In this varient of the problem, you should also switch. If Monty opens a goat door, then it doesn’t matter at all whether you keep your own door or switch to the other unopened door. But if Monty opens the door with the new car, then you should absolutely switch, to the door that Monty just opened.
Hm. Thanks for the extra input.
I guess I’m having trouble because I’m conceptualizing it as two separate games… in the first game, you have three doors to pick from, and choose one. In the second game, you’re down to two doors to pick from, and choose one. That model doesn’t seem to apply.
Cecil’s brother and sister example likewise hurts my brain, because it’s conflicting with the fact that the gender of the first child doesn’t affect the gender of the second, if one were to predict the second child’s gender before birth.
No conflict. If the question is stated, “The eldest child is a girl,” then the odds are 50/50. It is only when the question is stated, “One child is a girl,” that it changes to 33/67.
It’s conflicting IN MY BRAIN. 
What do you mean, “the first child”? Presumably, you mean “the one we know is a girl”, but that’s not a child. If they happen to both be girls, then saying “the one that’s a girl” isn’t really saying anything.
That said, similar to the questions about Monty’s motivations (does he always open a door, or only when he feels like it?), it depends on how you get the information. The puzzle basically assumes that you asked “How many children do you have?”, and “Is at least one of them a girl?”, and that the answers you got were “two” and “yes”, respectively. But this is somewhat of a contrived way of gaining information, and it’s a lot easier to come up with plausible situations where you do find out about one of the kids in particular (in which case the answer is the intuitive 50-50).
To recap, the key assumption in the problem is how often Monty opens another door. If he only does so when you’ve chosen the big prize, you should never switch. If he always opens another door, you should switch because you have twice as high a chance of winning (for the reasons stated above). If he opens another door, say, half the time, then Stranger is right, this ceases to be a probability problem and becomes instead a game theory problem.
To recap:
In the classic Monty Hall game, where Monty opens one of the other doors (after you pick a door), your chance of winning the prize is 2/3 if you switch. Your chance of winning is 1/3 if you do not switch. This is a fact.
To recap : I can read. What I was asking about was reconciling the truth with the more intuitive answer.
The generic “first child” of a couple who has two children. Its gender does not have any effect on the probability of the gender of the second child.
Ah, I tumbled to it, finally, from seeing the parallel thread : Boy/Girl probability in Monty Hall - Cecil's Columns/Staff Reports - Straight Dope Message Board
An already-born child getting a sibling has a 50% chance of either gender… and yet, if we know the gender of one child, we can predict the other with 66% probability… because in the first case I implicitly know the birth order of the children, and in the second, I do not.
Think of it this way; in the first round, you get to pick from three doors. In the second, you get to pick from a group of one door (#1, your original pick), or a group of two doors (#2 and #3, one of which Monty has already opened to display a goat). Going with the second group results in (effectively) getting to open two doors; hence, 2/3 likelyhood of success. (This isn’t the case when sticking with Door #1 because you still have the same probability that you started with – 1/3.)
Or think if it this way; if Monty didn’t open a door, when you switch you’d have a 1/3 chance of selecting the wrong door. By opening, say Door #2 with the goat, Monty has prevented you from switching to the wrong door; therefore, he gives you the chance to switch, but (assuming that the gold isn’t behind #1) not to make the wrong choice; effectively, you get to open both #2 and #3 without being penalized.
You are correct; the gender of one particular child doesn’t affect the gender of another, and the knowledge limits the number of permutations to 2, giving a 1/2 chance, i.e. if the younger child is female, it has no effect on whether the older child is male or female; the chance is 1/2 either way. However, if you only know that one of the children is female, this limits the number of permutations; out of the four possibilties (MM, MF, FM, FF) you have to eliminate the MM permutation, leaving three other choices of equal probability, and giving a 2/3 preference to a MF combination of children. Since our answer in the latter case is only concerned with the combinations, not the order, MF and FM have an equal likelyhood with FF.
Does that help?
Stranger
Of all the explanations of the Monty Hall probelem, I think borschevsky’s is the clearest. So, I’ll expand on it to muddy up the situation. 
Seriously, let’s approach it from the car’s perspective.
The car is equally likley to be behind door 1, 2 or 3. (p=1/3 for each)
So there are only 3 possible situations if you pick (for example) door #1.
Situation A
Car behind door 1 - you switch and always loose. (Happens 1/3 of the time)
Situation B
Car behind door 2 - you switch and always win. (Car is behind an unchosen door and Monty can only pick door #3) (Happens 1/3 of the time).
Situation C
Car behind door 3 - you switch and always win. (see above) (Happens 1/3 of the time).
Adding up the probabilities, we see that by always switching, you win 2/3 of the time and loose only 1/3 of the time.