As you may know, my Full Nonindexical Conditioning (FNC) approach (see http://www.cs.utoronto.ca/~radford/anth.abstract.html) uses the third-person perspective for all inference, while emphasizing the principle that all available information should be used when doing inference. In everyday problems, a third-person approach is not distinguishable from a first-person approach, since we all have an enormous amount of perceptions, both internal and external, that are with very, very high probability not the same as those of any other person. This approach leads one to dismiss the Doomsday Argument as invalid, and to adopt the Thirder position for Sleeping Beauty.

You argue against approaches like FNC by denying that one should always condition on all available information. You give an example purporting to show that doing so is sometimes wrong. But your example is simply mistaken - you make an error somewhat analogous to that made by many people in the Monte Hall problem.

Here is your example (with paragraph breaks added for clarity):

Imagine you are on an exotic island where all families have two children. The island is having their traditional festival of boys' day. On this day it is their custom for each family with a boy to raise a flag next to their door. Tradition also dictates in case someone knocks on the door then only a boy can answer. You notice about 3/4 of the families has raised a flag as expected. It should be obvious that if you randomly choose a family with flags then the probability of that family having two boys is 1/3. You also know if

you knock on the door a boy would come to answer it so by seeing him there is no new information.

But not so fast. When you see the boy you can ask him "are you the older or the younger child?". Say he is the older one. Then it can be stated that the older child of the family is a boy. This is new information since I could not know that just by seeing the flag. If both children are boys then the conditional probability of the older kid being a boy is one. If only one child is a boy then the conditional probability of the older kid being a boy is only half. Therefore this new evidence favours the former. As a result the probability of this family having 2 boys can be calculated by bayesian updating to increase to be 1/2. If the child is the younger kid the same reason can still be applied due to symmetry. Therefore even before knocking on the door I should conclude the randomly chosen family's probability of having 2 boys is 1/2 instead of 1/3.

This is absurd. This shows specifying the child base on ad hoc details is clearly wrong. For the same reason I should not specify today or this awakening by ad hoc details observed after waking up, such as the color of the paper.

Your mistake here is in asking the boy "are you the older or younger child?" and then reasoning as if an "older" answer to this question is the same as a "yes" answer to the question "is the older child in this family a boy?".

If you actually ask a neighbor "is the older child in that family a boy?", and get the answer "yes", then it WOULD be correct to infer that the probability of the younger child also being a boy is 1/2. But you didn't do that, or anything equivalent to that, as can be seen from the fact that the question you actually asked cannot possibly tell you that the older child is a girl.

The correct analysis is as follows. Before knowing anything about the family, there are four equally likely possibilities, which we can write as BB, BG, GB, GG, where the first B or G is the sex of the younger child, and the second is the sex of the older child. When you see the flag on the family's house, the GG possibility is eliminated, leaving BB, BG, GB, all having probability 1/3. When a boy answers the door, the probabilities stay the same. After you ask whether the boy is the younger or older child, and get the answer "older", the likelihood function over these three possibilities is 1/2, 0, 1, which when multiplied by 1/3, 1/3, 1/3 and renormalized gives probability 1/3 to BB and probability 2/3 to GB, with zero probability for BG (and GG). If instead the answer is "younger", the result is probability 1/3 for BB and 2/3 for BG.

There is nothing odd or absurd here. Conditioning on all available information is always the right thing to do (though one can ignore information if one knows that conditioning on it won't change the answer).