[Closed] The boy-girl puzzle

Oh - I get it now.

It is still 50:50.

What...? But but.... look at the spreadsheet... 😕

The probability of Smee making someone eat his dust, attacking someone for helping his child or refusing to admit he's wrong is 100%

Analogy I was given and roughly what was used on here.

If I tossed the coins, and hid them from you (one in each hand), and
said "at least one is a head", then you would say "OK, so I know that
one is a head, doesn't matter which one - I'll assume it's the one in
his left hand. So all that I don't know is what's in his right hand -
50/50 chance. Easy.". But that's not right. It's not obviously wrong,
but it *is* wrong. You could analyse it another way:

"What's in his left hand? Could be a head or a tail - I don't know.
Suppose it's a head, then there is an equal chance that his right hand
contains a head or a tail. So - *assuming* his left hand has a head,
then there we have two equally likely outcomes. Suppose his left hand
actually has a tail. Then his right hand *must* contain a head because
we know there's at least one head. There's a third outcome. Since the
coins were tossed fairly and randomly assigned to each hand, those
three outcomes are equally likely, and we're back to our 1/3, 2/3
probability.

Why are you all talking about probabilities when the question asks about odds?

Odds are 2:1 - which is not the same as a probability of 66%

Awwww! Bless look he's all confused.

Drac - show me where in the preceding 160 odd posts that someone has said that the odds are 2:1 in favour of there being one child of each gender.

2:1 is a ratio, which can also be expressed as a fraction or a percentage.

Why are you hiding behind picking tiny faults, rather than admitting being wrong?

Awwww! Bless look he's all confused.

It's the dust. It affects his thinking.

I would suggest that it is a major fault, not giving the answer in the correct format.

Just think of it like this: you go into the bank and ask them for £1000 and they give you it in $ - would you be happy?

I wouldn't care if they gave me it in £20 notes or £50 notes though, which is closer than $s and £s.

Or are you going to claim is was the use of percentages that confused you?

Why are you all talking about probabilities when the question asks about odds?

Ahhh the old [i]"try and wriggle out of it on a technicality" defence[/i]

[b][url= http://en.wikipedia.org/wiki/Odds ]From Wikipedia[/url]:[/b]
In probability theory and statistics the odds in favour of an event or a proposition are the quantity p / (1 ? p), where p is the probability of the event or proposition. The odds against the same event are (1 ? p) / p. For example, if you chose a random day of the week, then the odds that you would choose a Sunday would be 1/6, not 1/7. The odds against you choosing Sunday are 6/1. These 'odds' are actually relative probabilities.

So if we are saying that we think that the probability (expressed as a percentage) is 66.66..% then [i]p[/i] would be 0.666..

Expressed as [i]odds[/i] that would be

p / (1 ? p)
= 0.666.. / (1 - 0.666..)
= 0.666.. / 0.333..
= 2 / 1

So yes, odds of 2/1 are [u]exactly[/u] the same as a probability of 66.66..%

Fact remains - I am the first person to get the correct answer in the correct format.

1:1, 50:50 or 50% is still a perfectly valid interpretation though.

Actually I asked for the answer to be expressed in "percentage odds" in [url= http://www.singletrackworld.com/forum/topic/the-boy-girl-puzzle#post-62902 ]the sixth post[/url].

show me where in the preceding 160 odd posts that someone has said that the odds are 2:1 in favour of there being one child of each gender.

I did. [url= http://www.singletrackworld.com/forum/topic/the-boy-girl-puzzle/page/4#post-66891 ]back here[/url] where I said:

[i].."mixed sex" versus "both girls" is 50:25[/i]

Funnily enough 50:25 is the same as 2:1 or are you now going to argue about basic maths?

1:1, 50:50 or 50% is still a perfectly valid interpretation though.

Define "perfectly valid" for me. If you mean it is "perfectly valid" to come up with completely the wrong answer, which is empirically proven to be incorrect then yes, well done.

Don't feel bad Smee - it is a deliberately counter-intuitive problem - but there is only one correct answer and that is 66.66% (aka p=0.666.., aka odds of 2:1)

[i]Drac - show me where in the preceding 160 odd posts that someone has said that the odds are 2:1 in favour of there being one child of each gender. [/i]

Well Graham beat me to it but before you make wild claims I suggest your work out odds as percentages.

Well done everybody for making this thread last 175 posts so far.

And I realised my mistake on the first page! 😉

Probability is not an easy thing to explain in fairness though!

For Smee, perhaps another twist on explaining this:

If we DO NOT know the sex of any child, then the probability of there being a boy and a girl (Boy/Girl and Girl/Boy) is 2 opportunities out of 4, or 50%.

If we know the sex of the FIRST child is a girl, but not the sex of the second child, there are only 2 options... Girl/Boy and Girl/Girl. Therefore the probability of there being a boy and a girl is 1 opportunity out of 2, or 50%.

If we know the sex of the SECOND child, but not the first, then apply the same maths as above. Probability would be 1 in 2, or 50%.

Now, we only know the sex of ONE of the children, not whether it was the first or the second child. Because we don't know whether or not that the girl (that we know exists) was born first or second, we have to assume there were 3 possibilities of combinations of children being born... Boy 1st/Girl 2nd, Girl 1st/Boy 2nd, Girl 1st/Girl 2nd. As 2 of these opportunities contain a boy and a girl, in either order, and 1 doesn't, the probability of there being a boy and a girl is 2 opportunities out of 3, or 2/3, or 66.66%, or 2:1 etc.

I am now leaving the thread, never to return! 😆

Wow.

Smee, this really is incredible, so glad you came back. Here I was thinking you had walked off noble in defeat. This is much more fun!

Smee... humour me here would you by answering a simple question.

Which one of the following sets of probabilities is correct for a woman having two children, taking into account that the sum of all probabilities must equal 100%:-

Boy/Boy 25%
Boy/Girl 50%
Girl/Girl 25%

or

Boy/Boy 33.3%
Boy/Girl 33.3%
Girl/Girl 33.3%

Good one funky, here's sure to admit his mistake now

funkynick - the first one. neither are relevant to this one though.

Smee, I realise you reckon your reasoning comes from the fact you can discount the boy/girl combination of children, can you rationally justify that? I've not read a coherent response to this assertion yet.

Okay then, can you please explain why neither of those sets of probabilities are relevant?

Smee has already admitted that the "Odds are 2:1"

His only bone of contention was that odds of 2:1 are somehow different from a probability of 66%, which it isn't (as shown above).

Note that he also refuses to provide any working or answer any of the direct challenges to his "theories".

So I can only assume that A) he now agrees with us and/or B) he is just playing it dumb as a some mediocre form of trolling.

You have two children, you know that one or more is a girl. This means that at least one of them can't be a boy. This removes the options of boy/boy and boy/girl.

Why does it do this? Forget age. Lets call them kids 1 and 2. 1 is a girl - we know that - they may be the youngest, they may be the oldest it doesn't matter - what does matter is that they are not both. That is why you can remove one of the boy/girl, girl/boy permutations.

You have two children, you know that one or more is a girl.

Correct.

This means that at least one of them can't be a boy.

Yes.

This removes the options of boy/boy

True.

and boy/girl.

Wrong.

Sorry, stepping back in.

WRONG Smee

1 and 2 refers to the order in which they arrive. WE DO NOT KNOW that 1 is a girl, or that 2 is a girl, merely that at least one of them is.

Boy/Boy option is removed, Boy/Girl option is still valid as the lady said "at least one of them is a girl". It was not stated that the first child, nor indeed the second child is the girl, merely that "one of them is".

Therefore Boy/Girl, Girl/Boy and Girl/Girl are all valid options. In 2 of these there is one girl and one boy, therefore a 2 out of 3 chance of there being one of each, therefore a 66.66% chance of there being a boy and a girl, given that she has already told us that "at least one of them is a girl".

Ah! That explains it, you have a fundamental misunderstanding of the question.

You have two children, you know that one or more is a girl.

Yup

This removes the options of boy/boy and boy/girl.

But boy/girl has the same content as girl/boy.... i.e. one girl and one boy. So therefore they are the same.

But boy/girl has the same content as girl/boy.... i.e. one girl and one boy. So therefore they are the same.

Also wrong.

But, but, so many people have explained this subtletly so many times! And I was so proud of my version. Boo hoo hoo! 😥

What about the little tree diagrams! And the big calculation things!

Even the text descriptions were accurate, succintly worded and clear!

You just don't like us! That's it isn't it!

Dear me.

Mike - are you mental? Read what you have written.

I get the idea that most of you are using I just don't agree that it can be used here.

Mike - are you mental? Read what you have written.

I may be. But, I understand the question and the answer, which you seemingly do not.

I get the idea that most of you are using I just don't agree that it can be used here.

Ah, so the world and the laws of mathematics are wrong?

Ok then Mike. Boy/Girl and Girl/Boy - remove the ages of them and what is the difference?

How about this baby.

There are two towns.

Town A

Town B

Town A has two roads leading to it

Town B had only one

Pick a road purely at random

Probability you'll end in town A?

66%!

Think of town A as mixed kids

Think of town B as all girls

the two roads to mixed kids = G/B B/G

one road to all girls = G/G

Wiredchops - you can try answering that question too.

Until now, I have resisted looking at this thread, is it 5 pages of Smee making a fool of himself?

Ok, I'll try.

The position of the child has an inherent value with regards to order alone.

I.e. There is only one possible way of achieving a girl girl child combo when having two children.

There are two possibilities of achieving a mixed sex combo. Namely, G/B B/G

If you remove the ages of the children the only difference is the order in which you have written them down. WHICH IS NOT TO SAY IT RENDERS THEM THE SAME

THe problem is smee you're looking at the end result. Consider the possible ways of getting there, hence my awesome road/town analogy. THEY ARE NOT THE SAME

Ok then Mike. Boy/Girl and Girl/Boy - remove the ages of them and what is the difference?

There is no difference but there is a 50% of it being a girl/boy boy/girl combination compared to a 25% chance of a girl/girl combination, although I assume this attempted explanation will fail

Smee: if you really are serious (which I doubt) then lets do away with all those confusing formulae and look at the raw empirical data in [url= http://spreadsheets.google.com/ccc?key=p_H5o2Sep3PNxPhB1Cjb-Dg ]my fabulous spreadsheet[/url].

Take each row in turn, ask yourself "Does this row contain at least one girl?"
If it does then scratch a tally mark into the wall of your cell (or ask your therapist for a crayon).
Next ask yourself "Does this row contain one girl and one boy?"
If it does then scratch a mark into a different wall.

Once you have done this for all 100 rows you should find that your first wall has a a total of around 75. And the second wall has a total of around 50.

Now ask one of your carers to tell you what 50 out of 75 is as a percentage...

If there is no difference then they are the same. Why would you give one option double the chance of being the correct outcome?

Okay then... take 3 families

Family A has an older boy and younger girl
Family B has an older girl and younger boy
Family C has two girls

If you pick a family at random, what are the chances that you will pick a family with a boy?

If you believe you can discount either family A or family B from this, as you seem to be doing in your examples above, can you give a reason for which family you would discard, and why that one and not the other.

He must be trolling, this is pretty basic stuff.

Ok then Mike. Boy/Girl and Girl/Boy - remove the ages of them and what is the difference?

They are two separate 'routes' to having two children, where one is a boy and one is a girl. That's twice as many 'routes' as there are for having two children who are both girls.

How about:
Family A has one boy and one girl, or one girl and one boy if you prefer - they only have two kids so they can't have both.
Family B has two girls.

All permutations taken into account.

50:50

How about "djglover is correct" as a solution?

What about if they had a ladyboy?

Smee earlier:

[img] [/img]

Nope.

Smee.. if B/G is the same as G/B as you claim, and the chances of it occuring are not doubled, then you have either:-

B/B 25%
B/G or G/B 25%
G/G 25%

Which does not total 100%, or:-

B/B 33.3%
B/G or G/B 33.3%
G/G 33.3%

Which does equal 100%.

This is the logical extension of what you are arguing.

Smee.. so you are saying that an older boy/younger girl is the same as an older girl/younger boy?

Smee.. so you are saying that an older boy/younger girl is the same as an older girl/younger boy?

lol. Are you trying to get him to come out?

Yup, I'm off, smee admitted earlier it was 25:50:25 spread earlier, now he's denying it.

Trolling

Good fun though!

Funkynick

If you knew neither childs gender

b/b 25%
b/g 25%
g/b 25%
g/g 25%

You know the gender of one so you are left with

g/b 50%
g/g 50%

bless him

Can the mods link this post to all of George's subsequent posts as a warning either not to take them seriously (as they are trolls) or invalidate the argument being proposed?

I will not change my mind as a result of what others say. You should all know that by now.

Mat - go and do your presentation.

My guess is that he's either too stupid to realise he's looking stupid, or he'll "reveal" that he was trolling all along.

There's a probability in there somewhere....

Aaaah, so you are admitting you are wrong, but won't change your mind anyway.

Excellent.

Get got the result chaps!

😀

Not admitting anything. As a person capable of independent thought I'm happy with my arguement.

2 Mat's (M@'s) on 2 different forums, what's the probability of it being the same person? :p

Needless to say you've upset Tom's mathsist sensibilities

Pretty good I'd say.

Mathematicians, Engineers, Scientists and Midwives will probably all get different solutions the this puzzle. All will think their solution is correct. Can they all be correct? Yes.

I will not change my mind as a result of what others say. You should all know that by now.

Then. Do. The. Math!

Or try using the spreadsheet as I described above.

Or trying tossing two coins 100 times.
Every time you get two heads then cut you left arm.
Every time you get a head and a tail then cut your right arm.
Once you are finished simply measure the blood loss from each arm. You will find the right one has twice as many cuts as the left and will bleed out quicker.

The Mathematicians, Engineers and Scientists are all agreeing, are you a midwife? 😛

I have done the math. I just think you should all be doing different maths.

Smee... it doesn't matter what you say, I know that's what you meant. You lost, and are just too proud to admit it.

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

He is apparently an engineer, but for some reason he uses a different mathematical system from the rest of humanity. I hope you're not involved in anything critical Smee.

You have been given several clear empirical experiments that produce results which fit perfectly with our answer and you have been unable to find fault in them (and instead you've just quietly ignored them).

If your answer (the 50% answer, not the 2:1 answer) is correct then lets hear how we can conduct a simple experiment that backs it up.

Any experiment you like... in your own time...

[i]He is apparently an engineer, but for some reason he uses a different mathematical system from the rest of humanity. I hope you're not involved in anything critical Smee.[/i]

Guess that's why he became a driving instructor.

well i read my way thru that lot but going back to the original rules - how do we know the lady isn't a compulsive liar?

Anything yet Smee? Nope thought not. 😀

The coin toss backs it up.

The coin toss experiment that I outlined a few posts ago? That backs up [u]your[/u] answer?!?

Or have you devised your own perverted version of it?
Please explain in enough detail that we can try it too.

options for two throws
tt
th
ht
hh

when you know that one is a h you have:
one of ht or th - not both.
and
hh left as possible outcomes after the second throw.

The problem is the over-complicated way the question is being asked. As the gender of one child is already known the remaining and simplest unknown and therefore the only question which needs to be asked is what is the gender of the other child? The answer to this question will answer the original question and has a probability of 50%.

Andy.

Erm.. oh dear.

So you're saying "It is one of these four possibilities and if you get information that eliminates one of those possibilities then that leaves you with two."?

one of ht or th - not both

No I'm saying that once you know that one is a head then it removes the options that start with a tail.