Friday puzzle -- Tuesday's child

Re: Friday puzzle -- Tuesday's child

Dan Donaldson

1:3

No 

Re: Friday puzzle -- Tuesday's child

Larry Barrett

2:3

One boy eliminates the G/G possibility.

Re: Friday puzzle -- Tuesday's child

Dan Donaldson

1:2 ?

Sorry, no

Re: 

Larry Barrett

I meant 1:3, agreeing with Dan's first answer (which you already said was wrong). Seems like the possibilities are B/B, B/G, G/B.

I don't see what Tuesday has to do with this.

Re: 

Alex Y

The importance of "Tuesday" surprised me as well. But it is relevant.

Re:     *LINK*

Bill Earl

"A woman stepped forward and asked,

What is the strangest day?

Tuesday, the Master replied. "


The Profit, by Kehlog Albran

Interesting read

Alex Y

I was looking for some clever financial reference, explaining why the author used "profit" instead of "prophet", but I couldn't find it in a quick skim. Did I miss something?

Hint

Alex Y

In case you missed it in my reply to Larry, the "Tuesday" part of this puzzle is relevant.

I think Bill ...

Larry Barrett

and David must have exchanged reading lists.

And Tuesday remains a mystery.

Re: Interesting read *LINK*

Bill Earl

"I am here.

I am tired.

But I will answer your questions.

Bring me food, drink, and don't forget a little gold.

A little silver for an answer.

A drachma for a doubt, a penny for a thought.

For I am The Profit and what I have learned has cost me ten lifetimes.

What you are about to learn has cost you two dollars and fifty cents. "

For those not familiar with the reference, this book was a parody of "The Prophet", by Kahlil Gibran.


The Prophet

Doh!

Alex Y

totally missed the similarity in author's name, which made parody clear.

Hint 2

Alex Y

Let Bayes be your guide.

Re: I don't know if I even need to go to bayes

David Weaver

* day of the week doesn't cause any gender bias

* know already that first child is a boy, it is given

All we need to do is figure out what the probability of having one child as a boy could be.

And that is 1/2

Did I underthink this?

..so bayes is needed, don't post the solution yet!

David Weaver

So, tuesday is of some importance, huh?

I wanted to avoid bayes (out of laziness) and this problem has a little bit of the "lets make a deal" problem flavor.

I would set it up as follows, then.

There are the following possibilities:

Two children on tuesday:

* two boys (conditioned on the known one born on tuesday)

* one boy or one girl

One child on tuesday, one child on another day:

* two boys, one born on tuesday, one not (conditioned on it being known there is a boy born on tuesday)

* one girl, one boy - boy born on tuesday, girl born on another day of the week (conditioned on it being known a boy is born on tuesday)

Be back in a minute with the workup, see how close it is to 1/2

Bayes not needed

Alex Y

I suggested it in hint 2, since it is ONE way to approach the problem, and might help see what is going on. Once you understand why Tuesday is relevant, a straight-forward approach is probably better.

Re: I don't know if I even need to go to bayes

Alex Y

* know already that first child is a boy, it is given

No, that is NOT given.

Re: Bayes not needed

David Weaver

I come up with 0.4815, with *much* chance for arithmetic issues.

The probabilities that a boy is born on a tuesday (first or second child) are based on the following:

1) First child is born on a tuesday - it's a girl, and the second child is a boy born on tuesday = (1/7)(0.5)(1/7)(0.5) = 1/196

2) first child is born on a tuesday, it's a boy - second child is a boy (any day of the week)=(1/7)(0.5)(0.5) *we like this one* = (1/28) or 7/196

3) first child is born on a tuesday, it's a boy - second child is a girl (any day of the week)=(1/7)(0.5)(0.5) = (1/28) or 7/196

4) first child is born on any other day than tuesday, it's a girl, and the second child is a boy born on tuesday = (6/7)(0.5)(1/7)(0.5) = 6/196

5) first child is born on any other day, it's a boy, and the second is a boy is born on tuesday (6/7)(0.5)(1/7)(0.5) *we like this one, too*=6/196

So my workup is:

{P(2) + P(5)} / {p(1) + p(2) + p(3) + p(4)+p(5)} = (13/196) / (27/196)

=13/27

Correct! *LINK*

Alex Y

Kudos to David on this tough one.

I'll suggest another way to see it:

If we are given only gender information, then there are two possiblities for the description of each child, so there are 2^2 = 4 family possible family compositions. If we are told that one of the children is a boy, we have eliminated the the G/G possibility, leaving a 1/3 chance that they are both boys, the answer that both Dan and Larry got.

But we are given information about both gender and birth day of week. Using that information, each child could be one of 14 possibilities: BSu, BM, BTu, BW,... BSa, GSu, ... GSa, and the family could be be one of 14^2 possibilities. We are told that there is [at least] one boy born on Tuesday. That reduces the number of cases from 196 to 27: (BSu, BTu), (BM, BTu), ... (GSa, BTu), (BTu,BSu),... (not double-counting (BTu, BTu)). Of those 27 equally probable cases, 13 are families with two boys.

Graphically, the following picture shows both problems: The one where you are given only that one child is a boy lets you eliminate 1/4 of the cases (the pink square), while the Tuesday boy information lets you eliminate all but the bolded squares.

Here is a good write-up (from which I borrowed the graphical concept above) on the problem, including some other twists that I found of interest.


Some Thoughts on Tuesday's Child

Re: Correct!

David Weaver

Excellent question! Thanks for the heads up on it being here.

Haven't done these types of questions (other than on here) since about 1997, and they always were my favorites in college.

Have to really knock out some cobwebs every time a good one comes along.

Decimal answer came up first because I set up a matrix in excel for a check, and then did the fractional answer as a second check. It's easy to make a mistake or miss a possible combination if you don't set up something that encompasses all possible cases first (like the chart you showed) and prove total probability = 1.

Couldn't resist a lazy first shot at answering it without doing any work, though. Never can resist, unfortunately, laziness is my master (or maybe we can call it economical allocation of effort).

Re: Correct!

Larry Barrett

So if I modify the problem to ask "do you have a son born before noon on Tues, and you answer yes, what is the probability that you have a second son", I can modify David's equations to include 14 parts of the week (Su-Sa, before noon, after noon) and conclude that the prob = 27/55 = .4909.

So I conclude that by being more and more specific about when one son was born, I increase the probability of have a second son approaching .5.

I can't argue about the arithmetic, but this sure does not seem intuitive.

discussing on a completely non-intuitive basis

David Weaver

..hold on.. .have to go copy the equations, as intuition can be misleading....

OK...if you replace all of the 7s with n, so you can take a limit, and ignore the 0.5s (they are in every term, so they can be disregarded)

you get something that simplifies to (2n-1)/(4n-1)

Check that with the 7s, and you still get the right result - 13/27 - with the original numbers.

You can see as you take the limit N-infinity (haven't done limits in 14 years, someone correct me if I'm wrong) that this tends toward 2/4 or 1/2.

of course, taking the limit as N approaches infinity is the case where the exactness of the birth time gets more and more specific (or smaller).