Puzzle: I have two children ...

[Santino]

I actually had useful work-related things to do this morning. Extraordinary, I know. But I do occassionally have to make a token appearance. So I didn't spot it until now.

I'll be busy this afternoon too.

You used to be cool, man.

 

[kabbes]

I've done my work things now. That's me done for another week -- phew!

 

[Santino]

So, why does it SEEM as if I could affect the odds of someone having a girl by asking them what day their son was born?

 

[fractionMan]

it's a bit like the three doors thing then.

 

[kabbes]

So, why does it SEEM as if I could affect the odds of someone having a girl by asking them what day their son was born?

Because of philosophy.

 

[Santino]

Deep philosophy?

 

[mrsfran]

I think a geek of this magnitude has about a 0% probability of having sex, let alone children.

And yet the first person in the thread to answer the question correctly is going to be a dad soon. Some women appreciate smarts.

 

[fractionMan]

 

[kabbes]

Deep philosophy?

You know it.

The problem is a subtle one -- possibly too subtle even for my great mind. But primarily it's because probability is NOT about the actual odds of the universe. The universe is what it is, regardless of what our puny intelligence might say about it.

No, probability isn't about what state the universe is in. Instead, probability is about our knowledge of what state the universe is in. And that means that any piece of knowledge we factor in can have a surprising effect. Because it's about what we know, so if we know new stuff then it changes what we can say about what we know.

 

[Gromit]

Did anyone ask "Who the fuck cares?"

or exclaim

"Typical Patriachial society worrying about how probabilities of getting a boy, no doubt they'll be disappointed if its a girl who can't go on to be England Captain etc."

 

[mrsfran]

Did anyone ask "Who the fuck cares?"

or exclaim

"Typical Patriachial society worrying about how probabilities of getting a boy, no doubt they'll be disappointed if its a girl who can't go on to be England Captain etc."

Women's sports team have captains too. Racist.

 

[bi0boy]

Well, I know their silly example is wrong, so I presume the rest of it is also.

Let's look at their simplified example.

So, consider this preliminary question: "I have two children. One of them is a boy. What is the probability I have two boys?"

To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3.

They claim that if we eliminate GG the three remaining options are BG, GB and BB so that "the probability of the two boys is 1/3".

This is wrong. You need to eliminate one of either GB or BG also, since only one child can possibly be a girl. This gives odds of 1/2 for the second being a boy.

 

[Gromit]

Women's sports team have captains too. Racist.

Are you sure?

I thought they just had a male coach to explained to them patiently what they have to do, repeating lots just to be sure they get it.

Being careful not to shout it case they start to cry.

Then spying on them in the changing rooms later whilst they soap each other whilst snogging in the showers.

Please don't tell me that I've gotten it all wrong as I've just applied to manage Arsenal's Ladies team.

 

[kabbes]

Let's look at our state of knowledge, incidentally.

let's say that B1 is a boy born on the first day of the week, and so on. And we describe two children thus: B1_G3, for a boy born on Sunday followed by a girl born on Tuesday.

You can envisage your two-dimensional matrix running like so:

B1_B1, B1_B2, B1_B3, ... , B1_G5, B1_G6, B1_G7
B2_B1, B2_B2,        ...        , B2_G6, B2_G7
...
G6_B1, G6_B2,        ...        , G6_G6, G6_G7
G7_B1, G7_B2,        ...        , G7_G6, G7_G7

So when you know that one child is a boy, you knock out the bottom-right of the grid.
And when you know the day of the week of the same child, you then knock out all remaining rows and columns except for the critical row and column.

OK, but do it the other way around -- suppose you know the day of the week first, which means that you knock out the rows and columns first. At this point, you have 52 remaining cells.

You're only interested in the cells that have two boys, which are the ones remaining in the top-left of the grid. That's 13 cells.

So you have a 13/52 = 0.25 probability of two boys.

So the day of the week hasn't affected the probability of two boys at all!

But now we bring in the knowledge that one of them is a boy. That makes us knock out the bottom-right of the grid.

That leaves us with three identical cross-shapes in each of the three remaining quadrants. The probability of two boys is now 1/3. So the day of the week STILL hasn't affected anything!

But now we bring in the fact that the knowledge of the day of the week is in respect of the same child as the knowledge of the sex.

At this point we knock out an extra column and an extra row. Only now does the day of the week have an effect. So it's the convolution of the two facts that is the key.

 

[Santino]

This is a trick, right?

 

[kabbes]

Hang on, I think I know how to explain it.

 

[kabbes]

I just scrapped everything I wrote to replace it with this sentence:

When you find out that the boy was born on a Tuesday, you lose more second-child girl-options than second-child boy-options.

That's why the probability shifts.

And why do you lose more girl-options? Because you had more of them to begin with, because you were already down to a 1/3 to 2/3 ratio!

(I think that another way of looking at it is this: the different combinations for day of the week overwhelms the different combinations for boys and girls. But that might be unhelpful.)

 

[Santino]

Can we do this with born-in-summer and born-in-winter, assuming they are the only two seasons?

 

[beesonthewhatnow]

They claim that if we eliminate GG the three remaining options are BG, GB and BB so that "the probability of the two boys is 1/3".

This is wrong

No it's not.

 

[kabbes]

Can we do this with born-in-summer and born-in-winter, assuming they are the only two seasons?

Sure, here's your matrix:

[COLOR="Red"][B]B1_B1	B1_B2[/B]	B1_G1	B1_G2[/COLOR]
[COLOR="red"][B]B2_B1[/B][/COLOR]	B2_B2	B2_G1	B2_G2
[COLOR="red"]G1_B1[/COLOR]	G1_B2	G1_G1	G1_G2
[COLOR="red"]G2_B1[/COLOR]	G2_B2	G2_G1	G2_G2

Assuming a boy born in winter (winter=1), we keep the ones shaded red.

The ones that also have two boys are in bold.

There are three in bold red, out of 7 red in total, so the probability is 3/7.

 

[strung out]

and if we widen the possibilities even further (say, their birthday is on the 1st january), the probability gets closer and closer to 1/2, yes?

 

[kabbes]

So first of all we knocked out the bottom-right quadrant.

At this point, we have 4 boy-boys and 8 boy-girls. (1/3 and 2/3)

When we knock out the next 5 options, we lose more boy-girls than boy-boys because there are more boy-girls to lose! We actually lose four of them, compared with one boy-boy, so the odds shift back towards the 50/50 that they started from.

 

[kabbes]

and if we widen the possibilities even further (say, their birthday is on the 1st january), the probability gets closer and closer to 1/2, yes?

Yes, should be.

I think it's all because the starting probability is 50/50. To get to the 1:2 ratio, we had to knock out options in unequal quantities. More knowledge will then knock out more options, but the effect won't be symmetrical because we're no longer starting from a symmetrical distribution of options.

 

[littlebabyjesus]

I actually had useful work-related things to do this morning. Extraordinary, I know. But I do occassionally have to make a token appearance. So I didn't spot it until now.

I'll be busy this afternoon too.

Anyway, this is a favourite stats question of old. In fact, it is so well known that it is called the "boy-girl paradox". The day of the week thing just makes it more complicated but the principle is the same.

Firstly, just consider the boys and girls. Then you'll get the principle.

Two children can be BB, BG, GB and GG. So if I say that the eldest is a boy then you are left with BB and BG, which means that the younger is 50/50 for boy or girl.

But two children given only that you know that one is a boy can only be BB, BG and GB. Two of those have a girl as the second option, meaning that the probability that the other child is a girl is 2/3 (and its complement, which is two boys, is 1/3).

This question just takes the same approach and convolutes it with an equivalent question involving days of the week. It appears much more complicated because you have two dimensions, but it is the same principle at work.

This is wrong.

The two children are in two positions – the one whose sex I reveal and the one whose sex I do not reveal.

If you take 'the one whose sex I reveal' to be in position 1, then you are left with just two possibilities: BB and BG. The probability that the unstated child is a boy is 50/50.

 

[strung out]

no!

 

[kabbes]

How is that any different to what I said?

 

[kabbes]

No, I checked again. It's definitely not any different to what I said:

Two children can be BB, BG, GB and GG. So if I say that the eldest is a boy then you are left with BB and BG, which means that the younger is 50/50 for boy or girl.

 

[littlebabyjesus]

How is that any different to what I said?

You have not allowed for the fact that the declared child now has a fixed position. It is in position 1 now. Therefore, the chance is 50/50

This logic extends to the Tuesday example too. The chance is also 50/50 there. Any other answer is mistaken as to what you know at the outset.

Declaring one is equivalent to your saying youngest/eldest – there is now an order to them: declared; undeclared

 

[strung out]

no, you're wrong

 

[kabbes]

You have not allowed for the fact that the declared child now has a fixed position. It is in position 1 now. Therefore, the chance is 50/50

I suggest that you read again what I wrote!