Riddle: Thursday's Child

Thread Games
21

Not the answer I have.

(Note: This puzzle was an “icebreaker” at a math conference in 2010)

22

I never said I picked a child at random. I only give you two pieces of information:

  1. There are two children

  2. At least one of them is a child born on a Thursday.

23

Are you looking for 1/4? That would be the odds if we knew nothing.

24

There are 4 gender birth order option possibilities of equal likelihood:

  1. Boy/Boy
  2. Boy/Girl
  3. Girl/Boy
  4. Girl/Girl

We can eliminate the first option. That means that there’s a 1/3 option that the remaining child is a girl.

This ignores twins and gender non-conforming children.

25

That would be the odds if we know nothing. But we have information:

At least one of the children is a girl born on Thursday.

26

Ok, let’s try it this way.

You have two children. The possible permutations are:

  1. 1 B, 2 G
  2. 1 B, 2 B
  3. 1 G, 2 G
  4. 1 G, 2 B

You’ve told us one is a girl. That eliminates permutation 2. Of the remaining three permutations, the other child is a boy in two of them. In only one of the permutations are both children girls.

So, the chance that both are girls is 1/3.

ETA: ninja’d by @Telemark

27

This would correctly be the odds if we knew only that at least one was a girl.

But instead we know that at least one child was a girl who was born on a Thursday.

28

NM, will need to redo the calculations when I get back from lunch.

29

1/3. Of the four possibilities (MM,MF, FM, and FF) you’ve eliminated MM, so we need to look at only the remaining possibilities.

If you had an hundred families with exactly two children, and asked “Do you have at least one girl child” 75% of the parents would say yes - and of those 150 parents, 25% of them would say yes if you then asked “Are both your children girls?”

Double-ninja’d. I’ll reconsider the Thursday factor. I don’t think it matters. If the question was “is the oldest child a girl” that would be different.

30

But the key point is that we know that at least one of the children is a girl who was born on a Thursday. That changes things.

31

Martin Gardner did this once. I don’t recall his result.

32

Yes. He is the source for this riddle.

33

I have to go to work but I am hopeful someone will post the right answer soon.

34

Not sure if I should really get involved in this, as I do remember this riddle and its answer, but it should be noted that this riddle was criticised, in that there is actually a different situation if I ask a large group of people, “Who here has 2 children, one of whom is a girl born on a Thursday?” and you simply volunteering the information, as those two sets have a different odds for the other child’s gender.

35

True, and you never said you didn’t, either. There was some process that led you to give the information you did, and you never told us what that process was. And the calculation cannot be set up without knowing that process. In order to do any calculation at all, we must make assumptions about your unspecified process.

I mean, I could just say that the probability is 13/27 that both children are girls. But to say that, I would also have to assume something that you didn’t tell us. And the thing I would be assuming to reach that conclusion is much less plausible than the thing I did assume.

36

The surgeon is his mother!

37

13/27

With one girl born on a Thursday there are 7 equally likely cases where the boy in born on any of 7 days and the girl is the first born on a Thursday.

There are also 7 equally likely cases where the girl is born second.

And there are 13 cases where one (or two) of the two girls is born on a Thursday.

In those 13 cases there are two girls. In the first two scenarios the other child is a boy.

38

Ahhh, there are 14 kids of gender-days in a faulty probability dilemma.

39

No, it’s 13/27, and we can figure that out with the info given.

There are 13 ways to have two girls, since nothing is specified about the girl born on a Thursday, such as that she is the oldest. So if there are two girls, there are 13 ways to have them: A born on a Thurs, and B born on another day, or B born on a Thurs, and A born on another day, and finally, A & B both born on a Thursday.

For the girl & boy pairs, there are 14 possibilities. Girl is child A, born on any day, while boy is born on Thurs, or girl is child b, born any day, while boy is born on Thursday.

13 + 14 = 27 total possibilities. The target possibilities are 13.

13/27

40

But it’s not just just the count of the number of ways to have that scenario. You also have to consider how we know that scenario. If @Biotop had, in fact, chosen a child at random and decided to give us that child’s gender and birth day, the information we had would look exactly the same, but the answer to the calculation would be different.

Perhaps it would help if I framed this in terms of Bayes’ Theorem. Given that “Biotop has told us that at least one child was a girl born on a Thursday”, what is the probability of “both children are girls”? To answer that, we need to know what the probability would be of “Biotop has told us that at least one child was a girl born on a Thursday”. And we don’t know that, without knowing what the full range was of things Biotop could have told us.