Friday puzzle -- Tuesday's child

Re: discussing on a completely non-intuitive basis

Larry Barrett

I agree with your math, but this result does not seem at all intuitive to me. Why does being more specific about when one son was born increase the probability that the other child is a son.

I did not look at the link that Alex supplied; maybe that sheds some light on this.

Re: discussing on a completely non-intuitive basis

David Weaver

When the period gets shorter, the chance that the second child is born in a different period gets greater. When the point that the male child is born is chosen instead of an interval (this is the reason "no twins" is specified, i guess), then the chance that the second child is born in a separate interval

That means that you can toss out the chance that the male is the second child where two are born in the same period.

Then that tends toward a situation where you know there is only one child in the interval (which is a point in the limit scenario) and the only case that's relevant is the one where the first child *must* be male.

That case is my intellectually lazy first answer - 1/2.

An intuitive look

Alex Y

Larry, I feel your pain!

Think of it this way: As you get more specific about your questioning, you get closer to fully specifying one child, leaving the other at the basic 50/50 split. If you ask if one of George's children was a left-handed blond boy born on June 1, 1995, who is now 6'2", and he answered "yes", surely you would agree that the probability that his other child was a boy is 50%.

I do think that link will help.

When I originally heard this puzzle, I heard the version displayed at the top of that link, and answered "1/2", with the rationale that, lacking other information, I could only assume that it was puzzle 3. The version I posted here is puzzle 1 in that link.

I thought the dartboard discussion at the bottom was helpful.

Re: more non-theoretical description

David Weaver

If you map the function that I put up where n is very very small (i.e., the period is approaching a week), then you see the chance that there are two boys is 1/3 knowing that there is at least one boy, but not knowing when that boy was born in the order of children.

the other end is the instance I just described, where the chance that both children were born in the same period (and if you know both were and one is a boy, the chance both are boys is 1/3rd) becomes a smaller and smaller. You know the chance if they're born in separate periods is 1/2 since the first one must be a boy.

Therefore, the shorter the interval, the greater the chance the boy must be the child in the interval and the greater the chance the other child is a boy.

In college, this was the most static class..

David Weaver

.. and by that, I mean that it seemed to be the class that people either got the material or they didn't. Within two weeks, you could tell who wouldn't get it. People didn't go from "don't get it" to "get it" status like they might in a calculus class or something, or a class where you write proofs, but it is college these days and they pass everyone, anyway, as long as you show up.

So the folks who "got it", got As. The class was pretty easy, but like this problem, you still always had to be rigorous to in making sure you covered all possible cases so that you didn't knock yourself out of the right answer because of laziness.

The folks who didn't get it but tried really hard and went to office hours got Bs.

The folks who didn't get it and did their assignments wrong and handed them all in got Cs, maybe Ds if they were really slow...

and since it was a 400 level class, nobody got Fs unless they didn't come to class, I guess. That's how it works these days. Nobody wants to pay a college to get Fs, though I had some professors in other classes that loved to give Fs and keep students guessing until the end of the semester when they adjusted for a curve. I liked those classes, they were fun. I recall one where I had a 64% and nobody else was within 12% of that. I knew I would get an A. I could be lazy and just make sure I didn't let anyone else get ahead of me, but otherwise not push too hard.

The ones where you didn't know your standing in the class - that was a little less fun. Had a class one semester where I got a 41% and an A-, and I didn't know the distribution of scores. I was a little more anxious about that one.

But probability and statistics (as it was called) wasn't one of those. It was a "student friendly" class where no curve was needed.

(and that's not to suggest there weren't plenty of classes where I "never got it". Object oriented programming was one, one of the continuous mathematics proof classes, and art history was another...woof)

And one more

Alex Y

Census taker A asks a parent how many children he has and finds out there are two. "is at least one a boy" "yes". Probability of two boys =1/3.

Census taker B has the door answered by a boy and asks "do you have any siblings" "yes, one". Probability the other is a boy =1/2.

in both cases, you know that one of the children is a boy, but in the second case, you know that a specific child (the one who answered the door) is a boy.

In our case of a boy born on Tuesday, we are being more specific than "at lest one child is a boy", but less specific than "the child who answered the door is a boy"

Bayesian solution

Alex Y

I think the solution in the link I provided is clearer, but here's the same solution derived with the help of Bayes:

We want to find P(A|B) where

A is "Both children in the family are boys" and

B is "At least one of the two children in the family is a boy born on a Tuesday"

Bayes tells us that

P(A|B) = P(B|A)*P(A)/P(B)

P(A) = 1/4, I think we all would agree.

P(B|A) = 13/49: There are two boys. The probability that the first is born on Tuesday is 1/7. Same for the second. Add those two together, less the probability that they are both born on Tuesday = 1/& + 1/7 - 1/49 = 13/49.

P(B) = 27/196: This is the tough one. The probability that the oldest kid is a boy born on a Tuesday is 1/2 (for the "boy" part) * 1/7 (the "Tuesday" part)= 1/14. Same for the youngest child. So the probability that at least one of these meet the condition is 1/14 + 1/14 - (1/14)^2 = 27/196.

So back to the original formula, P(A/B) = (13/49) * (1/4) / (27/196) = 13/27.

Re: And one more

Larry Barrett

The link is quite interesting. Thanks for providing it.

This is starting to make more sense; just need to get my mind to bend a little.

A friend of mine used to say about things hard to understand - "it gives me a charley-horse in my brain".

Re: And one more

Dan Donaldson

I get a major charlie horse in mine Probability and statistics were never my strong suit.