I agree with this. I created a grid for myself to visualize it. In the base case we have 49 cases for Boy/Girl, 49 cases for Girl/Boy and 49 cases for Girl/Girl, one third each, as expected.
Add in the day restriction, we have what @RivkahChaya listed, namely that we lose 42 cases from Boy/Girl, lose 42 cases from Girl/Boy but only lose 36 cases from Girl/Girl since either child can be born on Thursday.
I’m still pretty sure that it depends on whether you have previously selected for someone with these qualifications, or if someone just tells you that they have these qualifications.
If someone says that they have 2 kids, what are the chances that both are girls, then the answer is 1/4. If they say that one of their kids is a girl, what are the chances the other one is a girl is now 1/2. If they add in that the girl was born on a Thursday, that doesn’t add any information as to the gender of the other child. If she was born on a Monday, Tuesday, Wednesday, Friday, Saturday, or Sunday, that would add in exactly the same information as if she were born on Thursday. It is only if I pick, out of a large group, only parents with 2 children, one of which is a girl born on a particular day, that that information factors into changing the probability.
Thursday is the bullseye, and it’s a question as to whether we are hitting the bullseye, or we are drawing a bullseye around where the dart hit.
It’s 1/3.
Some of these calculations are very circuitous!
It’s pretty simple. There are four types of families, all more-or-less equally represented in the population.
@Biotop 's family is in one of the first three groups. We don’t know which one. The chance of his family being in any one of those groups is 1/3. So 1/3 is the answer to the question
To put it another way, say I have 7 parents, each with one daughter born on different days of the week, and an unspecified other child.
Do each of these parents have a 13/27 chance of having two daughters?
@Aspidistra, that’s definitely not it. Why are you treating the gender of the children as relevant information, but not their day of birth? Considering both, there are 196 types of family, not 4.
The fun part of this riddle is … suppose you’re wandering around town on a Thursday and bump into @Biotop with his daughter. The daughter says “Today’s my birthday!”
You know that @Biotop has two children. What are the chances of them being two girls in this scenario?
1/2
Phrased that way, definitely no, because you’ve changed the problem by using the phrase “other child”.
It’s exactly the same problem, just with more complicated maths, and more opportunities to make math errors. If you do it right, all the times you multiply by 7 cancel each other out when you’re dividing.
It’s also the same problem as “One of my children is a daughter who’s 5’7” ". You don’t have to make a bunch of different categories of all the possible heights a girl could be
Does it? I’m not sure how. By “unspecified other child” I mean that they have two children, one of whom we do not know the gender.
I think I see the argument:
So if you set out to collect families with 2 children including a girl born on Thursday, you will find slightly higher numbers of two girl families because the odds they had a girl born on Thursday are better.
I’m not sure that is the right answer given the original formulation of the question, but I see the logic.
Yes, but there are only two types of families that have 2 children, one of whom is a girl
Girl, boy.
Girl, girl.
There’s twice as many families with girl/boy combinations as girl/girl cominations. Girl first then boy, and boy first then girl
(Is anyone else finding that the word “girl” starts to look odder and odder the more you type it?
girl
girl
girl
girl
I think it’s the “ir” combination that’s doing it to me…)
If I flip a coin twice, and I tell you that one of the results was heads, what are the odds that the other result was heads?
It’s not 1/3, it’s 1/2.
Same problem. Just add in the fact that I flipped one of the coins on a Thursday, and you get back to the OP.
Let’s say you have 196 parents with two children each. 49 have two boys, 49 have two girls and 98 have one of each.
When you ask the crowd “how many of you have a daughter who was born on a Thursday” Of the 49 who have two boys, 0 answer. Of the 98 who have one of each, 1/7th of them have a daughter born on Thursday, that’s 14. Of the 49 who have two girls, 7 of them have their eldest on Thurs, and 7 have their youngest on Thurs (14), but in the case where both daughters were born on Thurs, you only get one “yes”, so the total is 13.
Martin Gardner was quite well aware of ambiguous probability and logic puzzles, so if the question is due to him, it was deliberately formulated that way and many are missing the point if they think they know “the” answer. Cf. he famously asked once, if you randomly break a stick into three pieces, what is the probability they form a triangle?
This is not correct. If you said “the first flip was heads” then the answer is indeed 1/2. But if you said “one of flips was heads” then the answer is 1/3. It’s not intuitive but that’s the way the math works. Of the 4 possible 2-coin flip combinations you’ve eliminated one (TT), but the remaining three are (HT, TH, HH).
And if you say that HT and TH are the same result, that’s fine, but it occurs twice as often as HH. You get 1/3 for your result either way.
Again, it depends on how you know that one of the flips was heads. Without that information, the question is impossible to answer. And the answer that usually gets listed as “right” is counterintuitive, because it depends on an implausible way of getting the information.
Of course, to put a foot in the other camp slightly, it depends on the process by which you chose to formulate the riddle in the first place. If you flip your coins thinking “If I get a head, I’ll tell someone this riddle and see what they say”, then the two-heads probability is 1/3. If you think “Once I’ve flipped, I’ll pick one of the results I got and ask the riddle with that result” then the probability of two-the-same goes back to 1/2 (because you added an extra complication - the possibility that you might choose to say ‘I just rolled a tail, what are the chances of two tails?’ in the HT or TH cases)
They are two independent variables, the results of one coin flip do not affect the results of the other coin flip.
As others have said, it only makes a difference if I choose to only use a set where one of the results is heads.
So, if I get a bunch of people to flip coins twice, and I have the people that rolled at least one head to raise their hands, then about 1/3 of them will have two heads, two thirds will have a tail and a head.
If I simply ask one of the people what one of their flips was, then that gives me no information about the other flip, so the other flip is 50/50.
So, I guess in the OP, the gender of the child is another bullseye, as well as the day of birth. It makes a difference as to whether we hit the bullseye or whether we draw a bullseye around the dart as to whether or not it gives any information.
For instance, someone says, I have a child, what gender do you think it is? 50/50, right? If they then say, “I also have a daughter”, does that actually give you any information to add to the probability of the other one?