*The second question is much the same. The possible gender combinations for two children are:
(1) Child A is female and Child B is male.
(2) Child A is female and Child B is female.
(3) Child A is male and Child B is female.
(4) Child A is male and Child B is male.
We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.*
If you are presenting the possible cases in order then you should also discard case (3) as we already know the first kid is a girl. On the other hand, if you present the cases without an order (either A or B can be the first child) then case (1) and (3) are not different possible combinations: they’re the same. Wich gives the answer to the problem as 1/2 not 2/3 as suggested.
I know it sounds logical that, being the odds equal and having already a girl, it feels right to say that a boy would be more probable for the second kid. But… your reasoning, if “cleaned” from this wrong option, shows the opposite.
Except that we don’t know that the first child is a girl. All we know is that at least one child is a girl, but we don’t know which one, or both. Admittedly it’s difficult to construct real-world situations where we know that at least one child is a girl, without some identifying information, but if you do construct such a situation, Cecil’s answer is correct.
Your post is a bit muddled, and I think you threw Chronos off. Your argument starts off with saying we don’t know if the first child’s a girl.
True, case (1) and (3) do become two sides of the same case. However, that doesn’t mean that this new case occurs as frequently as the other, (2). Simple examples of probability often have all cases have equal likelihood, but that is not a rule. Saying that you now have two cases: (2) and (1+3) doesn’t give you any answers.
Funny you should say that, because I think most people will “intuitively” guess 1/2. However, your intuition that it’s more likely that a family has a boy and a girl instead of two girls is actually correct.
This is starting to resemble that old debate about how many times the moon rotates on its axis as it orbits the Earth; whether the answer is “once” or “twice” depends on how you interpret the question and some of the vague language within.
You know, dbird, I hope you sign up for this board. We need more people like you: capable of constructing a coherent argument, and also capable of immediately understanding what was wrong with it – AND having the courage to admit it. Class act.
I totally failed to get a date with a girl from Uraguay. Does that count?
Color me as glad that this thread didn’t degenerate into one of those shouting contests about this problem. And props to dbind for understanding and admitting his error, rather than doing the ego thing. He’s a good cobber in my book.
The non-responsive responses are offering examples of other debates that have occured on this Message Board over other columns of Cecil’s where the debate was over semantics more than substance.
Wouldn’t the Moon rotate about 1.073 times during each lunar cycle? After it does 1.000 rotations, it’s in the same spot compared to the background stars, but still has a little farther to go to be in the same spot relative to the Sun.
And back to the OP, here’s my favorite variation of the boy-girl problem. You meet an old acquaintance at the store for the first time in years. She is accompainied by her daughter. In conversation, you find out that she actually has two children. You do not know whether the kid you’re looking at is the older or younger one. What is the probability that both are girls?
The quote you are questioning was an attempt to reference the Straight Dope column on whether .9 repeating equals 1 or not. An issue over which many a battle has raged, the ignorant masses not comprehending the math involved in the equation.
Birth order doesn’t matter. All that matters is whether one particular child is specifically referenced. Here, one is: The one you see in the store. So the chance is 50-50.
