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(7×9)/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(7×9)/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 before 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:
with two children what is the probability of mixed sex versus same sex siblings? *

(*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%)

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.

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.

http://www.flickr.com/photos/mike_mc/3233045355/%5B/url%5D

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.

http://www.flickr.com/photos/mike_mc/3233062281/

She is certainly one of them

Yes, but she could be either 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 probability 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.