Is this actually true for the Bayesian model? Throw in a parameter for whether or not Bill and Lorena were trying to have children until they had a boy and a girl. Now your answer will depend quite heavily on your prior!
The posterior odds are computed ENTIRELY from the odds of the observed events under the prior beliefs. There is NO WAY in which might-have-beens and didn't-happens enter in. Therefore the posterior probabilities cannot depend on the knowledge of what they would have done if something different had happened.
Of course frequentist statistics are heavily affected by what would have happened if something different had happened.
In world one, I use bayesian update on the model P(# boys | Bias) = Mutltinomial(n = 7, p = Bias)
In world two, I use bayesian update on the model P(# children | Bias) = Geometric(p = Bias)
Might-have-beens and didn't happens do play in, in my choice of model. I should choose the one that I believe, and if I'm not certain, I should use an even more complicated model that incorporates my beliefs about what models might be appropriate.
One possible source of confusion for you is that Bayesian ideas have been a source of inspiration for a lot of ad hoc techniques (eg naive Bayes) which do NOT follow Bayesian rules of inference. The reason is that exact inference in Bayes nets is NP-hard. So you're used to hearing "Bayesian" applied to things that have nothing to do with Bayes' formula.
P(pb > 0.5 | "Six girls and one boy") = (P("Six girls and one boy" | pb > 0.5) * P(pb > 0.5)) / P("Six girls and one boy")
Applying Bayes theorem thus requires us to calculate P("Six girls and one boy" | pb > 0.5). How do you suggest that we do this? Why is your answer the unique correct solution?
Suppose our prior distribution of beliefs is 0.5 that the probability is exactly 1/2, versus 0.5 that the probability of a boy is some value P which is equally likely to be any value from 0 to 1.
In the first case, the probability of 6 boys and 1 girl is 0.5^7 = 1/2^7. In the second case the probability of 6 boys and 1 girl is P^6(1-P) = P^6 - P^7. The integral from 0 to 1 of P^6 - P^7 is 1/6-1/7 = 1/42. Each case also has a priori odds of 1/2 of holding true.
After observing 6 boys and 1 girl, the first case now has probability (0.5/2^7)/(0.5/2^7 + 0.5/42) = 1/(1 + 64/21) = 21/129 = 0.162790697674419. The second case now has probability 1 - this, which is 0.837209302325581. Furthermore if the second case is true, P is no longer uniformly distributed. In fact its density is now proportional to P^6-P^7.
So the posterior distribution is now going to be:
With probability 21/129, exactly 0.5. And otherwise any value P from 0 to 1 with a probability density of 108/129*(P^6-P^7)/42.
Given this prior and this set of observations, any other answer is wrong. Given a different prior you would get a different posterior, but as long as the prior gives a constant probability of male/female, the only fact that matters is how many boys and girls there are.
The order of births can only start to matter if you start with a prior that gives different probabilities of different genders based on prior events. Even then it is hard to come up with a realistic scenario in which the plans of the parents would make an order of magnitude difference in the posterior distribution.
If the set of prior theories does not include anything that depends on birth order, then birth order and the experimental design are irrelevant to the posterior conclusions.
We have 3 types of facts.
1. What is the set of prior theories? Probability theory says this should matter. Bayesians are explicit about its involvement. Frequentists ignore it.
2. Observed data. Probability theory says that this should matter. Everyone takes this into account.
3. Experimental design for what would have happened had something different than the observed actually happened. This matters a lot to frequentist approaches and does not matter at all to Bayes' theorem. Bayesian approaches generally do not care about it at all.
The difference between scenario 1 and scenario 2 is a fact of type 3, the conditions under which Bill and Lorena would have stopped having children. Unless you believe that Bill and Lorena's desire for one gender has a material impact on the probability of boys vs girls, this fact is irrelevant to any calculation of posterior probabilities. And is irrelevant in classical Bayesian approaches.
Yet, despite being irrelevant, it matters a lot for frequentist approaches.
Anyway, not sure if we're on the same page, but thanks for the discussion.
In particular 1 consists of exact statements about the likelihood of 7 births in a row being mmmmmmf.
By contrast 3 consists of statements about what Bill and Lorena's childbearing plans would have been if something different had happened.
Those are very different types of statement. There is no connection between statements of type 3 and statements of type 1 unless Bill and Lorena's state of mind makes a significant difference to the odds of the next child being a boy.
For Bayes' theorem, we need a theory of how the data is produced given the parameter of interest. Bill and Lorena's plans certainly influence what data I observe: in scenario two, I can never observe the data BBBBGGG, but in scenario one, I can. My point is that your first category is not "exact statements about the likelihood of 7 births in a row being mmmmmmmf", it is "exact statements about the likelihood of observing mmmmmmf", which is, in fact, quite different, if you admit the possibility of Bill and Lorena having particular childbearing plans.
You certainly DON'T wind up with anything like the factor of 8 difference that frequentist techniques will see!