>> The father says that at least one child is a boy.
>> That eliminates the option of GG. Are you somehow
>> suggesting that the probability distribution of
>> the other options is changed by this?
> Yes. ... Going (invisibly) from 4 possible worlds
> (including GG) to 3 changes the prob dist of the
> remaining 3, by which I mean 2. If Monty must
> guarantee you at least one boy, then there aren't
> really 3 remaining possibilities. The unguaranteed
> can be a boy, or it can be a girl. What is the
> third?
The standard reasoning is that there are two options, but they are not equally likely. Equally likely options are, labelling oder child first, BB, BG, GB. These were equally probable before we gained the extra information, and the standard interpretation is that they remain equally probable after.
In the Monty Hall case you choose a door, and it has probability 1/3 of being right. It remains 1/3 probable after he chooses a door. The other two doors together have probability 2/3 of concealing the car. It remains 2/3 after he has chosen and opened a door. The probabilities of the branches don't change in any of these cases.
Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still not be equi-probable?
> A: So now let's take another experiment.
> I tell you I'm going to flip two coins
> until at least one of them is heads. I
> do that. Now I ask you to bet on whether
> or not there is a tail. What do you
> think are fair odds?
And you say:
> I misread A. I agree that it's 1/3 if you
> flip pairs until at least one is heads.
OK. Now B:
> B: Now I hunt among couples with two children
> until I find one that doesn't have two girls,
> and I ask: what are the odds they have two
> boys?
Do you agree that it's 1/3?