That is just terminology.
The chance is 1 in 3. The odds are 2/1 against. Do you agree?
PMU I think we are just at cross purposes on terminology.You are correct. The question asked was “What are the odds she has 2 boys?” and the odds i.e. the ratio of winning events (BB) to losing events (BG, GB, GG), is 1 in 3. But the title of the thread is “'Two probability questions for nerds”, and the probability, i.e. the chance of the winning event (BB) occurring, is 1 in 4, i.e. 25%.
When you know she has two children, but not the gender of either, the possible combinations are BB, BG, or GG. Order is inconsequential so BG and GB are one and the same.I didn't say it was the first that was a boy. Maybe it was the second or maybe it was both.
Yes but BG is twice as likely as BB as it can happen in either order. You should really think of the space as BB, BG, GB, GG taking the eldest first.When you know she has two children, but not the gender of either, the possible combinations are BB, BG, or GG. Order is inconsequential so BG and GB are one and the same.
But you've already confirmed one of them is a boy, ruling out the GG option, so the only valid options remaining are BG or BB.
But in the context of the question. order is inconsequential. GB and BG are the same result and she would have 1 boy. Ultimately, someone with two kids can have two girls, two boys, or one of each, only three options. Knowing one is a boy eliminates the two girls option.Yes but BG is twice as likely as BB as it can happen in either order. You should really think of the space as BB, BG, GB, GG taking the eldest first.
How often are there 2 boys? 25% of the time. (Think tosses of 2 coins, how often 2 Heads?)But in the context of the question. order is inconsequential. GB and BG are the same result and she would have 1 boy. Ultimately, someone with two kids can have two girls, two boys, or one of each, only three options. Knowing one is a boy eliminates the two girls option.
But the question was asked after you told us one of them was a boy. At that point, the two girls option has already been ruled out and we're looking at the odds of a single child being a boy, and the rules provided state that is 50%. Were the question asked prior to sharing that one of them is a boy, then the probability of two boys is indeed 33%.How often are there 2 boys? 25% of the time. (Think tosses of 2 coins, how often 2 Heads?)
How often are there 2 girls? 25% of the time.
Yes that was the intended question. Do you think I framed it carelessly?So the question asked was really how often are there 2 boys when 1 of them is a boy.
I certainly looked at it taking all the information I had to hand at the point the question was asked, and at that point it's just down to the 50/50 of the other child being a boy assuming independent events. For it to be anything other than 50/50, you're introducing dependency.Yes that was the intended question. Do you think I framed it carelessly?
There is dependency. You know one of the children is a boy, so the question really is / should be "A lady tells you she has 2 children. You ask has she a boy and she says Yes. So, what is the probability she has 2 boys, given that one of the children is a boy?"I certainly looked at it taking all the information I had to hand at the point the question was asked, and at that point it's just down to the 50/50 of the other child being a boy assuming independent events. For it to be anything other than 50/50, you're introducing dependency.
That's a common mistake people make in the coin toss odds. Dependency implies that the first outcome affects the odds of the second event, like pulling a card from a deck affects the next draw as there are fewer cards remaining. We're told in the OP:There is dependency.
So this says there is no dependency on the odds of the second child being boy/girl based on the gender of the first.You may assume that it is 50% that a child is a boy and that if the eldest is a boy it is still 50% that the youngest is a boy.
Yep, but like a coin toss, each one is independent and 50/50. If the first one is heads, the odds of the second one being heads is still 50/50.You know one of the children is a boy, so the question really is / should be "A lady tells you she has 2 children. You ask has she a boy and she says Yes. So, what is the probability she has 2 boys, given that one of the children is a boy?"
Thanks, I have made that amendment to OP to make the question clear.There is dependency. You know one of the children is a boy, so the question really is / should be "A lady tells you she has 2 children. You ask has she a boy and she says Yes. So, what is the probability she has 2 boys, given that one of the children is a boy?"
Absolutely, but you have shifted the answer from Yes to No. If you get a Yes to a very remote possibility you do have, almost as an incidental, that there is at least 1 boy. But you are almost justified in talking about two separately identifiable children - the one that passed the test and the one that did not. The reason the second child did not pass is almost certainly because s/he didn't pass the extremely unlikely test and so we must reckon him/her to be 50% of being a boy, just as if we had no information on him/her.Duke
On question 1, I understand the 1 in 3 chance of her having two boys if we know that at least one of them is a boy: (BB) out of (BB, BG, GB).
On question 2, let's pose a different question. Suppose I ask her if she has a boy who was born between midnight and 1am on a New Year's Day. It will be astounding if I get a Yes. I will get No as the answer practically every time (probability 8765/8766 assuming equal probabilities - although I doubt the probabilities are equal: I would say that a heck of a lot fewer births are recorded between 11 and midnight on New Year's Eve than in the next hour, but that's a separate conversation). Therefore, getting the answer No to the second question is almost the same as not asking the question at all, so the probability of two boys is very close to 1 in 4. For other less challenging questions, the chance of two boys is greater than 25% but less than 33%. The less likely we are to get Yes, the closer the chance is to 1 in 4; the more likely we are to get Yes, the closer it is to 1 in 3. Am I on the right lines?
Hi Duke. You're right that, by shifting the answer from Yes to No, I've changed the problem. I leave it to others to give the correct answer!Absolutely, but you have shifted the answer from Yes to No.
But what you have shown is that a Yes answer must increase the probability if No reduces it.Hi Duke. You're right that, by shifting the answer from Yes to No, I've changed the problem. I leave it to others to give the correct answer!