end. It's not hard, there are only nine of them. Here, I'll start it for you:

1 0 0
0 1 0
0 0 1

That's three possibilities, where '1' is the door with the car, and the '0's are the doors with the goats. Assume that you always pick the first one.

That's three possibilities from your perspective. Now, cross-tab that with the three possibilities from Monty's accident perspective, and you have nine scenarios at the end. For the sake of discussion, daw this out as columns going across, where the first column is where Monty hits the first door, the second is where he hits the second door, and the third where he hits the third door. Because all three variables are independent, this is a valid and complete depiction of the problem space (with one of the three variables fixed; technically you could multiply the space by three, with identical setups where you chose the second door each time and then the third door each time but, again, being independent variables fixing one is legal).

100 -- 100 -- 100
010 -- 010 -- 010
001 -- 001 -- 001

Now, we know because the problem says it is so that the end state is such that Monty did not reveal your door nor the door with the car. That means that the entire first column gets eliminated from consideration, as well as the possibility at (2,2) and (3,3) (in both cases Monty would be revealing the car.

X00 -- 1-0 -- 10-
X10 -- 0X0 -- 01-
X01 -- 0-1 -- 00X

This leaves four outcomes:

1. (1,2) Monty reveals Door 2 and the car is behind door 1
2. (1,3) Monty reveals Door 3 and the car is behind door 1
3. (2,3) Monty reveals Door 3 and the car is behind door 2
4. (3,2) Monty reveals Door 2 and the car is behind door 3

50% of the time sticking your guns wins; 50% of the time switching wins. Thus, because the problem space has been reduced, we have a straightforward probabilities calculation.

Another way of thinking about this is this:

In the first starting condition, there's a 33% chance that Monty will mess up the whole scenario. In each of the other two starting conditions, there's a 66% chance that Monty will mess up the whole scenario (either by revealing the door you have already chosen or by revealing the car behind the other doors). This is the crux of it: since Monty has no guidance here, he is more likely to eliminate LOSING scenarios than winning scenarios.

Now, if Monty has INTENT, he is not just randomly reducing possibilities. He will always have two to choose from, and so what we need to look at is the pre-choice solution graph, not the post-choice one (because his choice is not an independent variable). Here, instead of a crosstab matrix, you have a tree.

The pre-choice graph looks, again, like:

100
010
001

The post-choice outcomes map from those starting points as:

100 -- 1-0 OR 10-
010 -- 01-
001 -- 0-1

In the first of the scenarios, sticking to your guns wins (no matter which Monty chooses to reveal). In two of the scenarios, switching to the other closed door wins.

Again, Monty isn't a random dummy here, and has exactly 0% chance of eliminating you in any scenario. Thus, the chance of your original choice selection being "right" remains at 33%, while the new information makes it twice as likely that the non-revealed other door is the right one.

Does that make more sense to you?

The fact that Monty can choose either of two outcomes in the first case does not make that case twice as likely to happen. Imagine giving Monty additional choices like opening the door with his left hand or right hand but only when you've picked the wrong door to start; these also would not make those possible outcomes any more likely.

Again, really: these problems are well-known and documented. If you think everyone has the Monty Hall or Monty Fall problems wrong, and you're not a professional statistician, it's like telling the world you know quantum mechanics better than Stephen Hawking. That's the realm of the crank.