Congratulations, Jeff, you win the Internet. Now please give someone else a turn.
I read through most of the thread, then drew a little diagram to help me think through it.
B1 50% G1 50%
/ \ / \
/ \ / \
B 50% G 50% G 50% B 50%
When a couple has a child, 50% that’s it a boy or a girl. Another child = 50% boy or girl. It doesn’t matter which is born first. When a couple has two children,there’s a 75% chance that one is a girl. When we know that one child is a girl, there is a 66.6 (2/3) chance that the other is a boy. This chart does not mean the the B1 and G1 on top are the first children. B1 and G1 could just as easily be the second children. Order does not really matter.
(sorry, didn’t read all the posts, this may have been correctly answered and explained already).
The answer is 2/3
This is similar to a problem in the all of statistics textbook by wasserman. The problem goes something like you have one card that is green on both sides, one that is red on both sides, and one that is red on one side, green on the other. You draw a card at random and see one side at random. If the side you see is green, what is the probability that the other side is also green?
Most people will intuitively (and incorrectly) answer 50%. Let’s say A is the event you see green, then P(A) = 1/2 (there’s 6 sides, 3 of them are green, 3/6 = 1/2).
Let B be the event you see green on the other side of the card, so the probability that we see green on the other side, given we saw green on the first side is (by bayes rule) P(B|A) = P(A and B)/P(A).
P(A and B) is 1/3, because there’s 3 cards and only one of them is green on both sides so:
P(B|A) = P(A and B)/P(A) = (1/3)/(1/2) = 2/3
None of this takes into account the fact that the sex of offspring tends to run in the family. So, given any set of two-child families, the number with a GG and BB combination is each higher than the number with a BG combination. Therefore, a couple that already has a girl is statistically more likely to produce another girl.
lt;gringt;
I think the chances are slim to NONE that the person can have a baby that is both a GIRL and a BOY…
@Jon Ericson: I’m not sure I follow.
I don’t see that it matters how Jeff chose to reveal what he did. Regardless of how random or planned it might have been, the probability of him revealing that there’s at least one girl is 1. Because that’s what he did.
Chickencha: It matters how Jeff chose to reveal what he did because, eg, if he was asked do you have at least one girl? then he would have definitely said yes, even if one of the children was a boy. But if he was asked Randomly pick one of your children and tell me its sex then he might have said boy even though he has a girl as well.
Classic Monty Hall problem: 66%
To the 66%ers mentioning that the possible combinations are BB, GB, BG, GG, and that knowing one is a girl removes the BB option, you do realise that knowing one is a girl also removes one of the BG/GB options as well? As such, I’m going for 50% based on my interpretation of the question and the assumption that child birth results in an even distribution of boys/girls over a large sample.
Its NOT Monty Hall.
Monty never opens the door on the car.
I’m switching back to 50%. Here’s why.
i’m going to make up my own rules (because I can). Pick a parent of two children at random. They pick one of their children at random. They tell me the gender of that child, and I guess whether they have one of each.
The probability that they have one of each is 50% (as established so many times above). The probability that they have one of each and say girl is 25%. They would choose their girl half the time.
So, by Bayesian theory, the probability that they have one of each GIVEN that they said girl is equal to .25/.50 = .50.
QED
0%. I just met this guy, and he’s probably lying. I’m going to check my pockets to ensure nothing’s missing.
The real problem is that the title of this post implies that he will never let this game finish…
Absolutely no way to tell from the information given.
Some possibilities besides the ones listed above:
The person SAYS they have two children, and SAYS that one of them is a girl. What if the person is delusional and has no children? Or is simply lying?
That’s a non-zero possibility. As such, it’s going to mess with the probabilities such that it’s not going to be 1, 1/2, or 2/3, or any of the other possibilites given.
If we assume that the information given is accurate – and that’s an assumption – we also have to consider HOW it was given. Was it phrased in such a way that this information was simply gathered incidentally? If so, then the fact that one of the children is a girl is almost irrelevant – I suppose that the fact that identical twins exist would SLIGHTLY push the chances of the other child being a girl up, but there aren’t that many identical twins, so it probably wouldn’t matter much. So the most likely situation is that the odds that the other child is a boy is the same odds that a random birth will be a boy, whatever those odds are.
In other words, if the situation is that you’ve established through conversation that the person has two children, and then, later, the parent mentions that they’re worried that their older child will become pregnant, then that information is irrelevant to the sex of the younger child.
If the person says, I’ve got two kids – one daughter, then, in almost all cases, that means that the person is saying that they have ONLY one daughter, and that the other child is either a son or intersexed or something – most likely a son, and the odds are NEARLY 100% that their other child is a boy – except that there is the chance that they’re lying, or that the other child is intersexed, or some other situation.
Chance of having a girl (.5) x Chance of having a boy (.5) = .25 = 25%
Chance of having a boy AND a girl = 25%
Appeal to authority is a logical fallacy. Just because wikipedia (or your statistics textbook) says this is the solution does not make it true.
There are three possibilities:
boy/boy
boy/girl (same as girl/boy)
girl/girl
thus it’s 33% because we can’t rule out boy/boy as we don’t know if he/she’s lying or not.
For the 50% crowd that says that order doesn’t matter or two possible outcomes (BG,GG), you have to take into account the how likely each of the outcomes is. Since there are two ways of getting a boy and a girl, and only one way of getting two girls. Having one of each is twice as likely.
The probability is 100%.
Why?
They told you they have two children. If they have two children, they can also have three. Or four. Or five. Or an infinite amount of children, in theory.
If they have an infinite amount of children, they will almost surely have a boy.
Do I win a prize?
I’d say 50%.