[Closed] The boy-girl puzzle

I'm agreeing on it being 50% and I'm also agreeing that those extra brackets are redundant...BIDMAS ordering - brackets, indices, division, multiplication, addition, subtraction

n(2+d(n-1))/2

equates to n(2+dn-d)/2
which is the same as what n(2+(d(n-1)))/2 equates to...n(2+(dn-d))/2

Go on yerself chvck, if you think it's 50% (despite the evidence) then let's see your working.

If i have a drawer with an infinite pair of socks, some blue, some red then if I pull a pair of socks out of the drawer then the probability of being blue first time is 50%, the probability of being blue the second time is 50%. I'm using infinite pairs of socks here as if I pull out a sock from a non-infinite drawer then the number of socks will go down. There is no quantity of boys and girls that exist can be given birth to....it's equal probability and if a girl is the first child then it doesnt reduce the probability of having a girl the next time.

Basically the point that I am trying to labour is that the first birth has absolutely no effect on the second birth meaning that they do not affect the probabilities of each other at all.

The probability of the first birth being a girl was 50% and so the probability of the second birth being a girl is 50%..they are independant probabilities.

I can see how the 66% comes about and it does make perfect sense to me but I do think that boy/girl is effectively the same as girl/boy...you can have a girl and a boy or a boy and a girl, it's the same thing

Anyway I've lost count of the number of times I've repeated myself in this and whether or not any of it is followable and it's too late to care anymore!

Yep socks works too.

Pick two pairs of your socks at random.
Tell me if you have at least one blue pair.
If you do then there is a 66.6% chance that your other pair is red.

Right, using the logic behind the 66% arguement you have to eliminate one of the possible events before calculating the probability

The probability of two independant events is found by multiplying the probabilities together

boy/girl=0.5*0.5=0.25
girl/boy=0.5*0.5=0.25
girl/girl=0.5*0.5=0.25
boy/boy=0.5*0.5=0.25

At this point we can see that girl and boy = 0.25 + 0.25 = 0.5, just because we already know that one is a girl doesn't mean that we can manipulate the probabilities, we can't remove events until after the calculations.

Also - look at it like this if I pick a blue sock then the probability is 50%, then my mate comes along and picks a sock...the colour of the sock that I picked is not going to influence the colour of sock that my mate picks out is it? Surely that's just logical!

note: I do see what you mean and tbh, it's hard for me to judge which is correct as I can actually see them both working depending on how you look at it

Indeed, the events (regardless of socks or child gender) are independent. No one is saying otherwise.

But your own figures contain the answer:

girl and boy = 0.25 + 0.25 = 0.5

girl/girl = 0.25

So a girl and boy is twice as likely as a girl/girl.

OK, I don't actually know anymore, I've royally confused the heck out of myself, I don't know which one is right as they both make sense to me depending on how I look at it - I daresay I could well be wrong. In my brain the logic and the numbers aren't fully agreeing! Either way it's certainly an interesting debate but my face hurts so I'm going to bed!

chvck - Member

OK, I don't actually know anymore, I've royally confused the heck out of myself, I don't know which one is right as they both make sense to me depending on how I look at it - I daresay I could well be wrong. In my brain the logic and the numbers aren't fully agreeing! Either way it's certainly an interesting debate but my face hurts so I'm going to bed!

Easy to do in fairness, but as was pointed out to me originally, we are talking about events that have already happened, not an event that is going to happen (in which case it would be 50/50).

Given that the woman had 2 children, there are 4 possible outcomes, of that we are all agreed.

Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl

It needs to be noted that Boy/Girl is NOT the same as Girl/Boy as these events have already happened. So given as stated, at least one of them is a girl, you can remove the Boy/Boy from the equation and take that there is 2/3 chance of there being a boy and a girl as there were 2 separate opportunities for a combination of the 2 to have happened, whereas there was only 1 opportunity for there to be 2 girls.

Talking about events that are going to happen in the future is totally different though.

Regarding the extra brackets, they DO make a difference.

If we say n=10 and d=5 then if you don't have the extra brackets in n(2+d(n-1))/2 you will get:

10x(2+5(10-1))/2 = 10x(7x9)/2 = 630/2 = 315 (INCORRECT)

Using the extra brackets we get:

10x(2+(5x(10-1)))/2 = 10x(2+45)/2 = 470/2 = 235 (CORRECT)

Sorry to be a pedant 😉

my mum is one of eight kids. she has one brother.

of those eight seven have had children. i've got 23 cousins. only three of us are boys.

of those 23 cousins 4 (all girls) have had children. 4 girls and 2 boys.

christmases used to be fun.

think i best get used to the idea of having daughters.....

Class.
Probability meets statistics again.

Sock drawer thing - that's probability.
Sex of child - that's statistics.
All that shiat on page 2 and most of page 3 of this post, no idea, skipped it.

10x(2+5(10-1))/2 = 10x(7x9)/2 = 630/2 = 315 (INCORRECT)

Because you've done it wrong.

10(2+5(10-1))/2 = 10(2+5(9))/2 = 10(2+45)/2 = 470/2 = 235

BODMAS - you have to multiply the 5 by 10-1 [u]before[/u] you add it to the 2

Sock drawer thing - that's probability.
Sex of child - that's statistics.

The sex of the child is probability when expressed as it is in this question. The could replace the child being a boy or a girl with any past event which has a 50/50 outcome:

You ask a woman how many coins she has just tossed and she says two.
Then for some odd reason you ask her "Is at least one a head?", to which she replies "Yes". So what are the odds that she has tossed a head and a tail?

It is still 50:50 for the chances of it being boy and a girl.

You know there is one girl:

So that must surely remove boy/boy and halve the probability of both boy/girl or girl/boy.

100% in my case - boy, boy, boy.

The x must be strong in this one.... 🙂

It is still 50:50 for the chances of it being boy and a girl.

You know there is one girl:

So that must surely remove boy/boy and halve the probability of both boy/girl or girl/boy.

Why would it half the probability of boy/girl and girl/boy?

I just tossed a coin twice. One of the times was a head. Does that change the 50/50 chance of getting a head on the other toss?

This thread is illustrating how peer pressure can make people believe errant nonsense.

This thread is illustrating how some people don't understand probability.

😉

It would halve the probability because you cannot have both of them - you know that one is a girl so you get to choose one of boy/girl or girl/boy - not both.

I understand probability perfectly well. Many years of doing engineering maths at uni has seen to that.

It would halve the probability because you cannot have both of them - you know that one is a girl so you get to choose one of boy/girl or girl/boy - not both.

The girl you know about could be the oldest or the youngest, so girl/boy and boy/girl are both still possible outcomes, as is girl/girl. The only outcome you know didn't happen is boy/boy.

I understand probability perfectly well. Many years of doing engineering maths at uni has seen to that.

Any bridges or machines I should keep away from?

I get where the flawed solution comes from, just dont agree that it should be implemented here.

Two coin tosses:

Chance at the beginning that you'll get two of the same side coming up = 50:50.
After the first coin toss chances of you getting a matching one = 50:50.

Why does the child scenario differ?

Its the old gamblers fallicy thing....

Okay Smee answer me this:
[b]with two children what is the probability of mixed sex versus same sex siblings?[/b] *

(*assuming of course that we are still operating in this nice predictable universe where, unlike reality, the probability of a girl or a boy being born is exactly 50%)

[i]Two coin tosses:

Chance at the beginning that you'll get two of the same side coming up = 50:50.
After the first coin toss chances of you getting a matching one = 50:50.[/i]

Your a mile out there.

if you don't know the sex of either then it is 50:50.

Drac - why a mile out?

before first throw your options are:
head/head
tail/tail
head/tail
tail/head

say you get a tail on the first throw your options are now limited to:
tail/tail
tail/head.

and what are the chances of each outcome?

0.25 chance of getting HH
0.25 chance of getting HT
0.25 chance of getting TH
0.25 chance of getting TT

So:

0.25 chance of getting no tails
0.5 chance of getting one tail
0.25 chance of getting two tails

So, you are twice as likely to get one tail as to get two tails.

[img] [/img]
[url= http://www.flickr.com/photos/mike_mc/3233045355/ ]http://www.flickr.com/photos/mike_mc/3233045355/[/url]

The other child is either a boy, a girl or a girl…

So, there's 2 in 3 chance of it being a girl.

mike - simply repeating something over and over again doesn't make it correct.

which one of the boy/girl options are you going for on your tree? you cant have them both. your girl must always be a girl she cant change sexes.

mike - read what you've just written.

So the other child is either a boy, a girl or a girl. Surely a girl is the same thing as a girl.

Why can I not have both girl options? Is the girl you know about the eldest child or the youngest?

She is certainly one of them.

[img] [/img]
[url] http://www.flickr.com/photos/mike_mc/3233062281/ [/url]

She is certainly one of them

Yes, but she [i]could[/i] be [i]either[/i] of them.

and certainly not the other one.

She could be either, but she has to be one of them - she can't be both.

She does have to be one of them. But there's a [b]probability[/b] that she's either 🙂

If she's the youngest, there's a 50:50 chance the other child is a boy. If she's the eldest, there's a 50:50 chance the other child is a boy.

Just like in my second diagram…

There is a 100% probability that she is not both of them though.

with two children what is the probability of mixed sex versus same sex siblings?

if you don't know the sex of either then it is 50:50.

Correct.

So if you now know that at least one child is a girl then you halve the probability of "same sex" as you know that boy/boy combination is not possible. So "mixed sex" versus "both girls" is 50:25

If she's the youngest that removes the option of her being the oldest and vice versa.

[i]Drac - why a mile out?[/i]

As explained by Mike, it's 2 coins so it's not as simple as H/H T/H T/T

It's

0.25 chance of getting HH
0.25 chance of getting HT
0.25 chance of getting TH
0.25 chance of getting TT

Took me all day to grasp yesterday I did get a little help from a friend also called Graham S who is rather exceptional at maths. He used various ways to explain it but it was the equation on here that made me see the answer.

Drac - go back to sleep.

[i]Drac - go back to sleep. [/i]

That won't make you right.

Ah ha, Smee has just unravelled the mess in my mind, I'm still with the 50% option.

As, if you get a girl first then you have the probabilities of

girl/boy
girl/girl

if you have a girl second then you have

boy/girl
girl/girl

in both scenario's the probability of having a girl and a boy is 50%....as Smee just pointed out you can't stick boy/girl and girl/boy together in this context and the two events are still entirely independent

God damn it I pasted the wrong bit, ok I'm for a nap.

Crikey,
Couldn't help but put my oar in here. To all those clinging on to 50% probability. The final order girl/boy boy/girl is indeed irrelevant the result is counted as the same. However there are two routes to get there. There is only one possible way to get a girl/girl combo. Hence the 2:1 ratio. Those tree diagrams posted earlier explain it all perfectly. Mind you I think it was explained perfectly on about the fourth post so what do I know.

I love these threads. It's better when the people who are wrong tell us how much maths they've done, or how their mum is a midwife so they must be right.

Good God it's catching...

Right.. yes, we all agree that the BIRTHS of these children are entirely independent events where the gender is 50:50 between boy and girl - just like tossing a coin or pulling a sock from an infinite drawer. That's fine.

But in this problem those events have already happened. The woman in the question already has two children and [i]without any extra information[/i] we know that the probabilities are:

all boys = 0.25, all girls = 0.25, mixed = 0.50

Hopefully we are all agreed on that point (if not then read back a bit as it has been shown multiple times)

We are then given additional information that allows us to exclude the all boys possibility, leaving all girls = 0.25 and mixed = 0.50

it also allows you to exclude one of the mixed sex options too. if you disagree tell me why.

Smee, how is that the case?
'At least one girl' accounts for G/B B/G surely? That includes both possible combinations of a mixed sex pair of kiddies?

Isn't that the crux of this whole thing?

So, basically what you're saying is that if you have no previous knowledge about genders but knowing that someone has two kids then the probabilities are

Having a boy and a girl - 50%
Having 2 girls - 25%
Having 2 boys - 25%

But yet if I wasn't to ask how many kids that they have but instead ask what is the gender of your first born (if you have one)

Having a boy - 50%
Having a girl - 50%

I then ask what is the gender of your second child (if you have one)

Having a boy - 50%
Having a girl - 50%

I see where you're coming from but I don't agree, if I have a daughter then there's no reason why my second born is more likely to be a boy

it also allows you to exclude one of the mixed sex options too. if you disagree tell me why.

Because both are still possible.

Wiredchops - you know that one IS a girl, so you cant have both options. It must either be one or the other, it cant be both.

Mike - nonsense. see above.

Sorry smee, that reasoning is entirely wrong. If you know one is a girl it clearly includes both options as both contain a girl. The only one it excludes is two boys, as neither is a girl?

B/G, G/B are two distinct possibilities and both meet the criteria of 'at least one girl'

[i]I then ask what is the gender of your second child (if you have one)[/i]

That's not the same question.

Wiredchops - you know that one IS a girl, so you cant have both options. It must either be one or the other, it cant be both.

It can't be both, but being either one is possible. How possible? 1 in 3...

Apart from repeating what has already been said, I can't see another way of changing your mind. The proof is all here, and it's been explained several times.

(Am I right in thinking Smee is Glupton? If so, there's a 100% chance that convincing him he's wrong is impossible.)

Yup Smee is Glupton.

miketually, I remember reasing something about the monty hall program and big arguments ensuing about it. One person was so massively convinced that he made some gigantic spreadsheet with 10,000 trials or something, relishing the thought of smashing it into the face of the smug maths guy. Only to see the probability drop out at 66%.

Then began the fiery e-mails to microsoft regarding the massive bugs in Excel....

Sometimes you just have to walk away.

If you choose B/G - how can you possibly also include G/B?

Smee,
Ok look at it this way, (I'm waiting on stuff to process on my PC so have time to kill)

There are three distinct outcomes, all boys, all girls, mixed.

For the last option (mixed) there are two distinct paths to this outcome.
G/B, B/G

You are choosing only one outcome (mixed)
This inherently includes the two 'possible' routes

one outcome is excluded completely (all boys)

the remaining outcome (all girls) has only one possible route to it, G/G

therefore, of the three possible routes, two of these lead to the mixed outcome.

You have a 2/3 probability of ending up with mixed kids. That's assuming your attitude to inter-racial relationships is open. (Guffaw, ignore that last awful quip)

If you choose B/G - how can you possibly also include G/B?

Who's choosing anything? BG is one possibility. GB is another possibility. GG is a third possibilty. You don't choose anything, you just state how likely each is to happen.

As a side note. Get the book by the dude who does the puzzles in the saturday guardian, Chris Maslanka. He does these awesome puzzles which you hack away at with algebra, probability and all this other foofurah only to flick to the answer and see that he's explained it eloquently in a single line. Awesome stuff. He has books of puzzles out, recommend.

[i]If you choose B/G - how can you possibly also include G/B? [/i]

It's the lack of specifics that keep the 2 options.

Have to go into work or I'd give you analogy that helped me see why but I'd be wasting my time with you as Glupton is never wrong.

miketually, I remember reasing something about the monty hall program and big arguments ensuing about it. One person was so massively convinced that he made some gigantic spreadsheet with 10,000 trials or something, relishing the thought of smashing it into the face of the smug maths guy. Only to see the probability drop out at 66%.

Then began the fiery e-mails to microsoft regarding the massive bugs in Excel....

A guy from here (BenKinetics, I think) took quite a lot of money off someone who was convinced that it made no difference whether you switched or not on the Month Hall problem, so he did it for real with three cups and some pound coins 🙂

If I look at this logically then I get 50%, if I look at it using numbers then I agree 66%....so basically I'm going to give up and revise for my exam tomorrow!....:(

Just for reference, can everyone give their answers to the various problems discussed here recently? I'll go first:

Plane On Conveyor: It takes off.
Monty Hall: Always switch doors.
Boy-girl: 2 in 3 chance it's one of each.

I see where you're coming from but I don't agree, if I have a daughter then there's no reason why my second born is more likely to be a boy

Correct. Those are independent events - as discussed.

But the case where you have a daughter and you are waiting for a second born is entirely different to knowing that one of the children is a daughter.

If the question was:

You ask a woman how many children she has and she says two.
Then for some odd reason you ask her "Is [b]the eldest one[/b] a girl?", to which she replies "Yes".

Then it would be 50:50

But it only asks [b]is either of them a girl[/b].

As I have said previously I understand how you are getting 66%, I simply dont agree that the method you are using is correct in this scenario.

You know you have one girl , that alone removes the options that start with a boy. End of thread.

take off
Switch
66%

smee, no, you know you have one girl AFTER she's had both kids so the second child could be a girl as much as the first.

You know you have one girl , that alone removes the options that start with a boy.

But the option that begins with a boy and ends with a girl (stop s****ing at the back) is still valid, if the girl you know about is the youngest daughter.

What you're saying here is only true if the woman tells you her eldest child is a daughter.

Ok, I think that the way I've been approaching this has been flawed and I see where my logic has gone tits up!

Now listen here Mr. Smee..

[b]Here is a spreadsheet (oh yes) showing a sample of 100 parents with two children:

http://spreadsheets.google.com/ccc?key=p_H5o2Sep3PNxPhB1Cjb-Dg [/b]

Now tell me what the probability is of a parent on that spreadsheet having mixed sex siblings?

GrahamS, not heard owt from him for a good old while, I reckon he's discovered he's wrong and scuttled off quietly (yes, this is a blatant lure)

But but.. I spent ages on that spreadsheet 😯

Hopefully he has realised the error of his ways, though his "End of thread" statement suggests he has instead run off crying because he's confused and scared by what the bigger boys are saying. (blatant lure number two)

(yes, this is a blatant lure)

(blatant lure number two)

Do we need to say his name three times in front of a mirror?

[i]Ok, I think that the way I've been approaching this has been flawed and I see where my logic has gone tits up! [/i]

Yup that's where I was going wrong too, until I realised as the order was never discussed it still leaves the B/G G/B option open.