Order doesn’t actually matter. And you don’t need to use playing cards. Just toss two coins 100 times and eliminate all the cases (about 25) with two heads. Of the about 75 remaining, about 25 will be two tails and about 50 will be one head and one tail.
I have only read to the point in the thread when this was posted. Just wanted to register that it’s a wonderful explanation. It makes concrete a distinction that can be really hard to wrap the head around!
Okay, I’m going to be doggedly stupid here. No matter how many times I read the “2/3” arguments, they sound like variations of the gambler’s fallacy. Each flip of a coin has a 50/50 chance; each sperm-egg dance has a 50/50 chance. Minor biological possibilities aside, no one flip/dance has any effect on any other in the series.
Just because one of two children is heads, how does that affect the odds of the other child being 50/50 heads/tails? The reasoning used for 2/3 seems to be erroneously reductive.
There is no “the other child”. Imagine:
“Do you have at least one girl?”
“Yes, I do.”
“What’s your other child’s name?”
If a parent with two girls is asked that, how should they answer?
So we’re back to semanto-babble.
It does matter. If a family has two children – or if you pick any two people at random, for that matter – there’s a 50% chance that they’ll both be the same sex, and a 50% chance that they’ll be different sexes. If they’re the same, that there’s a 50% chance (25% overall) that they’re both male, and 50% (again, 25% overall) that they’re both female.
If you don’t believe that, then try my experiment with the cards, or John W. Kennedy’s with the coins. (I like cards better, but that’s just me.)
It doesn’t. And that’s why the answer is 2/3.
You flip two coins. Each coin has a 50% chance of landing heads or tails. Both flips are independent of each other.
If coin A lands heads, there’s a 50% chance that B will be heads, and 50% that it will be tails.
If coin A lands tails, there’s still a 50% chance that B will be heads, and 50% that it will be tails.
So, because both flips are independent, there’s a 50% chance that they’ll come up different, a 25% chance that they’ll both be heads, and a 25% chance that they’ll both be tails.
Similarly, suppose you have one child. Whatever sex that child is, there’s a 50% chance that your next child will be the same sex, and 50% that it will be of the opposite sex.
Ok… real world example. I have a wife and a son. Let’s hypothetically say my wife’s currently pregnant.
Are you telling me we have 66% chance of having a daughter for our second child?
Just the opposite. You have a 50% chance of a son and a 50% chance of a daughter. As I said, there’s a 50% chance that your second child is the same sex as the first.
There is, however, a 0% chance that both of your children will be daughters.
Go grab a thousand couples who have a son and who are pregnant. I’ll wait.
Okay, got 'em? Now go give them all an ultrasound, and tell me what percentage of the fetuses are female.
About fifty percent. Okay. Now let’s do another.
Go grab a thousand couples who have two kids. Again, I’ll wait. (No cheating, make it random. Just screen for “having two kids” and nothing else.)
Alright, now ask them to raise a hand if they have at least one daughter. How many raised their hand? About seventy five percent, right? Okay, now dismiss the other twenty five percent. You’ve got 750 couples remaining. Now ask the 750 to raise their hand only if they have a son. How many raised their hand? About 500–67 percent of them.
There it is.
Are you doubting that any of the above would actually be true were the scenarios actually carried out?
I don’t get it. 
nvm
It doesn’t. It seems to, but it doesn’t. As you and Chronos point out, how you pose the question makes all the difference.
The gambler’s fallacy is that if I’ve just thrown 19 heads in a row, the chances of me throwing heads on my next toss are really small. In fact, they’re still 50/50. On the next toss. But if instead of asking “What are the odds that this next toss comes up heads?” you ask, “What are the odds that I can throw 20 heads in a row?”, then you get the teeny tiny odds you’re looking for.
Similarly, this question isn’t asking, “If I have one daughter, what are the chances that the second child is a boy?” That would be indeed be 50/50, or 1:2. It’s asking, “Of all the families that have two kids, let’s eliminate those without at least one girl. Now, of the remaining families, what are the odds that both kids are girls?”
Or if you want to put it in coin flips, it’s asking, “Consider all possible two-flip scenarios. Eliminate those scenarios that have no heads. Now, of the remaining scenarios, what are the odds that you have two heads?”
It’s tricky because it sounds straightforward, but it’s a really weird question, when you get down to it. And that’s why it’s interesting. It sheds light on flaws in common-sense approaches to problem solving. A more standard question would be, “Consider all possible two-flip scenarios. What are the odds of getting two of a kind?” Your odds are 1:2. Obviously, right? Because whatever your first throw was, the next one will either be the same or different. So you have a 1:2 chance of getting the same thing on the next throw. Or perhaps, “Consider all possible two-flip scenarios. What are the odds of getting two heads?” Your odds are 1:4. Now, even here, some people might say, “It should be 1:3! I can get two tails, a head and a tail, or two heads!” But again, not all possibilities are equally likely. The only way to correctly calculate the odds is consider what can actually happen at each juncture and multiply it out. The first throw has a 50% chance of being heads, and so does the second. This yields four different equally likely scenarios - TT, TH, HT, and HH - each with a 25% or 1:4 chance. This means that throwing a head and a tail (in any order) is twice as likely as throwing two heads.
But this question is basically saying, “Consider all possible two-flip scenarios. Half are two of a kind, and half are two different sides. Then, from the half that are two of a kind, eliminate half of that group: all the two-tails. Now, you’re left with 3/4 of your original scenarios: the half that were two different sides, and the two-heads half of the half that were two of a kind. Of these remaining scenarios, how many also have a tail? How many have two heads?” At a glance, it doesn’t sound like nearly as complicated an operation as it actually is.
And that’s why stuff like this is interesting to me: we can make huge errors in logic because of small changes in language, and it behooves us to be on our guard for things like this that we encounter in everyday life. “Wait… what is this graph really saying?”
His example started with “Ok… real world example. I have a wife and a son.”
If he specifies that one of his two children is a son, then we can eliminate the possibility that both children are daughters.
For that matter, we’re also eliminating the daughter/son combination, leaving two equal possibilities: son/daughter or son/son.
I’m not sure what the smiley face means here. Are you saying “I don’t get it, but I remain friendly?” Or are you saying “I don’t get it, but actually I’m just kidding and I do get it?”
In that case, that’s the difference between the two statements: in one, you know they’re speaking of a specific girl. In the other, they could be speaking of either of their two children.
Powers &8^]
No, you have a 50% chance of having a daughter, because in this situation, you’ve already eliminated two of the four different ways of having two kids:
A. boy, girl
B. girl, boy
C. girl, girl
D. boy, boy
Each of these is equally likely, but you’ve eliminated B and C by telling us your first child was a boy. That leaves A and D, and each has equal probability, or 50%.
Powers &8^]
This isn’t the Monty Hall question, and that’s why I think applying that logic to this one is wrong.
You’re right in my example that I’ve ruled out B and C, but in the OP’s example, there’s no difference between B and C because it doesn’t matter who came first.
Just knowing that a couple has a daughter doesn’t change the odds that they may also have a son, assuming that sons and daughters are born with equal frequency.
Explain how my question is different from the OP’s question. There seems to be no difference from a logical and practical standpoint.
In your question, you know the sex of one particular child (your son). In the OP’s question, you only know that one of the two children is a girl. You don’t know the sex of a particular child. That’s the difference.
Then later you wrote:
So to further answer this:
Cases 1 and 2 are in no way equivalent for you. You have zero chance of ending up with Case 1, with a girl for your first child and a boy for your second, and a 50% chance of ending up with an older boy and a younger girl. But in the OP’s question, when all we know is that at least one child a girl, Case 1 is possible, and just as likely as Case 2 or Case 4. That’s how the questions are different.