Right, you have to very clearly define how you do the sampling. Here's some possible cases:

Assume all houses have 2 children, and each child has equal probability of being either a boy or a girl. It will help to treat children as distinguishable, e.g. by eldest vs youngest.

* You randomly choose a house, and a random child answers the phone. You question the person who answers the phone about his/her gender and that of his/her sibling. You repeat this experiment. Given that the person who answered the phone was a girl, the conditional probably she has a sister is 1/2. Note that this is not bertrand’s box, as the 1 girl house can occur as either eldest vs youngest girl (unlike Bertrand box where only 1 box has exactly 1 gold) so they cancel out and so a girl you spoke with is equally likely to have been from either a 2 girl or 1 girl house.

* You categorize houses into 2 girl, 1 girl, and 0 girl houses. You randomly pick a category, then pick a house from that category, then phone them. As before a random child answers, and you question them. given that the person who answered the phone was a girl, the conditional probably she has a sister is 2/3. This is bertrand’s box, you’re more likely to have spoken with a girl from from a 2 girl house than a 1 girl house. Explicitly grouping by # of girls first before sampling breaks the previous symmetry.

* You randomly choose a house, and ask for the eldest child. You question him/her. You repeat this experiment. Given that the person whom you spoke with was a girl, the probability she has a sister is 1/2. Nothing new here, as seen in case (1) you were already equally likely to speak with a girl from a 2 girl vs 1 girl house, so asking for the eldest person (which by symmetry is equally likely to be a boy or a girl) doesn’t change anything here.

* You categorize houses into 2 girl, 1 girl, and 0 girl houses. You randomly pick a category, then pick a house from that category, then phone them and ask for the eldest child. Given that the person who answered the phone was a girl, the conditional probably she has a sister is 1/2. Explicitly choosing the eldest disrupts the asymmetry in bertrand’s box: since every house has only 1 eldest which is the one you speak with, being from a two girl house no longer makes that girl more likely to have spoken with the caller.

* You randomly choose a house, and a random child answers the phone. You question the person who answers the phone. You repeat this experiment. Given that EITHER the child you spoke with OR their sibling is a girl, the probability you spoke with someone from a 2 girl house is 1/3. It might seem counterintuitive at first that loosening the criteria _reduces_ the probability of speaking with someone from a 2 girl house. But this makes sense, since there’s still only 2 ways you can speak with someone from a 2 girl house (either the eldest or youngest sister), but now 4 ways you can speak with someone from a 1 girl house, since you’re allowed to speak with the boys of that house as well.

* You randomly choose a house, and ask for the eldest child, and question them. Given that EITHER the child you spoke with OR their sibling is a girl, the probability you spoke with someone from a 2 girl house is again 1/3. Explicitly speaking with the eldest doesn’t make a difference here because we’re already conditioning on either the eldest or youngest being a girl.