Isn't the probability still 2/3 even in the proposed alternative? The fact that we "arbitrarily announce the gender of one of the children" provides new information for calculating the probability, which is ignored if you assume that the original 50/50 distribution is unchanged by the announcement.

In an extreme example, consider this algorithm, analogous to the one in the article:

1. Choose a random parent that has exactly two children

2. Announce the gender of _both_ children

3. Ask about the odds that the parent has both a boy and a girl

In this situation, it would be ridiculous to say that the odds calculated at step 3 are still 50/50. Because of step 2, it is either certainly true (100%) or not true (0%) that the parent had both a boy and a girl. Granted, on average with repeated trials, half the time the step-3 value will be 100% and the other half 0%. But still, the step 2 information changes the value at step 3.