I’ll just note in passing that this puzzle is discussed in this post, so you may find it or the associated comments helpful.
I think the specific issue is that in the first case, you’re assuming that each of the three possible orderings yields the same chance of your observation (the son out walking with him is a boy). If you assume that his choice of which child to go walking with is random, then the fact that you see a boy makes the (girl, boy) possibilities each less likely, so together they are equally likely to the (boy, boy) one.
Let’s define (imagining, for the sake of simplicity, that Omega descended from the heavens and informed you that the man you are about to meet has two children who can both be classified into ordinary gender categories):
h1 = "Boy then Girl"
h2 = "Girl then Boy"
h3 = "Girl then Girl"
h4 = "Boy then Boy"
o = "The man is out walking with a boy child"
Our initial estimates for each should be 25% before we see any evidence. Then if we make the aforementioned assumption that the man doesn’t like one child more than the other:
The denominator is a constant factor which works out to 0.5 (meaning “before making that observation I would have assigned it 50% probability”), and overall the math works out to:
So the result in the former case is the same as in the latter, seeing one child offers you no information about the gender of the other (unless you assume that the man hates his daughter and never goes walking with her, in which case you get the original 1⁄3 chance of it being a boy).
The lesson to take away here is the same lesson as the usual bayesian vs frequentist debate, writ very small: if you’re getting different answers from the two approaches, it’s because the frequentist solution is slipping in unstated assumptions which the bayesian approach forces you to state outright.
Thanks. I see why the probability of H1|o and H2|o need to be taken as 25% each. In that case, it seems like Sarah can say that it is 50% likely a boy and 50% likely a girl (at home). Why is the answer to the question then given as 66%?
The standard formulation of the problem is such you are the one making the bizarre contortions of conditional probabilities by asking a question. The standard setup has no children with the person you meet, he tells you only that he has two children, and you ask him a question rather than them revealing information. When you ask “Is at least one a boy?”, you set up the situation such that the conditional probabilities of various responses are very different.
In this new experimental setup (which is in very real fact a different problem from either of the ones you posed), we end up with the following situation:
h1 = "Boy then Girl"
h2 = "Girl then Boy"
h3 = "Girl then Girl"
h4 = "Boy then Boy"
o = "The man says yes to your question"
With a different set of conditional probabilities:
And it’s relatively clear just from the conditional probabilities why we should expect to get an answer of 1⁄3 in this case now (because there are three hypotheses consistent with the observation and they all predict it to be equally likely).
I’ll just note in passing that this puzzle is discussed in this post, so you may find it or the associated comments helpful.
I think the specific issue is that in the first case, you’re assuming that each of the three possible orderings yields the same chance of your observation (the son out walking with him is a boy). If you assume that his choice of which child to go walking with is random, then the fact that you see a boy makes the (girl, boy) possibilities each less likely, so together they are equally likely to the (boy, boy) one.
Let’s define (imagining, for the sake of simplicity, that Omega descended from the heavens and informed you that the man you are about to meet has two children who can both be classified into ordinary gender categories):
Our initial estimates for each should be 25% before we see any evidence. Then if we make the aforementioned assumption that the man doesn’t like one child more than the other:
And then we can apply bayes theorem to figure out the posterior probability of each hypothesis:
The denominator is a constant factor which works out to 0.5 (meaning “before making that observation I would have assigned it 50% probability”), and overall the math works out to:
So the result in the former case is the same as in the latter, seeing one child offers you no information about the gender of the other (unless you assume that the man hates his daughter and never goes walking with her, in which case you get the original 1⁄3 chance of it being a boy).
The lesson to take away here is the same lesson as the usual bayesian vs frequentist debate, writ very small: if you’re getting different answers from the two approaches, it’s because the frequentist solution is slipping in unstated assumptions which the bayesian approach forces you to state outright.
Thanks. I see why the probability of H1|o and H2|o need to be taken as 25% each. In that case, it seems like Sarah can say that it is 50% likely a boy and 50% likely a girl (at home). Why is the answer to the question then given as 66%?
The standard formulation of the problem is such you are the one making the bizarre contortions of conditional probabilities by asking a question. The standard setup has no children with the person you meet, he tells you only that he has two children, and you ask him a question rather than them revealing information. When you ask “Is at least one a boy?”, you set up the situation such that the conditional probabilities of various responses are very different.
In this new experimental setup (which is in very real fact a different problem from either of the ones you posed), we end up with the following situation:
With a different set of conditional probabilities:
And it’s relatively clear just from the conditional probabilities why we should expect to get an answer of 1⁄3 in this case now (because there are three hypotheses consistent with the observation and they all predict it to be equally likely).
That makes a lot of sense, thank you.