A Probability Question

Since 3 is impossible (Sarah knows there is at least one boy) that leaves three options. Two of those options imply a girl, the other implies a boy. Therefore, she can conclude that her probability estimate must be that it is 66.6% likely that there is a girl at home, and 33.3% likely that there is a boy.

You've taken a piece of information - you observe the man with a son - and half-applied it, by crossing off two daughters as a possibility, but you forgot to update the relative probability of having two boys, since you were more likely to see him with a son if he had two sons than if he had a son and a daughter.

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:

P(o | h1) = 0.5
P(o | h2) = 0.5
P(o | h3) = 0.0
P(o | h4) = 1.0

And then we can apply bayes theorem to figure out the posterior probability of each hypothesis:

P(h1 | o) = P(h1) * P(o | h1) / P(o)
P(h2 | o) = P(h2) * P(o | h2) / P(o)
P(h3 | o) = P(h3) * P(o | h3) / P(o)
P(h4 | o) = P(h4) * P(o | h4) / P(o)
(where P(o) = P(o | h1)*P(h1) + P(o | h2)*P(h2) + P(o | h3)*P(h3) + P(o | h4)*P(h4))

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:

P(h1 | o) = P(h1) * P(o | h1) / 0.5 = 0.25
P(h2 | o) = P(h2) * P(o | h2) / 0.5 = 0.25
P(h3 | o) = P(h3) * P(o | h3) / 0.5 = 0.0
P(h4 | o) = P(h4) * P(o | h4) / 0.5 = 0.5

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.

h1 = "Boy then Girl"
h2 = "Girl then Boy"
h3 = "Girl then Girl"
h4 = "Boy then Boy"
o = "The man says yes to your question"

P(o | h1) = 1.0
P(o | h2) = 1.0
P(o | h3) = 0.0
P(o | h4) = 1.0

wgd is correct as to the logic, but not as to the biology of the problem. In fact, the other kid is more likely than not to be male.

These problem types tend to assume an equal chance of a boy and a girl being born, which is a false assumption. (See: http://www.infoplease.com/ipa/A0005083.html)

I realize this may seem petty, but this is roughly like calculating the chance of picking the three of clubs as a random card from a deck is one in fifty. It's close, but it's wrong. An implicit assumption otherwise seems misguided; it should be made explicit (to make a logic problem rather than a logic and biology problem.)

EDIT: As wgd points out below, my answer here is wrong in its particulars (I didn't take into account all of the information available to Sarah and George in the puzzle as stated). The general principles invoked are sound, though.

George does actually know more. I think you're getting thrown by the fact that his 50-50 probability distribution seems more equivocal (less concentrated) than Sarah's 66-33 distribution. But remember that these distributions are defined over a space of four elements (boy-boy, boy-girl, girl-boy and girl-girl), so the actual distributions are 0.5-0.5-0-0 for George and 0.33-0.33-0.33-0 for Sarah. When you see it this way, it becomes a bit more plausible that Sarah's distribution is actually more "spread out", more equivocal.

To be more precise, suppose you have been given the task of conveying information about the genders of this man's children. You decide that you will transmit a 0 to represent a boy and a 1 to represent a girl. If the receiver has absolutely no information about the man's children, apart from the fact that there are two of them and neither is genderqueer, you will need to send two bits of information -- one for the elder child's gender, and one for the younger one's -- in order to convey full information about the genders. On the other hand, if the receiver already knows the children's genders, you have to send zero bits of information in order to convey full information. So you can think of the number of bits you need to transmit as a measure of the lack of knowledge of the receiver. The fewer bits you need to send, the more the receiver knows.

Now let's compare George and Sarah. George already knows the elder child's gender, so you only need to send one further bit, representing the younger child's gender, in order to convey full information. Sarah's case is trickier. She knows that one of the children is a boy, but she doesn't know which one. If it turns out that both children are boys, then your task is easy: you need to send just one bit of information, a 0, representing the gender of the child she hasn't seen. Once she gets this bit, Sarah will know the genders of both children. But if the other child is a girl, you can't just say that. You will also need to tell Sarah whether the order of birth is boy-girl or girl-boy. So besides sending a 1 to represent a girl, you'll need to send one more bit of information in order to distinguish between boy-girl and girl-boy. This means that the number of bits you will have to send Sarah is either 1 or 2, depending on whether the other child is a boy or a girl. If you did this experiment over and over again, with a bunch of different groups of siblings, the average number of bits you send Sarah will be greater than 1 but less than 2.

So with George you only need 1 bit to convey full information, while with Sarah you need (on average) more than 1 bit. This means Sarah does indeed know less about the situation, and there is no paradox. All of this can be made a lot more rigorous using the concept of Shannon entropy, if you're interested.