Once again, good job. I'm going to provide some alternate explanations now. Oh, and btw deadly, real men use fractions.
First, to clarify what deadly said, there are three permutations:
1. guy girl
2. girl guy
3. girl girl
If the other one is a girl then we have two girls, and only one out of three of the possible permutations gives us three girls.
So you may be asking, why isn't there just two options: 1. girl guy 2. girl girl. I said that one of the two children is a girl, not the first one. The first child OR the second child could have been a boy. If you say that the first children was a girl, then there's only two permutations: girl guy and guy girl.
Another way to think about this is that saying "one of them is a girl" is not an event, since you don't know when it happened. Let's say you have three marbles in a bag, two white and one red. If you say you picked out two marbles and one of them was white then you can't determine the probability of you picking out that white marble. If the first one you picked was red, then you have a 100% chance. If the first one you picked was white, then you have a 50% chance. An event requires a timeframe, and you can only apply the rule "prior events don't affect future events" to independent events, and an independent event must be an event in the first place.
You can also do a simulation of this. Keep a variable for the sum of all the outcomes. Let 0 be boy and 1 be girl. Let the sum of the two numbers be the outcome.
0 = two boys
1 = one girl one boy OR one boy one girl
2 = two girls
Generate two numbers in Z & [0, 1] (set of integers intersected with the set of numbers from 0 to 1 inclusive). If both of them are 0's, add one to the sum (since one of them has to be a girl). After you get one outcome add it to the variable that's the sum of all outcomes. You don't need an array for this, an O(n) operation for finding out the fraction of girl-boy/boy-girl combos is inefficient.
So here's the algorithm for finding out the overall result. Let n be your number of trials. If your sum variable == 2n then you got 100% 2-girl permutations. If your sum variable == n then you got 100% mixed permutations. To simplify things, we'll find out the fraction of mixed permutations first. You can find out your fraction of mixed permutations by n / 1 : S / A. S is your sum variable and A is the answer. Thus A = S / n. And the fraction of two-girl combos is 1 - A.
